NCERT Solutions Class 11th Maths Chapter 9: Sequences and SeriesExercise 9.1Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are: 1. an = n (n + 2) SOLUTION nth term of the give sequence = n (n + 2) First term = a1 = 1 (1 + 2) = 3 Second term = a2 = 2 (2 + 2) = 8 Third term = a3 = 3 (3 + 2) = 15 Fourth term = a4 = 4 (4 + 2) = 24 Fifth term = a5 = 5 (5 + 2) = 35 Hence, the first five terms of the given sequence are 3, 8, 15, 24, 35. 2. an = n/(n + 1) SOLUTION nth term of the give sequence = n/(n + 1) First term = a1 = 1/(1 + 1) = 1/2 Second term = a2 = 2/(2 + 1) = 2/3 Third term = a3 = 3/(3 + 1) = 3/4 Fourth term = a4 = 4/(4 + 1) = 4/5 Fifth term = a5 = 5/(5 + 1) = 5/6 Hence, the first five terms of the given sequence are 1/2, 2/3, 3/4, 4/5 and 5/6. 3. an = 2n SOLUTION nth term of the give sequence = 2n First term = a1 = 21 = 2 Second term = a2 = 22 = 4 Third term = a3 = 23= 8 Fourth term = a4 = 24 = 16 Fifth term = a5 = 25 = 32 Hence, the first five terms of the given sequence are 2, 4, 8, 16, 32. 4. an = (2n - 3)/6 nth term of the give sequence = (2n - 3)/6 First term = a1 = (2(1) - 3)/6 = -1/6 Second term = a2 = (2(2) - 3)/6 = 1/6 Third term = a3 = (2(3) - 3)/6 = 1/2 Fourth term = a4 = (2(4) - 3)/6 = 5/6 Fifth term = a5 = (2(5) - 3)/6 = 7/6 Hence, the first five terms of the given sequence are -1/6, 1/6, 1/2, 5/6 and 7/6. 5. an = (-1)n - 1 5n + 1 SOLUTION nth term of the give sequence = (-1)n - 1 5n + 1 First term = a1 = (-1)1 - 1 51 + 1 = 25 Second term = a2 = (-1)2 - 1 52 + 1 = -125 Third term = a3 = (-1)3 - 1 53 + 1 = 625 Fourth term = a4 = (-1)4 - 1 54 + 1 = -3125 Fifth term = a5 = (-1)5 - 1 55 + 1 = 15625 Hence, the first five terms of the given sequence are 25, -125, 625, -3125, and 15625. 6. an = n (n2 + 5)/4 SOLUTION nth term of the give sequence = n (n2 + 5)/4 First term = a1 = 1 (12 + 5)/4 = 3/2 Second term = a2 = 2 (22 + 5)/4 = 9/2 Third term = a3 = 3 (32 + 5)/4 = 21/2 Fourth term = a4 = 4 (42 + 5)/4 = 21 Fifth term = a5 = 5 (52 + 5)/4 = 75/2 Hence, the first five terms of the given sequence are 3/2, 9/2, 21/2, 21 and 75/2. Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are: 7. an = 4n - 3; a17, a24 SOLUTION nth term of the given sequence = an = 4n - 3 17th term of the sequence = a17 = 4(17) - 3 = 65 24th term of the sequence = a24 = 4(24) - 3 = 93 8. an = n2/2n; a7 SOLUTION nth term of the given sequence = an = n2/2n 7th term of the sequence = a7 = 72/27 = 49/128 9. an = (-1)n - 1n3; a9 SOLUTION nth term of the given sequence = an = (-1)n - 1n3 9th term of the sequence = a9 = (-1)9 - 193 = 729 10. an = n(n - 2)/(n + 3); a20 SOLUTION nth term of the given sequence = an = n(n - 2)/(n + 3) 20th term of the sequence = a20 = 20(20 - 2)/(20 + 3) = 20(18)/23 = 360/23 Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 11. a1 = 3, an = 3an - 1 + 2 for all n > 1 SOLUTION nth term of the give sequence = 3an - 1 + 2 First term = a1 = 3 Second term = a2 = 3a1 + 2 = 3(3) + 2 = 11 Third term = a3 = 3a2 + 2 = 3(11) + 2 = 35 Fourth term = a4 = 3a3 + 2 = 3(35) + 2 = 107 Fifth term = a5 = 3a4 + 2 = 3(107) + 2 = 323 Therefore, the first five terms of the given sequence are 3, 11, 35, 107, and 323. Hence, the corresponding series is 3 + 11 + 35 + 107 + 323 + … 12. a1 = - 1, an = an - 1/n, n ≥ 2 SOLUTION nth term of the give sequence = an - 1/n First term = a1 = -1 Second term = a2 = a1/2 = -1/2 Third term = a3 = a2/3 = -1/6 Fourth term = a4 = a3/4 = -1/24 Fifth term = a5 = a4/5 = -1/120 Therefore, the first five terms of the given sequence are -1, -1/2, -1/6, -1/24, and -1/120. Hence, the corresponding series is -1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + … 13. a1 = a2 = 2, an = an - 1 - 1, n > 2 SOLUTION nth term of the give sequence = an - 1 - 1 First term = a1 = 2 Second term = a2 = 2 Third term = a3 = a2- 1 = 2 - 1 = 1 Fourth term = a4 = a3 - 1 = 1 - 1 = 0 Fifth term = a5 = a4 - 1 = 0 - 1 = -1 Therefore, the first five terms of the given sequence are 2, 2, 1, 0, and -1. Hence, the corresponding series is 2 + 2 + 1 + 0 + (-1) + … 14. The Fibonacci sequence is defined by 1 = a1 = a2 and an = an - 1 + an - 2, n > 2. Find an + 1/an, for n = 1, 2, 3, 4, 5. SOLUTION a1 = a2 = 1 nth term of the Fibonacci Sequence = an - 1 + an - 2 Third term = a3 = a2 + a1 = 1 + 1 = 2 Fourth term = a4 = a3 + a2 = 2 + 1 = 3 Fifth term = a5 = a4 + a3 = 3 + 2 = 5 Sixth term = a6 = a5 + a4 = 5 + 3 = 8 Vales of an + 1/an when: n = 1 ⇒ a2/a1 = 1/1 = 1 n = 2 ⇒ a3/a2 = 2/1 = 2 n = 3 ⇒ a4/a3 = 3/2 n = 4 ⇒ a5/a4 = 5/3 n = 5 ⇒ a6/a5 = 8/5 Exercise 9.21. Find the sum of odd integers from 1 to 2001. SOLUTION The sequence of odd integers from 1 to 2001 is 1, 3, 5, 7, …, 1997, 1999, 2001. This sequence is in the form of an AP. First term = a = 1 Common difference = d = 2 The last term = a + (n - 1)d = 2001 1 + (n - 1)2 = 2001 1 + 2n - 2 = 2001 2n - 2 = 2000 2n = 2002 n = 1001 Therefore, the sequence of odd numbers from 1 to 2001 has 1001 terms. Sum of the AP = Sn = n/2 × [2a + (n - 1)d] = 1001/2 × [2(1) + (1000)2] = 1001/2 × [2 + 2000] = 1001/2 × 2002 = 10012 = 1002001 Hence, the sum of odd numbers from 1 to 2001 is 1002001. 2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5. SOLUTION The sequence of natural numbers lying between 100 and 1000, which are multiples of 5 is 105, 110, 115, …, 990, 995. This sequence is in the form of an AP. First term = a = 105 Common difference = d = 5 The last term = a + (n - 1)d = 995 105 + (n - 1)5 = 995 105 + 5n - 5 = 995 5n + 100 = 995 5n = 895 n = 179 Therefore, the sequence of natural numbers lying between 100 and 1000, which are multiples of 5 has 179 terms. Sum of the AP = Sn = n/2 × [2a + (n - 1)d] = 179/2 × [2(105) + (178)5] = 179/2 × [210 + 890] = 179/2 × 1100 = 179 × 550 = 98450 Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5 is 98450. 3. In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112. SOLUTION First term of the AP = a = 2 Let the common difference of the AP be d. So, the AP is 2, 2 + d, 2 + 2d, …, Sum of first five terms = S5 = 5/2 × [2(2) + (4)d] = 5/2 × [4 + 4d] = 5/2 × 2(2 + 2d) = 10 + 10d Sum of the next five terms = Sum of the first ten terms - Sum of the first five terms = 10/2 × [2(2) + 9d] - (10 + 10d) = 5 × (4 + 9d) - 10 - 10d = 20 + 45d - 10 - 10d = 10 + 35d It is given that the sum of the first five terms is one fourth of the sum of the next five terms of the given AP. Therefore, 10 + 10d = 1/4 × (10 + 35d) 4(10 + 10d) = 10 + 35d 40 + 40d = 10 + 35d 30 + 5d = 0 5d = -30 d = -6 20th term of the AP = a20 = 2 + 19(-6) = 2 - 114 = -112 Hence, proved that the 20th term of the AP is -112. 4. How many terms of the A.P. - 6, -11/2, - 5, … are needed to give the sum -25? SOLUTION Required sum = Sn = -25 where n is the number of terms of the given AP required to give the sum -25. First term = a = -6 Common difference = -11/2 - (-6) = -11/2 + 6 = 1/2 Sn = n/2 × [2a + (n - 1)d] -25 = n/2 × [2(-6) + (n - 1)/2] -50 = n [-12 + (n - 1)/2] -50 = -12n + n(n - 1)/2 -50 = [-24n + n2 - n]/2 -100 = n2 - 25n n2 - 25n + 100 = 0 n2 - 5n - 20n + 100 = 0 n(n - 5) - 20(n - 5) = 0 (n - 5) (n - 20) = 0 Therefore, (n - 5) = 0 ⇒ n = 5 OR (n - 20) = 0 ⇒ n = 20 Hence, either 20 or 5 terms of the AP are required to give the sum -25. 5. In an A.P., if pth term is 1/q and the qth term is 1/p, prove that the sum of first pq terms is (pq + 1)/2, where p ≠ q. SOLUTION General term of an AP = an = a + (n - 1)d pth term of the AP = ap = a + (p - 1)d = 1/q qth term of the AP = aq = a + (q - 1)d = 1/p ap - aq = a + (p - 1)d - (a + (q - 1)d) 1/q - 1/p = a + (p - 1)d - a - (q - 1)d 1/q - 1/p = (p - 1 - q + 1)d (p - q)/pq = (p - q)d d = 1/pq Now, ap = a + (p - 1)d 1/q = a + (p - 1)/pq 1 = q (a + (p - 1)/pq) 1 = aq + (p - 1)/p 1 = (apq + p - 1)/p p = apq + p - 1 1= apq a = 1/pq Now, sum of first pq terms: Spq = pq/2 × [2(1/pq) + (pq - 1)/pq] = pq/2 × 1/pq × (2 + pq - 1) = 1/2 × (pq + 1) Hence, proved that the sum of first pq terms of the AP is 1/2 (pq + 1). 6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term. SOLUTION First term = a = 25 Common difference = d = 22 - 25 = -3 Let the number of terms of the AP whose sum is 116 be n. Sn = n/2 × [2a + (n - 1)d] 116 = n/2 × [2(25) + (n - 1)(-3)] 232 = n × [50 - 3n + 3] 232 = 53n - 3n2 3n2 - 53n + 232 = 0 3n2 - 24n - 29n + 232 = 0 3n(n - 8) - 29(n - 8) = 0 (3n - 29) (n - 8) = 0 Therefore, (n - 8) = 0 ⇒ n = 8 OR (3n - 29) = 0 ⇒ n = 29/3 n can only be an integral value, so n = 29/8 is rejected. n = 8 Therefore, the sum 116 is obtained with 8 terms of the AP. Last term = a8 = a + 7d = 25 + 7(-3) = 25 - 21 = 4 Hence, the last term is 4. 7. Find the sum to n terms of the A.P., whose kth term is 5k + 1. SOLUTION kth term of the AP = ak = 5k + 1 a + (k - 1)d = 5k + 1 a + dk - d = 5k + 1 On comparing the coefficients of k, we get, d = 5. a + 5k - 5 = 5k + 1 a - 5 = 1 a = 6 Sum of n terms of the AP = Sn = n/2 × [2a + (n - 1)d] = n/2 × [2(6) + (n - 1)5] = n/2 × [12 + 5n - 5] = n/2 × [5n + 7] = (5n + 7)/2 8. If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference. SOLUTION Sum of n terms of the AP = Sn = n/2 × [2a + (n - 1)d] (pn + qn2) = n/2 × [2a + (n - 1)d] 2qn2 + 2pn = 2an + dn(n - 1) 2qn2 + 2pn = dn2 + 2an - dn 2qn2 + 2pn = dn2 + n(2a - d) Compare the coefficients of n2 2q = d Hence, common difference of the given AP is 2q. 9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms. SOLUTION Let a be the first term and d be the common difference of the first AP. Let A be the first term and D be the common difference of the second AP. Sum of n terms of the first AP = n/2 × [2a + (n - 1)d] Sum of n terms of the second AP = n/2 × [2A + (n - 1)D] It is given that the ratio of the sum of n terms of the first and second AP is 5n + 4 : 9n + 6. Therefore, [n/2 × [2a + (n - 1)d]]/[n/2 × [2A + (n - 1)D]] = (5n + 4)/(9n + 6) [2a + (n - 1)d]/[2A + (n - 1)D] = (5n + 4)/(9n + 6) Put n = 35 [2a + (35 - 1)d]/[2A + (35 - 1)D] = (5(35) + 4)/(9(35) + 6) [2a + 34d]/[2A + 34D] = (175 + 4)/(315 + 6) 2[a + 17d]/2[A + 17D] = 179/321 (a + 17d)/(A + 17D) = 179/321 a18/A18 = 179/321 Hence, the ratio of the 18th terms of both the AP is 179 : 321. 10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms. SOLUTION Sum of the first p terms of the AP = Sp = p/2 × [2a + (p - 1)d] Sum of the first q terms of the AP = Sq = q/2 × [2a + (q - 1)d] It is given that the sum of first p terms of the AP is equal to the sum of the first q terms. Therefore, Sp = Sq p/2 × [2a + (p - 1)d] = q/2 × [2a + (q - 1)d] p [2a + (p - 1)d] = q [2a + (q - 1)d] 2ap + p(p - 1)d = 2aq + q(q - 1)d 2ap - 2aq + p(p - 1)d - q(q - 1)d = 0 2a (p - q) + d(p2 - p - q2 + q) = 0 2a (p - q) + d(p2 - q2 - p + q) = 0 2a (p - q) + d[(p + q) (p - q) - (p - q)] = 0 2a (p - q) + d(p - q)[(p + q) - 1] = 0 (p - q) [2a + (p + q - 1)d] = 0 2a + (p + q - 1)d = 0 Multiply both sides by (p + q - 1)/2 (p + q - 1)/2 × [2a + (p + q - 1)d] = 0 Sp + q = 0 Hence, the sum of first p + q terms of the AP is 0. 11. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that a(q - r)/p + b(r - p)/q + c(p - q)/r = 0. SOLUTION Let the first term of the AP = A and the common difference = d Sum of first p terms of the AP = Sp = p/2 × [2A + (p - 1)d] a = p/2 × [2A + (p - 1)d] 2a/p = 2A + (p - 1)d … Equation I Sum of first q terms of the AP = Sq = q/2 × [2A + (q - 1)d] b = q/2 × [2A + (q - 1)d] 2b/q = 2A + (q - 1)d … Equation II Sum of first r terms of the AP = Sr= r/2 × [2A + (r - 1)d] c = r/2 × [2A + (r - 1)d] 2c/r = 2A + (r - 1)d … Equation III Subtract Equation II from Equation I 2a/p - 2b/q = 2A + (p - 1)d - {2A + (q - 1)d} 2(a/p - b/q) = 2A + (p - 1)d - 2A - (q - 1)d 2(aq - bp)/pq = d(p - 1 - q + 1) 2(aq - bp)/pq = d(p - q) d = 2(aq - bp)/pq(p - q) Subtract Equation III from Equation II 2b/q - 2c/r = 2A + (q - 1)d - {2A + (r - 1)d} 2(b/q - c/r) = 2A + (q - 1)d - 2A - (r - 1)d 2(br - cq)/qr = d(q - 1 - r + 1) 2(br - cq)/qr = d(q - r) d = 2(br - cq)/qr(q - r) Now, 2(aq - bp)/pq(p - q) = 2(br - cq)/qr(q - r) (aq - bp)/p(p - q) = (br - cq)/r(q - r) r(q - r) (aq - bp) = p(p - q) (br - cq) (q - r) (aqr - bpr) = (p - q) (bpr - cpq) Divide both sides by pqr (q - r) (aqr - bpr)/pqr = (p - q) (bpr - cpq)/pqr (q - r) (a/p - b/q) = (p - q) (b/q - c/r) a(q - r)/p - b(q - r)/q = b(p - q)/q - c(p - q)/r a(q - r)/p + c(p - q)/r - b(q - r)/q - b(p - q)/q = 0 a(q - r)/p - b(q - r + p - q)/q + c(p - q)/r = 0 a(q - r)/p - b(p - r)/q + c(p - q)/r = 0 Hence, proved. 12. The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m - 1) : (2n - 1). SOLUTION Let the first term of the AP = a and the common difference = d Sum of m terms = Sm = m/2 × [2a + (m - 1)d] Sum of n terms = Sn = n/2 × [2a + (n - 1)d] It is given that the ratio of the sum of m and terms of the AP is m2 : n2. Therefore, Sm/Sn = m/2 × [2a + (m - 1)d]/[n/2 × [2a + (n - 1)d]] m2/n2 = m {2a + (m - 1)d}/n {2a + (n - 1)d} Put m = 2m - 1 and n = 2n - 1 (2m - 1)/(2n - 1) = [2a + (2m - 1 - 1)d]/[2a + (2n - 1 - 1)d] (2m - 1)/(2n - 1) = [2a + (2m - 2)d]/[2a + (2n - 2)d] (2m - 1)/(2n - 1) = 2[a + (m - 1)d]/2[a + (n - 1)d] (2m - 1)/(2n - 1) = [a + (m - 1)d]/[a + (n - 1)d] (2m - 1)/(2n - 1) = am/an am: an = (2m - 1) : (2n - 1) Hence, proved. 13. If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m. SOLUTION Let the first term of the AP = a and the common difference = d Sum of n terms of the AP = Sn = n/2 × [2a + (n - 1)d] 3n2 + 5n = n/2 × [2a + (n - 1)d] n(3n + 5) = n/2 × [2a + (n - 1)d] 2(3n + 5) = 2a + (n - 1)d 6n + 10 = 2a + dn - d On comparing the coefficients of on both sides, we get: 6 = d 2a - d = 10 2a - 6 = 10 2a = 16 a = 8 mth term of the AP = am = a + (m - 1)d 164 = 8 + (m - 1)6 156 = (m - 1)6 26 = m - 1 m = 27 Hence, the value of m is 27. 14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P. SOLUTION Let the five numbers between 8 and 26 be A1, A2, A3, A4 and A5. The AP will be 8, A1, A2, A3, A4, A5, 26. Firs term = a = 8 a7 = 26 a + 6d = 26 8 + 6d = 26 6d = 18 d = 3 A1 = a2 = a + d = 8 + 3 = 11 A2 = a3= a + 2d = 8 + 2(3) = 14 A3 = a4 = a + 3d = 8 + 3(3) = 17 A4 = a5 = a + 4d = 8 + 4(3) = 20 A5 = a6 = a + 5d = 8 + 5(3) = 23 Therefore, five numbers to between 8 and 26 in order to make the sequence an AP are 11, 14, 17, 20, and 23. 15. If (an+ bn)/(an - 1 + bn - 1) is the A.M. between a and b, then find the value of n. SOLUTION Arithmetic Mean between a and b = (a + b)/2 It is given that that Arithmetic Mean between a and b is (an+ bn)/(an - 1 + bn - 1). Therefore, (a + b)/2 = (an+ bn)/(an - 1 + bn - 1) (a + b) (an - 1 + bn - 1) = 2(an + bn) an + an - 1b + bn + abn - 1 = 2an + 2bn an - 1b + abn - 1 = an + bn abn - 1 - bn = an - an - 1b bn - 1 (a - b) = an - 1 (a - b) bn - 1 = an - 1 (a/b)n - 1 = 1 (a/b)n - 1 = (a/b)0 Compare the exponents n - 1 = 0 n = 1 Hence, value of n is 1. 16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m - 1)th numbers is 5 : 9. Find the value of m. SOLUTION Let the m numbers between 1 and 31 be A1, A2, A3, …, Am. The AP will be 1, A1, A2, A3, …, Am, 31. First term = a = 1 an = 31 a + (n - 1)d = 31 1 + (m + 2 - 1)d = 31 (m + 1)d = 30 d = 30/(m + 1) A1 = a2 = a + d = 1 + 30/(m + 1) A2 = a3 = a + 2d = 1 + 60/(m + 1) … A7 = a8 = a + 7d = 1 + 210/(m + 1) Am - 1 = am = a + (m - 1)d = 1 + 30(m - 1)/(m + 1) It is given that the ratio of the 7th and the (m - 1)th number is 5 : 9. Therefore, a8/am = 5/9 [1 + 210/(m + 1)]/[1 + 30 (m - 1)/(m + 1)] = 5/9 9 [1 + 210/(m + 1)] = 5 [1 + 30 (m - 1)/(m + 1)] 9 (m + 1 + 210)/(m + 1) = 5 [(m + 1) + 30 (m - 1)]/(m + 1) 9 (m + 211) = 5 (m + 1 + 30m - 30) 9m + 1899 = 5 (31m - 29) 9m + 1899 = 155m - 145 2044 = 146m m = 14 Hence, m = 14 numbers were inserted between 1 and 31. 17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment? SOLUTION First instalment of the loan = Rs 100 Second instalment of the loan = Rs 100 + 5 = Rs 105 Third instalment of the loan = Rs 105 + 5 = Rs 110 The instalments paid by the man each month form an AP: 100, 105, 110, … First term of the AP = a = 100 Common difference = d = 5 Amount he will pay in the 30th instalment = 30th term of the AP = a30 = a + 29d = 100 + 29(5) = 100 + 145 = 245 Hence, the man will pay Rs 245 in the 30th instalment. 18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120° , find the number of the sides of the polygon. SOLUTION The difference between two consecutive angles is 5°. Therefore, the sequence of the angles will form an AP. 120, 125, 130, ... First term of the AP = Smallest angle of the polygon = a = 120 Common difference = d = 5 We know that the sum of all angles of a polygon is 180° (n - 2) where n is the number of sides of the polygon. Sum of the AP = Sn = n/2 × [2a + (n - 1)d] 180 (n - 2) = n/2 × [2(120) + (n - 1)5] 360 (n - 2) = n × [240 + 5n - 5] 360n - 720 = n × [5n + 235] 360n - 720 = 5n2 + 235n 5n2 - 125n + 720 = 0 5(n2 - 25n + 144) = 0 n2 - 25n + 144 = 0 n2 - 16n - 9n + 144 = 0 n (n - 16) - 9 (n - 16) = 0 (n - 16) (n - 9) = 0 (n - 16) = 0 ⇒ n = 16 OR (n - 9) = 0 ⇒ n = 9 Hence, the required polygon has either 9 sides or 16 sides. Exercise 9.31. Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, … SOLUTION First term = a = 5/2 Common ratio = r = (5/4)/(5/2) = 1/2 20th term of the given GP = a20 = ar19 = 5/2 × (1/2)19 = 5/2(2)19 = 5/220 nth term of the given GP = an = arn - 1 = 5/2 × (1/2)n - 1 = 5/2(2)n - 1 = 5/2n 2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. SOLUTION Common ratio of the GP = r = 2 First term = a 8th term of the GP = a8 = ar7 192 = a(2)7 192 = 128a 3/2 = a 12th term of the GP = a12 = ar11 = 3/2 × (2)11 = 3 × 210 = 3072 Hence, the 12th term of the GP is 3072. 3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps. SOLUTION Let the first term of the GP = a and the common ratio = r 5th term of the GP = a5 = ar4 p = ar4 8th term of the GP = a8 = ar7 q = ar7 11th term of the GP = a11 = ar10 s = ar10 Now, a8/a5 = ar7/ar4 q/p = r3 AND a11/a8 = ar10/ar7 s/q = r3 Therefore, q/p = s/q q2 = ps Hence, proved. 4. The 4th term of a G.P. is square of its second term, and the first term is - 3. Determine its 7th term. SOLUTION Let the first term of the GP = a = -3 and the common ratio = r It is given that 4th term of the GP is the square of 2nd term of the GP. Therefore, a4 = (a2)2 ar3 = (ar)2 ar3 = a2r2 -3r3 = 9r2 -3r = 9 r = -3 7th term of the GP = a7 = ar6 = (-3) (-3)6 = (-3)7 = -2187 Hence, the 7th term of the given GP is -2187. 5. Which term of the following sequences: (a) 2, 2√2, 4, … is 128? (b) √3, 3, 3√3, … is 729? (c) 1/3, 1/9, 1/27, … is 1/19683? SOLUTION (a) First term = a = 2 Common ratio = r = 2√2/2 = √2 an = 128 arn - 1 = 128 2 (√2)n - 1 = 128 (√2)2 (√2)n - 1 = ((√2)2)7 (√2)n - 1 + 2 = (√2)14 (√2)n + 1 = (√2)14 Compare the exponents n + 1 = 14 n = 13 Hence, 128 is the 13th term of the given GP. (b) First term = a = √3 Common ratio = r = 3/√3 = √3 an = 729 arn - 1 = 729 √3 (√3)n - 1 = 729 (√3)n - 1 + 1 = ((√3)2)6 (√3)n = (√3)12 Compare the exponents n = 12 Hence, 729 is the 12th term of the given GP. (c) First term = a = 1/3 Common ratio = r = 1/9 × 3/1 = 1/3 an = 1/19683 arn - 1 = 1/19683 1/3 × (1/3)n - 1 = 1/19683 1/(3)n - 1 + 1 = 1/(3)9 (3)n = (3)9 Compare the exponents n = 9 Hence, 1/19683 is the 9th term of the given GP. 6. For what values of x, the numbers -2/7, x, -7/2 are in G.P.? SOLUTION It is given that the number -2/7, x, -7/2 are in GP. Therefore, Common ratio = r = x/(-2/7) = -7x/2 AND r = (-7/2)/x = -7/2x So, -7x/2 = -7/2x x2 = 1 x = ± 1 Hence, the given numbers will be in GP when x = ± 1. Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 7. 0.15, 0.015, 0.0015, ... 20 terms. SOLUTION First term = a = 0.15 Common ratio = r = 0.015/0.15 = 1/10 = 0.1 We know that sum of n terms of a GP = Sn = a(1 - rn)/(1 - r) Sum of 20 terms of the given GP = S20= 0.15(1 - (0.1)20)/(1 - 0.1) = 0.15(1 - (0.1)20)/(0.9) = 1/6 × {1 - (0.1)20} 8. √7, √21, 3√7, … n terms. SOLUTION First term = a = √7 Common ratio = r = √21/√7 = √3 Sum of n terms of the GP = Sn = a(1 - rn)/(1 - r) = √7(1 - (√3)n)/(1 - √3) Rationalise √7(1 - (√3)n)/(1 - √3) × (1 + √3)/(1 + √3) = √7(1 - (√3)n)(1 + √3)/(1 - 3) = √7(1 - (√3)n)(1 + √3)/(-2) = √7((√3)n - 1)(1 + √3)/2 9. 1, - a, a2, - a3, ... n terms (if a ≠ - 1). SOLUTION First term = 1 Common ratio = r = -a/1 = -a Sum of n terms of the GP = Sn = a(1 - rn)/(1 - r) = 1(1 - (-a)n)/(1 - (-a)) = (1 - (-a)n)/(1 + a) 10. x3, x5, x7, … n terms (if x ≠ ± 1). SOLUTION First term = a = x3 Common ratio = r = x5/x3 = x2 Sum of n terms of the GP = Sn = a(1 - rn)/(1 - r) = x3(1 - x2n)/(1 - x2) 11. Evaluate 11∑k = 1 (2 + 3k). SOLUTION 11∑k = 1 (2 + 3k) = 11∑k = 1 (2) + 11∑k = 1 (3k) 11∑k = 1 (2) = 2 + 2 + 2 + … 11 times = 2 (11) = 22 11∑k = 1 (3k) = 31 + 32 + 33 + … + 311 The expansion of 11∑k = 1 (3k) forms a GP. First term = a = 31 = 3 Common ratio = r = 32/31 = 31 = 3 We know that sum of n terms of a GP = Sn = a(1 - rn)/(1 - r) Sum of 11 terms of the given GP = S11 = 3(1 - 311)/(1 - 3) = 3(1 - 311)/(-2) = 3/2 × (311 - 1) Hence, 11∑k = 1 (2 + 3k) = 11∑k = 1 (2) + 11∑k = 1 (3k) = 22 + 3/2 × (311 - 1) 12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms. SOLUTION Let the first three terms of the GP be a/r, a and r. Sum of first three terms = a/r + a + ar = 39/10 Product of the first three terms = a/r × a × ar = 1 ⇒ a3 = 1 ⇒ a = 1 Therefore, a/r + a + ar = 39/10 1/r + 1 + r = 39/10 (1 + r + r2)/r = 39/10 10 (1 + r + r2) = 39r 10r2 + 10r + 10 = 39r 10r2 - 29r + 10 = 0 10r2 - 25r - 4r + 10 = 0 5r (2r - 5) - 2 (2r - 5) = 0 (2r - 5) (5r - 2) = 0 (2r - 5) = 0 ⇒ 2r = 5 ⇒ r = 5/2 OR (5r - 2) = 0 ⇒ 5r = 2 ⇒ r = 2/5 Hence, the three terms of the GP are 5/2, 1 and 2/5 OR 2/5, 1 and 5/2. 13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120? SOLUTION First term = a = 3 Common ratio = r = 32/3 = 3 Sn = a(1 - rn)/(1 - r) 120 = 3 (1 - 3n)/(1 - 3) 40 = (1 - 3n)/(-2) -80 = 1 - 3n -81 = -3n 3n = 81 3n = 34 Compare the exponents n = 4 Hence, 4 terms of the given GP are needed in order to get the sum 120. 14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P SOLUTION Let the GP be a, ar, ar2, … First term = a Common ratio = r It is given that the sum of first three terms of the GP is 16. Therefore, a + ar + ar2 = 16 a(1 + r + r2) = 16 a = 16/(1 + r + r2) It is also given that the sum of next three terms of the GP is 128. Therefore, ar3 + ar4 + ar5 = 128 ar3 (1 + r + r2) = 128 Substitute a = 16/(1 + r + r2) 16/(1 + r + r2) × r3 (1 + r + r2) = 128 16r3 = 128 r3 = 8 r3 = 23 Compare the exponents r = 2 Now, a = 16/(1 + r + r2) a = 16/(1 + 2 + 22) = 16/(3 + 4) a = 16/7 Sum of n terms of the GP = Sn = a(1 - rn)/(1 - r) = 16(1 - 2n)/7(1 - 2) = -16(1 - 2n)/7 Sn = 16/7 × (2n - 1) Hence, first term of the GP = 16/7, common ratio = 2 and sum of n terms = 16/7 × (2n - 1). 15. Given a G.P. with a = 729 and 7th term 64, determine S7. SOLUTION First term = a = 729 7th term = a7 = 64 ar6 = 64 729r6 = 64 r6 = 64/729 r6= (2/3)6 Compare the exponents r = 2/3 S7 = a(1 - r7)/(1 - r) = 729(1 - (2/3)7)/(1 - 2/3) = 729(1 - 128/2187)/(1/3) = 2187 (2187 - 128)/2187 = 2059 Hence, S7 = 2059. 16. Find a G.P. for which sum of the first two terms is - 4 and the fifth term is 4 times the third term. SOLUTION Let the first term of the GP = a and the common ratio = r. It is given that the sum of first two terms of the GP is -4. Therefore, S2 = a(1 - r2)/(1 - r) -4 = a (12 - r2)/(1 - r) -4 = a (1 - r)(1 + r)/(1 - r) -4 = a(1 + r) It is also given that the fifth term is 4 times the third term. Therefore, 4a3 = a5 4ar2 = ar4 4r2 = r4 4 = r2 r = ± 2 Now, -4 = a(1 + r) For r = 2, -4 = a(1 + 2) -4 = 3a a = -4/3 For r = -2, -4 = a(1 - 2) -4 = -a a = 4 Hence, the GP will be either 4, -8, 16, … OR -4/3, -8/3, -16/3, … 17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P. SOLUTION Let the first term of the GP = a and the common ratio = r. 4th term of the GP = a4 = ar3 = x 10th term of the GP = a10 = ar9 = y 16th term of the GP = a16 = ar15 = z Now, a10/a4 = ar9/ar3 y/x = r6 a16/a10 = ar15/ar9 z/y = r6 Therefore, y/x = z/y. Hence, x, y and z are in GP with the common ratio r6. 18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… . SOLUTION Sn = 8 + 88 + 888 + 8888 + … n terms This can be rewritten as Sn = 8/9 × [9 + 99 + 999 + 9999 + … n terms] = 8/9 × [(10 - 1) + (100 - 1) + (1000 - 1) + (10000 - 1) + … n terms] = 8/9 × [(10 - 1) + (102 - 1) + (103 - 1) + (104 - 1) + … n terms] = 8/9 × [10 + 102 + 103 + 104 + … n terms - (1 + 1 + 1 + … n terms)] = 8/9 × [10(1 - 10n)/(1 - 10) - n] = 8/9 × [10(1 - 10n)/(-9) - n] = 8/9 × [10(10n - 1)/9 - n] = 8/9 × [10(10n - 1) - 9n]/9 = 80(10n - 1)/81 - 8n/9 Hence, the sum of give sequence is 80(10n - 1)/81 - 8n/9. 19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2. SOLUTION Sum of products of the corresponding terms of the given sequences = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 × 1/2 = 64 [2 × 2 + 2 × 1 + 1 × 1 + 1/2 × 1 + 1/2 × 1/2] = 64 [4 + 2 + 1 + 1/2 + 1/4] It can be observed that the terms in 4 + 2 + 1 + 1/2 + 1/4 are in GP. First term = a = 4 Common ratio = r = 2/4 = 1/2 4 + 2 + 1 + 1/2 + 1/4 = Sn = a(1 - rn)/(1 - r) S5 = 4(1 - (1/2)5)/(1 - 1/2) = 4(1 - 1/32)/(1/2) = 8 (32 - 1)/32 = 31/4 Hence, the required sum of products of the corresponding terms of given sequences = 64 × 31/4 = 496. 20. Show that the products of the corresponding terms of the sequences a, ar, ar2, …arn - 1 and A, AR, AR2, … ARn - 1 form a G.P, and find the common ratio. SOLUTION Sequence formed by the products of corresponding terms of the given two sequences: aA, arAR, ar2AR2, arn - 1ARn - 1, … Second term/First term = arAR/aA = rR Third term/Second term = ar2AR2/arAR = rR Hence, the obtained sequence forms a GP with common ratio rR. 21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18. SOLUTION Let the first term of the GP = a and the common ratio = r. It is given that the third term is greater than the first term by 9. Therefore, a3 = a + 9 ar2 = a + 9 ar2 - a = 9 a(r2 - 1) = 9 a = 9/(r2 - 1) It is also given that the second term is greater than the 4th term by 18. Therefore, ar2 = ar4 + 18 ar = ar3 + 18 -18 = ar3 - ar -18 = ar(r2 - 1) -18/r(r2 - 1) = a Now, -18/r(r2 - 1) = a = 9/(r2 - 1) -18/r(r2 - 1) = 9/(r2 - 1) -18/r = 9 -18 = 9r r = -2 Thus, a = 9/(r2 - 1) = 9/(4 - 1) = 9/3 ⇒ a = 3 The four terms will be 3, 3(-2), 3(-2)2, 3(-2)3 3, -6, 12, -24 Hence, the first four terms of the given GP are 3, -6, 12, -24. 22. If the pth , qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq - r br - p cp - q = 1 SOLUTION Let the first term of the GP = A and the common ratio = R. pth term of the GP = Ap = ARp - 1 = a qth term of the GP = Aq = ARq - 1 = b rth term of the GP = Ar = ARr - 1 = c Now, aq - r br - p cp - q = (ARp - 1)q - r (ARq - 1)r - p (ARr - 1)p - q = Aq - r R(p - 1) (q - r) Ar - p R(q - 1) (r - p) Ap - q R(r - 1) (p - q) = Aq - r + r - p + p - q Rpq - q - pr + r + qr - r - pq + p + pq - p - qr + p = A0 R0 = 1 × 1 = 1 Hence, proved that aq - r br - p cp - q = 1. 23. If the first and the n th term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n. SOLUTION First term = a Let common ratio = r nth term of the GP = an = arn - 1 = b Product of n terms = P = a × ar × ar2 × ar3 × … arn - 1 = an r1 + 2 + 3 + … n - 1 The term in 1 + 2 + 3 + ... (n - 1) form an AP with first term 1 and common difference 1. Sum of the AP = 1 + 2 + 3 + ... (n - 1) = (n - 1)/2 × [2(1) + (n - 1 - 1)1] = (n - 1)/2 × [2 + n - 2] = n(n - 1)/2 Therefore, P = an rn(n - 1)/2 Square both sides P2 = a2n rn(n - 1) P2 = (a2rn - 1)n P2 = (a × arn - 1)n P2 = (a × b)n = (ab)n Hence, proved. 24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1/rn. SOLUTION Let the first term of the GP = a and the common ratio = r. Sum of the first n terms of the GP = a(1 - rn)/(1 - r) From the (n + 1)th term to the (2n)th OR (n + n)th term there are a total of n terms. Therefore, sum of the terms from (n + 1)th to (2n)th = an + 1(1 - rn)/(1 - r) Now, an + 1 = arn Ratio of the sum of first n terms to the sum of terms from (n + 1)th to (2n)th term = [a(1 - rn)/(1 - r)]/[an + 1(1 - rn)/(1 - r)] = a(1 - rn)/(1 - r) × (1 - r)/arn(1 - rn) = 1/rn Hence, proved. 25. If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2. SOLUTION It is given that a, b, c and d form a GP. Therefore, b/a = c/b = d/c b2 = ac bc = ad c2 = bd On the RHS, we have (ab + bc + cd)2 = (ab + ad + cd)2 = (ab + d(a + c))2 = a2b2 + [d(a + c)]2 + 2abd(a + c) = a2b2 + [d2(a2 + c2 + 2ac)] + 2a2bd + 2abcd = a2b2 + a2d2 + c2d2 + 2acd2 + 2a2bd + 2acbd = a2b2+ 2a2c2 + 2b2c2 + a2d2 + c2d2 + 2b2d2 = a2b2+ a2c2 + a2c2 + b2c2 + b2c2 + a2d2 + c2d2 + b2d2 + b2d2 = a2b2 + a2c2 + a2d2 + a2c2 + b2c2 + b2d2 + b2c2 + b2d2 + c2d2 = a2(b2 + c2 + d2) + b4 + b2c2 + b2d2 + b2c2 + c4 + c2d2 = a2(b2 + c2 + d2) + b2(b2 + c2 + d2) + c2(b2 + c2 + d2) = (b2 + c2 + d2) (a2 + b2 + c2) = (a2 + b2 + c2) (b2 + c2 + d2) = LHS Hence, proved. 26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P. SOLUTION Let x and y be two numbers between 3 and 81 such that the resulting sequence is a GP. 3, x, y, 81 First term = a = 3 Common ratio = r 4th term of the GP = a4 81 = ar3 81 = 3r3 27 = r3 3 = r 2nd term of the GP = x = a2 = ar = 3(3) = 9 3rd term of the GP = y = a3 = ar2 = 3(3)2 = 27 Hence, 9 and 27 are the two required numbers to be inserted between 3 and 81 in order to make the resulting sequence a GP. 27. Find the value of n so that (an + 1 + bn + 1)/(an + bn) may be the geometric mean between a and b. SOLUTION We know that the geometric mean between a and b is given by √ab. Therefore, (an + 1 + bn + 1)/(an + bn) = √ab Square both sides (an + 1 + bn + 1)2/(an + bn)2 = ab (an + 1 + bn + 1)2 = ab (an + bn)2 (an + 1)2 + (bn + 1)2 + 2an + 1bn + 1 = ab (a2n + b2n + 2anbn) a2n + 2 + b2n + 2 + 2an + 1bn + 1 = a2n + 1b + b2n + 1a + 2an + 1bn + 1 a2n + 2 + b2n + 2 = a2n + 1b + b2n + 1a a2n + 2 - a2n + 1b = b2n + 1a - b2n + 2 a2n + 1(a - b) = b2n + 1(a - b) a2n + 1 = b2n + 1 a2n + 1/b2n + 1 = 1 (a/b)2n + 1 = (a/b)0 Compare the exponents 2n + 1 = 0 2n = -1 n = -1/2 Hence, n is -1/2. 28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 - 2√2). SOLUTION Let the two numbers be a and b. It is given that the sum of the two numbers is 6 times greater than the geometric mean. Therefore, a + b = 6√ab … Equation (I) Square both sides (a + b)2 = 62 (√ab)2 a2 + b2 + 2ab = 36ab Subtract 4ab from both sides a2 + b2 - 4ab + 2ab = 36ab - 4ab a2 + b2 - 2ab - 2ab + 2ab = 32ab (a - b)2 = 32ab a - b = √(32ab) a - b = 4√2ab … Equation (II) Add Equation (I) and (II) a + b + (a - b) = 6√ab + 4√2ab 2a = (√ab) (6 + 4√2) a = (√ab) (3 + 2√2) Substitute a = (√ab) (3 + 2√2) in Equation (I) (√ab) (3 + 2√2) + b = 6√ab b = 6√ab - 3√ab - 2√2ab b = 3√ab - 2√2ab b = (3 - 2√2) (√ab) Ratio of the two numbers = a/b = (√ab) (3 + 2√2)/(√ab) (3 - 2√2) = (3 + 2√2)/(3 - 2√2) Hence, proved that the ratio of the two numbers is (3 + 2√2) : (3 - 2√2). 29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A ± √(A + G)(A - G). SOLUTION Let the two required positive numbers be a and b. It is given that the AM of the two numbers is A and their GM is G. Therefore, (a + b)/2 = A a + b = 2A AND √ab = G Square both sides ab = G2 We know that (x - y)2 = (x + y)2 - 4xy. Therefore, (a - b)2 = (a + b)2 - 4ab (a - b)2 = (2A)2 - 4(G2) (a - b)2 = 4A2 - 4G2 (a - b)2 = 4(A2 - G2) (a - b)2 = 4(A + G) (A - G) Take square root on both sides (a - b) = √4(A + G) (A - G) a - b = 2√(A + G) (A - G) a = 2√(A + G) (A - G) + b Now, a + b = 2A 2√(A + G) (A - G) + b + b = 2A 2√(A + G) (A - G) + 2b = 2A 2(√(A + G)(A - G) + b) = 2A √(A + G)(A - G) + b = A b = A - √(A + G)(A - G) Thus, a = 2√(A + G) (A - G) + b a = 2√(A + G) (A - G) + A - √(A + G)(A - G) a = A + √(A + G)(A - G) Hence, proved that the two numbers are A ± √(A + G)(A - G). 30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour? SOLUTION Bacteria present in the first hour = 30 Bacteria present in the second hour = 30 × 2 = 60 Bacteria present in the third hour = 60 × 2 = 120 and so on until the nth hour. Thus, it can be observed that the sequence of the number of bacteria at the end of every hour form a GP. First term of the GP = a = 30 Common ratio of the GP = r = 2 Number of bacteria at the end of 2nd hour = a3 = ar2 = 30 × 22 = 30 × 4 = 120 Number of bacteria at the end of the 4th hour = a5 = ar4 = 30 × 24 = 480 Number of bacteria at the end of the nth hour = an + 1 = arn = 30 × 2n = 30(2n) 31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually? SOLUTION Initial deposited amount = Rs 500 Interest rate after each year = 10% Amount at the end of the 1st year = 500 + 500 × 10/100 = 500 (1 + 0.1) = 500 (1.1) Amount at the end of the 2nd year = 500 (1.1)(1.1) = 500 (1.1)2 Amount at the end of the 3rd year = 500 (1.1)2(1.1) = 500 (1.1)3 and so on until the 10th year Thus, it can be observed that the sequence of the amount of money at the end of every year form a GP. First term of the GP = a = 500 Common ratio of the GP = r = 1.1 10th term of the GP = a10 = ar10 = 500 × (1.1)10 Hence, the Rs 500 will amount to Rs 500 × (1.1)10 after 10 years. 32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation. SOLUTION Let the roots of the quadratic equation be a and b. It is given that the AM of a and b is 8 and their GM is 5. Therefore, (a + b)/2 = 8 a + b = 16 AND √ab = 5 Square both sides ab = 25 We can form the quadratic equation with the following format: x2 - (Sum of roots)x + (Product of roots) = x2 - (a + b)x + ab = x2 - 16x + 25 Hence, the required quadratic equation is x2 - 16x + 25. Exercise 9.4Find the sum to n terms of each of the series in Exercises 1 to 7. 1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … SOLUTION The series can be rewritten as 1 × (1 + 1) + 2 × (2 + 1) + 3 × (3 + 1) + 4 × (4 + 1) + … Thus, the nth term in the series will be an = n (n + 1) Sum of the given series = n∑k = 1 ak = n∑k = 1 k(k + 1) = n∑k = 1 (k2 + k) = n∑k = 1k2+ n∑k = 1k = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)/2 × [(2n + 1)/3 + 1] = n(n + 1)/2 × [2n + 4]/3 = n(n + 1)/2 × 2(n + 2)/3 = n(n + 1)(n + 2)/3 2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … SOLUTION The series can be rewritten as 1 × (1 + 1) × (1 + 2) + 2 × (2 + 1) × (2 + 2) + 3 × (3 + 1) × (3 + 2) + … Thus, the nth term in the series will be an = n(n + 1)(n + 2) = (n2 + n)(n + 2) = n3 + 2n2 + n2 + 2n = n3 + 3n2 + 2n Sum of the given series = n∑k = 1 ak = n∑k = 1 (k3 + 3k2 + 2k) = n∑k = 1 k3 + 3 n∑k = 1 k2 + 2 n∑k = 1 k = [n(n + 1)/2]2 + 3n(n + 1)(2n + 1)/6 + 2n(n + 1)2 = [n(n + 1)/2]2 + n(n + 1)(2n + 1)/6 + n(n + 1) = n(n + 1)/2 × [n(n + 1)/2 + (2n + 1) + 2] = n(n + 1)/2 × [n2 + n + 4n + 2 + 4]/2 = n(n + 1)/4 × [n2 + 5n + 6] = n(n + 1)/4 × [n2 + 2n + 3n + 6] = n(n + 1)/4 × [n(n + 2) + 3(n + 2)] = n(n + 1)/4 × (n + 2)(n + 3) = n(n + 1)(n + 2)(n + 3)/4 3. 3 × 12 + 5 × 22 + 7 × 32 + … SOLUTION The series can be rewritten as (2(1) + 1) × 12 + (2(2) + 1) × 22 + (2(3) + 1) × 32 + … Thus, the nth term in the series will be an = (2n + 1)n2 = 2n3 + n2 Sum of the given series = n∑k = 1 ak = n∑k = 1 (2k3 + k2) = 2 n∑k = 1 k3 + n∑k = 1 k2 = 2 × [n(n + 1)/2]2 + n(n + 1)(2n + 1)/6 = n2(n + 1)2/2 + n(n + 1)(2n + 1)/6 = n(n + 1)/2 × [n(n + 1) + (2n + 1)/3] = n(n + 1)/2 × [(3n2 + 3n + 2n + 1)/3] = n(n + 1)/2 × (3n2 + 5n + 1)/3 = n(n + 1)(3n2 + 5n + 1)/6 4. 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + … SOLUTION The series can be rewritten as 1/(1 × (1 + 1)) + 1/(2 × (2 + 1)) + 1/(3 × (3 + 1)) + … Thus, the nth term in the series will be an = 1/n(n + 1) = 1/n - 1/(n + 1) Sum of the given series = n∑k = 1 ak = n∑k = 1 (1/k - 1/(k + 1)) = n∑k = 1 1/k - n∑k = 1 1/(k + 1) = [1/1 + 1/2 + 1/3 + … + 1/n] - [1/2 + 1/3 + … + 1/(n + 1)] = 1 - 1/(n + 1) = (n + 1 - 1)/(n + 1) = n/(n + 1) 5. 52 + 62 + 72 + … + 202 SOLUTION The series can be rewritten as (1 + 4)2 + (2 + 4)2 + (3 + 4)2 + … + (16 + 4)2 Thus, the nth term in the series will be an = (n + 4)2 = n2 + 8n + 16 Sum of the given series = n∑k = 1 ak = n∑k = 1 (k2 + 8k + 16) = n∑k = 1 k2 + 8 n∑k = 1 k + n∑k = 116 = n(n + 1)(2n + 1)/6 + 8 × n(n + 1)/2 + 16n Now, the last term of the series = 202 = (16 + 4)2 So, n = 16 Sum of the given series = S16 = 16(16 + 1)(2 × 16 + 1)/6 + 8 × 16(16 + 1)/2 + 16 × 16 = 16(17)(33)/6 + 4 × 16(17) + 256 = 8(17)(11) + 1088 + 256 = 1496 + 1344 = 2840 6. 3 × 8 + 6 × 11 + 9 × 14 + … SOLUTION The series can be rewritten as 1 × 3 × (1 × 3 + 5) + 2 × 3 × (2 × 3 + 5) + 3 × 3 × (3 × 3 + 5) + … Thus, the nth term in the series will be an = 3n(3n + 5) = 9n2 + 15n Sum of the given series = n∑k = 1 ak = n∑k = 1 (9k2 + 15k) = 9 n∑k = 1 k2 + 15 n∑k = 1 k = 9 × n(n + 1)(2n + 1)/6 + 15 × n(n + 1)/2 = 3n(n + 1)(2n + 1)/2 + 15n(n + 1)/2 = 3n(n + 1)/2 × [(2n + 1) + 5] = 3n(n + 1)/2 × [2n + 6] = 3n(n + 1)/2 × 2(n + 3) = 3n(n + 1)(n + 3) 7. 12 + (12 + 22) + (12 + 22 + 32) + … SOLUTION First term of the series = 12 Second term of the series = 12 + 22 Third term of the series = 12 + 22 + 32 Thus, the nth term in the series will be an = 12 + 22 + 32 + … n2 = n(n + 1)(2n + 1)/6 = n(2n2 + 2n + n + 1)/6 = n(2n2 + 3n + 1)/6 = (2n3 + 3n2 + n)/6 = n3/3 + n2/2 + n/6 Sum of the given series = n∑k = 1 ak = n∑k = 1 (k3/3 + k2/2 + k/6) = 1/3 × n∑k = 1 k3 + 1/2 × n∑k = 1 k2 + 1/6 × n∑k = 1 k = 1/3 × [n(n + 1)/2]2 + 1/2 × n(n + 1)(2n + 1)/6 + 1/6 × n(n + 1)/2 = n2(n + 1)/12 + n(n + 1)(2n + 1)/12 + n(n + 1)/12 = n(n + 1)/12 × [n(n + 1) + (2n + 1) + 1] = n(n + 1)/12 × [n(n + 1) + 2n + 2] = n(n + 1)/12 × [n(n + 1) + 2(n + 1)] = n(n + 1)/12 × (n + 1)(n + 2) = n(n + 1)2(n + 2)/12 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by 8. n (n + 1) (n + 4). SOLUTION nth term of the series = an = n(n + 1)(n + 4) = n(n2 + 4n + n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n Sum of the series = Sn = n∑k = 1 ak = n∑k = 1 (k3 + 5k2 + 4k) = n∑k = 1 k3 + 5 × n∑k = 1 k2 + 4 × n∑k = 1 k = [n(n + 1)/2]2 + 5 × n(n + 1)(2n + 1)/6 + 4 × n(n + 1)/2 = n2(n + 1)2/4 + 5n(n + 1)(2n + 1)/6 + 4n(n + 1)/2 = n(n + 1)/2 × [n(n + 1)/2 + 5(2n + 1)/3 + 4] = n(n + 1)/2 × [3n(n + 1) + 10(2n + 1) + 24]/6 = n(n + 1)/2 × [3n2 + 3n + 20n + 10 + 24]/6 = n(n + 1)/2 × [3n2 + 23n + 34]/6 = n(n + 1)(3n2 + 23n + 34)/12 9. n2 + 2n SOLUTION nthterm of the series = an = n2 + 2n Sum of the series = Sn = n∑k = 1 ak = n∑k = 1 (k2 + 2k) = n∑k = 1 k2+ n∑k = 1 2k Now, n∑k = 1 k2 = n(n + 1)(2n + 1)/6 n∑k = 1 2k = 21 + 22 + 23 + … + 2n The terms of 21 + 22 + 23 + … + 2n form a GP. First term of the GP = a = 2 Common ratio = r = 22/2 = 2 Sum of the n terms of the GP = a(1 - rn)/(1 - r) = 2(1 - 2n)/(1 - 2) = 2(1 - 2n)/(-1) = 2(2n - 1) Hence, sum of the given series = n∑k = 1 k2+ n∑k = 1 2k = n(n + 1)(2n + 1)/6 + 2(2n - 1) 10. (2n - 1)2 SOLUTION nthterm of the series = an = (2n - 1)2 = 4n2 - 4n + 1 Sum of the series = Sn = n∑k = 1 ak = n∑k = 1 (4k2 - 4k + 1) = 4 n∑k = 1 k2 - 4 n∑k = 1 k + n∑k = 1 1 = 4 × n(n + 1)(2n + 1)/6 - 4 × n(n + 1)/2 + n = 2n(n + 1)(2n + 1)/3 - 2n(n + 1) + n = n × [2(n + 1)(2n + 1)/3 - 2(n + 1) + 1] = n × [2(2n2 + 2n + n + 1)/3 - 2n - 2 + 1] = n × [2(2n2 + 3n + 1)/3 - 2n - 1] = n × [4n2 + 6n + 2 - 6n - 3]/3 = n/3 × (4n2 - 1) = n/3 × [(2n)2 - 12] = n/3 × (2n + 1)(2n - 1) = n(2n + 1)(2n - 1)/3 Miscellaneous Exercise1. Show that the sum of (m + n)th and (m - n)th terms of an A.P. is equal to twice the mth term. SOLUTION Let there be an AP with first term = a and common difference = d. (m + n)th term of the AP = am + n = a + (m + n - 1)d (m - n)th term of the AP = am - n = a + (m - n - 1)d mth term of the AP = am= a + (m - 1)d Sum of (m + n)th and (m - n)th terms of the AP = am + n + am - n = a + (m + n - 1)d + a + (m - n - 1)d = 2a + d(m + n - 1 + m - n - 1) = 2a + d(2m - 2) = 2a + 2d(m - 1) = 2 × [a + (m - 1)d] = 2 × am Hence, proved that the sum of (m + n)th and (m - n)th terms of an A.P. is equal to twice the mth term. 2. If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers. SOLUTION Let the three numbers in AP be a - d, a, and a + d. It is given that the sum of the three numbers is 24 and their product is 440. Therefore, a - d + a + a + d = 24 3a = 24 a = 8 AND (a - d) × a × (a + d) = 440 a (a2 - d2) = 440 8 (64 - d2) = 440 64 - d2 = 55 d2 = 9 d = ± 3 When d = 3, the three numbers are 8 - 3, 8, 8 + 3 ⇒ 5, 8, 11 When d = -3, the three numbers are 8 - (-3), 8, 8 + (-3) ⇒ 11, 8, 5. Hence, the three numbers are 5, 8, and 11 OR 11, 8 and 5. 3. Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3(S2 - S1). SOLUTION Let there be an AP with first term = a and common difference = d. Sum of the first n terms of the AP = S1 = n/2 × [2a + (n - 1)d] Sum of the first 2n terms of the AP = S2 = 2n/2 × [2a + (2n - 1)d] = n [2a + (2n - 1)d] Sum of the first 3n terms of the AP = S3 = 3n/2 × [2a + (3n - 1)d] Now, S2 - S1 = n [2a + (2n - 1)d] - n/2 × [2a + (n - 1)d] S2 - S1 = n × [2a + (2n - 1)d - 1/2 × {2a + (n - 1)d}] S2 - S1 = n × [4a + (2n - 1)2d - 2a - (n - 1)d]/2 S2 - S1 = n/2 × [2a + (4n - 2)d - (n - 1)d] S2 - S1 = n/2 × [2a + d(4n - 2 - n + 1)] S2 - S1 = n/2 × [2a + (3n - 1)d] Multiply both sides by 3 3(S2 - S1) = 3n/2 × [2a + (3n - 1)d] 3(S2 - S1) = S3 Hence, proved. 4. Find the sum of all numbers between 200 and 400 which are divisible by 7. SOLUTION The numbers between 200 and 400 that are divisible by 7 are: 203, 210, 217, … , 399 This sequence forms an AP where First term = a = 203 Common difference = d = 7 Last term = an = a + (n - 1)d 399 = 203 + (n - 1)7 196 = (n - 1)7 28 = n - 1 n = 29 Therefore, there are 29 numbers between 200 and 400 that are divisible by 7. Sum of the AP = Sn = n/2 × [2a + (n - 1)d] S29 = 29/2 × [2(203) + 28(7)] = 29/2 × [406 + 196] = 29/2 × 602 = 29 × 301 = 8729 Hence, the sum of all numbers between 200 and 400 that are divisible by 7 is 8729. 5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5. SOLUTION The integers from 1 to 100 that are divisible by 2 are: 2, 4, 6, …, 100 This sequence forms an AP where First term = a = 2 Common difference = d = 2 Last term = an = a + (n - 1)d 100 = 2 + (n - 1)2 98 = (n - 1)2 49 = n - 1 n = 50 Therefore, there are 50 numbers from 1 to 100 that are divisible by 2. Sum of the AP = Sn = n/2 × [2a + (n - 1)d] S50 = 50/2 × [2(2) + 49(2)] = 50/2 × [4 + 198] = 25 × 102 = 2550 The integers from 1 to 100 that are divisible by 5 are: 5, 10, 15, …, 100 This sequence forms an AP where First term = a = 5 Common difference = d = 5 Last term = an = a + (n - 1)d 100 = 5 + (n - 1)5 95 = (n - 1)5 19 = n - 1 n = 20 Therefore, there are 20 numbers from 1 to 100 that are divisible by 5. Sum of the AP = Sn = n/2 × [2a + (n - 1)d] S20 = 20/2 × [2(5) + 19(5)] = 10 × [10 + 95] = 10 × 105 = 1050 The integers from 1 to 100 that are divisible by 2 and 5 are: 10, 20, 30, …, 100 This sequence forms an AP where First term = a = 10 Common difference = d = 10 Last term = an = a + (n - 1)d 100 = 10 + (n - 1)10 90 = (n - 1)10 9 = n - 1 n = 10 Therefore, there are 10 numbers from 1 to 100 that are divisible by 2 and 5. Sum of the AP = Sn = n/2 × [2a + (n - 1)d] S10 = 10/2 × [2(10) + 9(10)] = 5 × [20 + 90] = 5 × 110 = 550 Required Sum = 2550 + 1050 - 550 = 3050 Hence, the sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050. 6. Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder. SOLUTION The two digit numbers that yield 1 as the remainder when divided by 4 are: 13, 17, 21, …, 97 This sequence forms an AP where First term = a = 13 Common difference = d = 4 Last term = an = a + (n - 1)d 97 = 13 + (n - 1)4 84 = (n - 1)4 21 = n - 1 n = 22 Therefore, there are 22 two digit numbers that yield 1as the remainder when divided by 4. Sum of the AP = Sn = n/2 × [2a + (n - 1)d] S22 = 22/2 × [2(13) + 21(4)] = 11 × [26 + 84] = 11 × 110 = 1210 Hence, the sum of all two digit numbers that yield 1as the remainder when divided by 4 is 1210. 7. If f is a function satisfying f (x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and n∑x = 1 f(x) = 120, find the value of n. SOLUTION f (x + y) = f(x) f(y) (Given) f(1) = 3 (Given) Put x = 1 and y = 1 in f (x + y) = f(x) f(y) f (1 + 1) = f(1) f(1) f (2) = 3 × 3 f (2) = 9 Similarly, f (1 + 1 + 1) = f (3) = f (1 + 2) = f(1) f(2) = 3 × 9 = 27 f (1 + 1 + 1 + 1) = f(4) = f(1 + 3) = f(1) f(3) = 3 × 27 = 81 Now, the sequence f(1), f(2), f(3), … OR 3, 9, 27, … forms a GP where First term = a = 3 Common ratio = r = 9/3 = 3 Sum of the GP = Sn = a(1 - rn)/(1 - r) It is given that n∑x = 1 f(x) = 120. Thus, 120 = 3(1 - 3n)/(1 - 3) 40 = (1 - 3n)/(-2) 80 = 3n - 1 3n = 81 3n = 34 Compare the exponents n = 4 8. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. SOLUTION First term = a = 5 Common ratio = r = 2 Let the number of terms that make up the sum 315 be n. Sum of n terms of the GP = a(1 - rn)/(1 - r) 315 = 5(1 - 2n)/(1 - 2) 63 = (1 - 2n)/(-1) 63 = 2n - 1 2n = 64 2n = 26 Compare the exponents n = 6 Last term of the GP = a6 = ar5 = 5 × 25 = 160 Hence, the number of terms is 6 and the last term of the GP is 160. 9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P. SOLUTION First term = a = 1 Let the common ratio of the GP be r. Third term of the GP = a3 = ar2 = (1)r2 = r2 Fifth term of the GP = a5 = ar4 = r4 It is given that the sum of the fifth and the third terms of the GP is 90. Therefore, a3 + a5 = 90 r2 + r4 = 90 r4+ r2 - 90 = 0 Using the quadratic formula r2 = [-1 ± √(1 + 4(1)(90))]/2(1) = [-1 ± √361]2 = (-1 ± 19)/2 When r2 = (-1 + 19)/2 r2 = 18/2 r2 = 9 r = ± 3 When r2 = (-1 - 19)/2 r2 = -20/2 r2 = -10 The roots are not real. Therefore, r2 = (-1 + 19)/2, which implies that r = ± 3. Hence, common ratio of the GP is ± 3. 10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. SOLUTION Let the three numbers in GP be a, ar, and ar2. It is given that the sum of these numbers is 56. Therefore, a + ar + ar2 = 56 a(1 + r + r2) = 56 a = 56/(1 + r + r2) It is also given that upon subtracting 1, 7 and 21 from the three numbers respectively, an AP is formed. Therefore, a - 1, ar - 7, ar2 - 21 is in AP Common difference of the AP = ar - 7 - (a - 1) = ar2 - 21 - (ar - 7) ar - 7 - a + 1 = ar2 - 21 - ar + 7 ar - a - 6 = ar2 - ar - 14 ar2 - ar - ar + a = 8 ar2 - 2ar + a = 8 a(r2 - 2r + 1) = 8 56/(1 + r + r2) × (r2 - 2r + 1) = 8 7 (r2 - 2r + 1) = 1 + r + r2 7r2 - 14r + 7 = 1 + r + r2 6r2 - 15r + 6 = 0 3(2r2 - 5r + 2) = 0 2r2 - 5r + 2 = 0 2r2 - 4r - r + 2 = 0 2r(r - 2) - 1(r - 2) = 0 (r - 2) (2r - 1) = 0 (r - 2) = 0 ⇒ r = 2 (2r - 1) = 0 ⇒ 2r = 1 ⇒ r = 1/2 Therefore, When r = 2: a = 56/(1 + 2 + 22) = 56/(3 + 4) = 56/7 a = 8 The three numbers are 8, 8(2), 8(2)2 ⇒ 8, 16, 32 When r = 1/2: a = 56/(1 + 1/2 + 1/4) = 56/[(4 + 2 + 1)/4] = 56 (4)/7 a = 32 The three numbers are 32, 32(1/2), 32(1/2)2 ⇒ 32, 16, 8 Hence, the required three numbers are 8, 16 and 32 OR 32, 16, and 8. 11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. SOLUTION Let the terms of the GP be T1, T2, T3, …, T2n. It is given that the sum of all terms is 5 times that sum of terms which occupy the odd places in the GP. Therefore, T1 + T2 + T3 + … + T2n = 5 × [T1 + T3 + T5 + … + T2n - 1] [T1 + T2 + T3 + … + T2n] - 5 [T1 + T3 + T5 + … + T2n - 1] = 0 [T2 + T4 + T6 + … + T2n] + [T1 + T3 + T5 + … + T2n - 1] - 5 [T1 + T3 + T5 + … + T2n - 1] = 0 [T2 + T4 + T6 + … + T2n] - 4 [T1 + T3 + T5 + … + T2n - 1] = 0 [T2 + T4 + T6 + … + T2n] = 4 [T1 + T3 + T5 + … + T2n - 1] ar(rn - 1)/(r - 1) = 4ar(rn - 1)/(1 - r) r = 4 Hence, the common ratio of the GP is 4. 12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms. SOLUTION Let the AP be a, a + d, a + 2d, …, a + (n - 2)d, a + (n - 1)d. First term = a = 11 Common difference = d Sum of the first four terms = S4 = 4/2 × [2a + 3d] 56 = 2 × [2a + 3d] 28 = 2a + 3d 3d = 28 - 2a d = (28 - 2(11))/3 d = (28 - 22)/3 d = 2 Sum of the last four terms = a + (n - 4)d + a + (n - 3)d + a + (n - 2)d + a + (n - 1)d 112 = 4a + (n - 4 + n - 3 + n - 2 + n - 1)d 112 = 4a + (4n - 10)d 112 = 4(11) + 2(2n - 5)(2) 112 = 4 × [11 + (2n - 5)] 28 = 11 + 2n - 5 28 = 6 + 2n 2n = 22 n = 11 Hence, the number of terms in the AP is 11. 13. If (a + bx)/(a - bx) = (b + cx)/(b - cx) = (c + dx)/(c - dx) (x ≠ 0), then show that a, b, c, and d are in GP. SOLUTION (a + bx)/(a - bx) = (b + cx)/(b - cx) (a + bx) (b - cx) = (b + cx) (a - bx) ab + b2x - acx - bcx2 = ab + acx - b2x - bcx2 2b2x - 2acx = 0 2b2x = 2acx b2x = acx b2 = ac b/a = c/b (b + cx)/(b - cx) = (c + dx)/(c - dx) (b + cx) (c - dx) = (c + dx) (b - cx) bc + c2x - bdx - cdx2 = bc + bdx - c2x - cdx2 2c2x - 2bdx = 0 2c2x = 2bdx c2 = bd c/b = d/c Therefore, b/a = c/b = d/c Hence, a, b, c, and d are in GP. 14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn. SOLUTION Let the terms in the GP be a, ar, ar2, …, arn - 1. First term = a Common ratio = r Sum of the n terms of the GP = Sn = a(1 - rn)/(1 - r) S = a(1 - rn)/1(1 - r) Product of the n terms of the GP = a × ar × ar2 × … × arn - 1 P = an × r1 + 2 + 3 + … + (n - 1) = an × rn(n - 1)/2 Sum of the reciprocals of n terms of the GP = 1/a + 1/ar + 1/ar2 + … + 1/arn - 1 R = 1/a × [1 + 1/r + 1/r2 + 1/r3 + … + 1/rn - 1] = 1/a × (rn - 1 + rn - 2 + rn - 3 + … + 1)/rn - 1 The terms in rn - 1 + rn - 2 + rn - 3 + … + 1 form a GP. Therefore, rn - 1 + rn - 2 + rn - 3 + … + 1 = 1(1 - rn)/(1 - r) Thus, R = 1/arn - 1 × 1(1 - rn)/(1 - r) = (1 - rn)/arn - 1(1 - r) Now, P2Rn = (an × rn(n - 1)/2)2 × [(1 - rn)/arn - 1(1 - r)]n = a2n rn(n - 1) × (1 - rn)n/anrn(n - 1)(1 - r)n = an(1 - rn)n/(1 - r)n = [a(1 - rn)/(1 - r)]n = Sn Hence, proved. 15. The pth , qth and rth terms of an A.P. are a, b, c, respectively. Show that (q - r )a + (r - p )b + (p - q )c = 0. SOLUTION Let the first term of the AP be A and its common difference be D. pth term of the AP Ap= A + (p - 1)D a = A + (p - 1)D qth term of the AP Aq= A + (q - 1)D b = A + (q - 1)D rth term of the AP Ar= A + (r - 1)D c = A + (r - 1)D Aq - Ar = A + (q - 1)D - [A + (r - 1)D] b - c = A + (q - 1)D - A - (r - 1)D b - c = D(q - 1 - r + 1) b - c = D(q - r) D = (b - c)/(q - r) Ap - Aq = A + (p - 1)D - [A + (q - 1)D] a - b = A + (p - 1)D - A - (q - 1)D a - b = D(p - 1 - q + 1) a - b = D(p - q) D = (a - b)/(p - q) Therefore, (b - c)/(q - r) = (a - b)/(p - q) (p - q) (b - c) = (a - b) (q - r) bp - bq - cp + cq = aq - bq - ar + br bp - cp + cq - aq + ar - br = 0 -aq + ar + bp - br - cp + cq = 0 -a(q - r) - b(r - p) - c(p - q) = 0 a(q - r) + b(r - p) + c(p - q) = 0 Hence, proved. 16. If a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in A.P., prove that a, b, c are in A.P. SOLUTION a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP. Therefore, Common difference = b(1/c + 1/a) - a(1/b + 1/c) = c(1/a + 1/b) - b(1/c + 1/a) b(a + c)/ac - a(c + b)/bc = c(b + a)/ab - b(a + c)/ac [(ab + bc)b - a(ac + ab)]/abc = [c(bc + ac) - b(ab + bc)]/abc ab2 + b2c - a2c - a2b = bc2 + ac2 - ab2 - b2c ab2 - a2b + b2c - a2c = ac2 - ab2 + bc2 - b2c ab(b - a) + c(b2 - a2) = a(c2 - b2) + bc(c - b) ab(b - a) + c(b - a) (b + a) = a(c - b) (c + b) + bc(c - b) (b - a) [ab + c(b + a)] = (c - b) [a(c + b) + bc] (b - a) [ab + bc + ac] = (c - b) [ac + ab + bc] b - a = c - b Hence, a, b and c are in AP. 17. If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P. SOLUTION Given, a, b, c, and d are in GP b2 = ac … (i) c2 = bd … (ii) ad = bc … (iii) Required to prove (an + bn), (bn + cn), (cn + dn) are in GP i.e., (bn + cn)2 = (an + bn) (cn + dn) Taking LHS (bn + cn)2 = b2n2 + 2bncn + c2n2 = (b2)n2+ 2bncn + (c2) n = (ac)n2 + 2bncn + (bd)n2 [Using (i) and (ii)] = ancn + bncn + bncn + bndn = ancn + bncn + andn + bndn [Using (iii)] = cn (an + bn) + dn (an + bn) = (an + bn) (cn + dn) = RHS Therefore, (an + bn), (bn + cn), and (cn + dn) are in GP. Hence proved. 18. If a and b are the roots of x2 - 3x + p = 0 and c, d are roots of x2 - 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q - p) = 17:15. SOLUTION Given, a and b are the roots of x2 - 3x + p = 0 So, we have a + b = 3 and ab = p … (i) Also, c and d are the roots of x2 - 12x + q = 0 So, c + d = 12 and cd = q … (ii) And given a, b, c, d are in G.P. Let's take a = x, b = xr, c = xr2, d = xr3 From (i) and (ii), we get x + xr = 3 x (1 + r) = 3 And, xr2 + xr3 =12 xr2 (1 + r) = 12 On dividing, we get xr2(1 + r)/x(1 + r) = 12/3 r2 = 4 r = ± 2 When r = 2, x = 3/(1 + 2) = 3/3 = 1 When r = -2, x = 3/(1 - 2) = 3/-1 = -3 Case I: When r = 2 and x =1, ab = x2r = 2 cd = x2r5 = 32 (q + p)/(q - p) = (32 + 2)/(32 - 2) = 34/30 = 17/15 (q + p) : (q - p) = 17 : 15 Case II: When r = -2, x = -3, ab = x2r = -18 cd = x2r5 = -288 (q + p)/(q - p) = (-288 - 18)/(-288 + 18) = -306/-270 = 17/15 (q + p) : (q - p) = 17 : 15 Hence, proved. 19. The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a : b = (m + √(m2 - n2)) : (m - √(m2 - n2)). SOLUTION AM of a and b = (a + b)/ 2 and GM of a and b = √ab It is given that the ratio of the AM and GM of a and b is m : n. Therefore, (a + b)/2√ab = m/n Square both sides (a + b)2/4ab = m2/n2 (a + b)2 = 4abm2/n2 a + b = 2m√ab/n … (i) We know that (a - b)2 = (a + b)2 - 4ab (a - b)2 = 4abm2/n2 - 4ab (a - b)2 = 4abm2/n2 - 4abn2/n2 (a - b)2 = 4ab × (m2 - n2)/n2 Take square root on both sides a - b = 2√ab × √(m2 - n2)/n … (ii) Add (i) and (ii) a + b + a - b = 2m√ab/n + 2√ab × √(m2 - n2)/n 2a = [2m√ab + 2√ab√(m2- n2)]/n 2an = 2√ab (m + √(m2 - n2)) a = √ab [m + √(m2 - n2)]/n Substitute the value of a in (i) a + b = 2m√ab/n b = 2m√ab/n - a b = 2m√ab/n - √ab [m + √(m2 - n2)]/n b = √ab × [2m - m - √(m2 - n2)]/n b = √ab [m - √(m2 - n2)]/n Now, a/b = [√ab [m + √(m2 - n2)]/n]/[√ab [m - √(m2 - n2)]/n] a/b = [m + √(m2 - n2)]/[m - √(m2 - n2)] a : b = [m + √(m2 - n2)] : [m - √(m2 - n2)] Hence, proved. 20. If a, b, c are in A.P.; b, c, d are in G.P. and 1/c, 1/d, 1/e are in A.P. prove that a, c, e are in G.P. SOLUTION It is given that a, b, c are in AP. Therefore, Common difference = b - a = c - b 2b = a + c b = (a + c)/2 It is also given that b, c, d are in GP. Therefore, Common ratio = c/b = d/c d = c2/b Also, 1/c, 1/d, 1/e are in AP. Therefore, Common difference = 1/d - 1/c = 1/e - 1/d 2/d = 1/e + 1/c 2/(c2/b) = 1/e + 1/c 2b/c2 = 1/e + 1/c 2(a + c)/2c2= 1/e + 1/c (a + c)/c2 = (e + c)/ec (a + c)/c = (e + c)/e (a + c)e = (e + c)c ae + ce = ec + c2 ae = c2 Hence, a, c, and e are in GP. 21. Find the sum of the following series up to n terms: (i) 5 + 55 +555 + … (ii) .6 +. 66 +. 666 + … SOLUTION (i) Sum of the given series = S = 5 + 55 + 555 + … upto n terms S = 5/9 × [9 + 99 + 999 + … upto n terms] = 5/9 × [(10 - 1) + (100 - 1) + (1000 - 1) + … upto n terms] = 5/9 × [(101 - 1) + (102 - 1) + (103 - 1) + … upto n terms] = 5/9 × [(101 + 102 + 103 + … + 10n) - (1 + 1 + 1 + … n times)] The terms in 101 + 102 + 103 + … + 10n form a GP where first term = a = 10 and common ratio = r = 10. Sum of n terms of the GP = 101 + 102 + 103 + … + 10n = a(1 - rn)/(1 - r) = 10(1 - 10n)/(1 - 10) Therefore, S = 5/9 × [10(1 - 10n)/(1 - 10) - n × 1] = 5/9 × [10(1 - 10n)/(-9) - n] = 5/9 × [10(10n - 1)/9 - n] = 50(10n - 1)/81 - 5n/9 (ii) Sum of the given series = S = 0.6 + 0.66 + 0.666 + … upto n terms S = 6 × [0.1 + 0.11 + 0.111 + … upto n terms] = 6/9 × [0.9 + 0.99 + 0.999 + … upto n terms] = 6/9 × [(1 - 1/10) + (1 - 1/100) + (1 - 1/1000) + … upto n terms] = 6/9 × [(1 - 1/101) + (1 - 1/102) + (1 - 1/103) + … upto n terms] = 6/9 × [(1 + 1 + 1 + … upto n terms) - (1/101 + 1/102 + 1/103+ … 1/10n)] The terms in 1/101 + 1/102 + 1/103 + … + 1/10n form a GP where first term = a = 1/10 and common ratio = r = 1/10. Sum of n terms of the GP = 1/101 + 1/102 + 1/103 + … + 1/10n = a(1 - rn)/(1 - r) = 1/10 × (1 - 1/10n)/(1 - 1/10) Therefore, S = 6/9 × [n × 1 - 1/10 × (1 - 1/10n)/(1 - 1/10)] = 6/9 × [n - 1/10 × (1 - 1/10n)/(0.9)] = 6/9 × [n - (1 - 1/10n)/9] = 2n/3 - 2(1 - 1/10n)/27 22. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms. SOLUTION The series can be rewritten as 2(1) × (2(1) + 2) + 2(2) × (2(2) + 2) + 2(3) × (2(3) + 2) + … + upto n terms Thus, the nth term in the series will be an = 2n × (2n + 2) = 4n2 + 4n 20th term of the series = a20 = 4(202) + 4(20) = 4(400) + 80 = 1600 + 80 = 1680 Hence, the 20th term of the given series is 1680. 23. Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + … SOLUTION Sum of the given series = S = 3 + 7 + 13 + 21 + 31 + … + an - 1 + an S = 3 + 7 + 13 + 21 + … + an - 1 + an Subtract both equations S - S = [3 + 7 + 13 + 21 + 31 + … + an - 1 + an] - [3 + 7 + 13 + 21 + … + an - 2 + an - 1 + an] 0 = [3 + (7 + 13 + 21 + 31 + … + an - 1 )] - [3 + 7 + 13 + 21 + … + an - 1 + an] 0 = 3 + [(7 - 3) + (13 - 7) + (21 - 13) + (31 - 21) + … + (an - an - 1)] - an an = 3 + [(4) + (6) + (8) + (10) + … + upto (n - 1) terms] The terms in 4 + 6 + 8 + 10 + … + upto (n - 1) terms form an AP. Therefore, an = 3 + [(n - 1)/2 × {2(4) + (n - 2)2}] an = 3 + [(n - 1)/2 × 2{4 + n - 2}] an = 3 + [(n - 1) × (n + 2)] an = 3 + (n2 - n + 2n - 2) an = 3 + (n2 + n - 2) an = n2 + n + 1 Now, Sum of the series = S = n∑k = 1 ak = n∑k = 1 (k2 + k + 1) = n∑k = 1 k2 + n∑k = 1 k + n∑k = 1 1 = n(n + 1)(2n + 1)/6 + n(n + 1)/2 + n = n × [(n + 1)(2n + 1)/6 + (n + 1)/2 + 1] = n × [(2n2+ 2n + n + 1) + 3(n + 1) + 6]/6 = n/6 × [2n2 + 3n + 1 + 3n + 3 + 6] = n/6 × [2n2 + 6n + 10] = n/6 × 2[n2 + 3n + 5] = n(n2 + 3n + 5)/3 24. If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9 S22 = S3 (1 + 8S1). SOLUTION According to the question, it is given that: S1 = Sum of first n natural numbers = n(n + 1)/2 S2 = Sum of the squares of first n natural numbers = n(n + 1)(2n + 1)/6 S3 = Sum of the cubes of first n natural numbers = [n(n + 1)/2]2 We need to prove that 9 S22 = S3 (1 + 8S1). Taking LHS = 9 S22 = 9 × [n(n + 1)(2n + 1)/6]2 = 9 × [n(n + 1)(2n + 1)]2/36 = [n(n + 1)(2n + 1)]2/4 = [n(n + 1)(2n + 1)/2]2 Taking RHS = S3 (1 + 8S1) = [n(n + 1)/2]2 × [1 + 8 × n(n + 1)/2] = [n(n + 1)/2]2 × [1 + 4n(n + 1)] = [n(n + 1)/2]2 × [1 + 4n2 + 4n] = [n(n + 1)/2]2 × [(2n)2 + 2(2n)(1) + 12] = [n(n + 1)/2]2 × (2n + 1)2 = [n(n + 1)(2n + 1)/2]2 = LHS Since, LHS = RHS. Hence, proved that 9 S22 = S3 (1 + 8S1). 25. Find the sum of the following series up to n terms: 13/1 + (13 + 23)/(1 + 3) + (13 + 23 + 33)/(1 + 3 + 5) + … SOLUTION The nth term in the series will be an = (13 + 23 + 33 + … + n3)/(1 + 3 + 5 + … + (2n - 1)) = [n(n + 1)/2]2/(1 + 3 + 5 + … + (2n - 1)) The terms in 1 + 3 + 5 + … + (2n - 1) form an AP. Sum of the AP = 1 + 3 + 5 + … + (2n - 1) = n/2 × [2(1) + (n - 1)2] = n/2 × 2[1 + n - 1] = n(n) = n2 So, an = [n(n + 1)/2]2/n2 = n2(n + 1)2/4n2 = (n + 1)2/4 = (n2 + 2n + 1)/4 Sum of the given series = S = n∑k = 1 ak = n∑k = 1 (k2 + 2k + 1)/4 = 1/4 n∑k = 1 k2 + 2/4 n∑k = 1 k + 1/4 n∑k = 1 1 = 1/4 × n(n + 1)(2n + 1)/6 + 1/2 × n(n + 1)/2 + n/4 = n(n + 1)(2n + 1)/24 + n(n + 1)/4 + n/4 = n/4 × [(n + 1)(2n + 1)/6 + (n + 1) + 1] = n/4 × [(n + 1)(2n + 1) + 6(n + 1) + 6(1)]/6 = n/24 × [2n2 + 2n + n + 1 + 6n + 6 + 6] = n/24 × [2n2 + 9n + 13] = n(2n2 + 9n + 13)/24 26. Show that [1 × 22 + 2 × 32 + … + n × (n + 1)2]/[12 × 2 + 22 × 3 + … + n2 × (n + 1)] = (3n + 5)/(3n + 1) SOLUTION nth term of the numerator = n × (n + 1)2 = n × (n2 + 2n + 1) = n3 + 2n2 + n nthterm of the denominator = n2 × (n + 1) = n3+ n2 LHS = [1 × 22 + 2 × 32 + … + n × (n + 1)2]/[12 × 2 + 22 × 3 + … + n2 × (n + 1)] = n∑k = 1 ak/n∑k = 1 ak = n∑k = 1 (k3 + 2k2 + 1)/n∑k = 1 (k3 + 2k2) = [n∑k = 1 k3 + 2 n∑k = 1 k2 + n∑k = 1 k]/[n∑k = 1 k3 + n∑k = 1 k2] Numerator = n∑k = 1 k3 + 2 n∑k = 1 k2 + n∑k = 1 k = [n(n + 1)/2]2 + 2 × n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)/2 × [n(n + 1)/2 + 2(2n + 1)/3 + 1] = n(n + 1)/2 × [3n(n + 1) + 4(2n + 1) + 6]/6 = n(n + 1)/2 × [3n2 + 3n + 8n + 4 + 6]/6 = n(n + 1)/12 × [3n2 + 11n + 10] = n(n + 1)/12 × [3n2 + 6n + 5n + 10] = n(n + 1)/12 × [3n(n + 2) + 5(n + 2)] = n(n + 1)/12 × (n + 2)(3n + 5) = n(n + 1)(n + 2)(3n + 5)/12 Denominator = n∑k = 1 k3 + n∑k = 1 k2 = [n(n + 1)/2]2 + n(n + 1)(2n + 1)/6 = n(n + 1)/2 × [n(n + 1)/2 + (2n + 1)/3] = n(n + 1)/2 × [3n(n + 1) + 2(2n + 1)]/6 = n(n + 1)/12 × [3n2 + 3n + 4n + 2] = n(n + 1)/12 × [3n2 + 7n + 2] = n(n + 1)/12 × [3n2 + 6n + n + 2] = n(n + 1)/12 × [3n(n + 2) + (n + 2)] = n(n + 1)/12 × (n + 2)(3n + 1) = n(n + 1)(n + 2)(3n + 1)/12 Therefore, LHS = [1 × 22 + 2 × 32 + … + n × (n + 1)2]/[12 × 2 + 22 × 3 + … + n2 × (n + 1)] = [n(n + 1)(n + 2)(3n + 5)/12]/[n(n + 1)(n + 2)(3n + 1)/12] = (3n + 5)/(3n + 1) = RHS Since, LHS = RHS. Hence, proved. 27. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him? SOLUTIONS Original cost of the tractor = Rs 12000 Amount paid by the farmer in cash = Rs 6000 Remaining balance to be paid = Rs 12000 - 6000 = Rs 6000 The farmer has agreed to pay Rs 500 + 12% interest on the remaining amount. Therefore, the interest paid annually will be: 12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500 Thus, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500 = 12% of (6000 + 5500 + 5000 + … + 500) = 12% of (500 + 1000 + 1500 + … + 6000) The terms in 500 + 1000 + 1500 + … + 6000 form an AP where first term = a = 500 and common difference = d = 500 Last term of the AP = an = a + (n - 1)d 6000 = 500 + (n - 1)500 5500 = (n - 1)500 11 = n - 1 n = 12 Sum of the AP = Sn = 500 + 1000 + 1500 + … + 6000 = n/2 × [2a + (n - 1)d] = 12/2 × [2(500) + 11(500)] = 6 × [1000 + 5500] = 6 × 6500 = 39000 Thus, the total interest to be paid = 12% of 39000 = 12/100 × 39000 = Rs 4680 Total cost of the tractor = Cost + Interest = Rs 12000 + 4680 = Rs 16680 Hence, the tractor will cost the farmer Rs 16680. 28. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him? SOLUTION Original cost of the scooter = Rs 22000 Amount paid by Shamshad Ali in cash = Rs 4000 Remaining balance to be paid = Rs 22000 - 4000 = Rs 18000 Shamshad Ali has agreed to pay Rs 1000 + 10% interest on the remaining amount. Therefore, the interest paid annually will be: 10% of 18000, 10% of 17000, 10% of 16000, …, 10% of 1000 Thus, the total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000 = 10% of (18000 + 17000 + 16000 + … + 1000) = 10% of (1000 + 2000 + 3000 + … + 18000) The terms in 1000 + 2000 + 3000 + … + 18000 form an AP where first term = a = 1000 and common difference = d = 1000 Last term of the AP = an = a + (n - 1)d 18000 = 1000 + (n - 1)1000 17000 = (n - 1)1000 17 = n - 1 n = 18 Sum of the AP = Sn = 1000 + 2000 + 3000 + … + 18000 = n/2 × [2a + (n - 1)d] = 18/2 × [2(1000) + 17(1000)] = 9 × [2000 + 17000] = 9 × 19000 = 171000 Thus, the total interest to be paid = 10% of 171000 = 10/100 × 171000 = Rs 17100 Total cost of the scooter = Cost + Interest = Rs 22000 + 17100 = Rs 39100 Hence, the scooter will cost Shamshad Rs 39100. 29. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed. SOLUTION Number of letters mailed by the person originally = 4 Number of letters mailed by the person's friends = 4 × 4 = 42 Number of letters mailed by the next recipients = 4 × 42 = 43 … and so on. The number of letters mailed in the chain at every step form a GP 4, 42, 43, … First term = a = 4 Common ratio = r = 4 Letters sent until the 8th set = Sum of first terms of the GP = S8 = 4(1 - 48)/(1 - 4) = 4(1 - 65536)/(-3) = 4(-65535)/(-3) = 4(21845) = 87380 Cost of mailing one letter = 50 paisa = Rs 0.5 Cost of mailing 87380 letters = Rs 0.5 × 87380 = 43690 Hence, the amount spent on the postage when the 8th set of the letter is mailed = Rs 43690. 30. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years. SOLUTION Amount deposited by the man = Rs 10000 Interest in each year = 5% of 10000 = 5/100 × 10000 = Rs 500 The sequence of amount after each year: 10000, 10500, 11000, … This sequence is an AP. First term = a = 10000 Common Difference = d = 500 Amount in the 15th year = a15 = a + 14d = 10000 + 14(500) = 10000 + 7000 = Rs 17000 Total amount after 20 years = a21 = a + 20d = 10000 + 20(500) = 10000 + 10000 = 20000 Hence, the amount in 15th year is Rs 17000 and total amount after 20 years is Rs 20000. 31. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years. SOLUTION Original cost of the machine = Rs 15625 The value of the machine will depreciate by 20% every year. Therefore, its value will be 80% of the cost after every year. Value after the end of 5 years = 15625 × 80/100 × 80/100 × 80/10 × … 5 times = 15625 × 4/5 × 4/5 × 4/5 × … 5 times = 15625 × 45/55 = 56 × 1024/ 55 = 5 × 1024 = Rs 5120 Hence, the value of the machine after the end of 5th year will be Rs 5120. 32. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed. SOLUTION Number of workers engaged originally = 150 Let the number of days in which 150 workers complete the job be x. It is given that each day 4 workers would drop out. Therefore, 150x = 150 + 146 + 142 + … + upto (x + 8) terms The terms in 150 + 146 + 142 + … + upto (x + 8) terms form an AP. First term = a = 150 and common difference = d = -4 Number of terms in the AP = (x + 8) Sum of the AP = 150 + 146 + 142 + … + upto (x + 8) = (x + 8)/2 × [2(150) + (x + 7)(-4)] Therefore, 150x = (x + 8)/2 × [2(150) + (x + 7)(-4)] 300x = (x + 8) × [300 - 4x - 28] 300x = (x + 8) × [272 - 4x] 300x = (x + 8) × 4(68 - x) 75x = (x + 8) (68 - x) 75x = 68x - x2 + 544 - 8x x2 + 75x = 60x + 544 x2 + 15x - 544 = 0 x2 + 32x - 17x - 544 = 0 x(x + 32) - 17(x + 32) = 0 (x + 32) (x - 17) = 0 (x + 32) = 0 ⇒ x = -32 OR (x - 17) = 0 ⇒ x = 17 x depicts number of days which cannot be negative, so x = -32 is rejected. Thus, x = 17. Hence, the number of days in which the job was completed due to workers dropping out = 17 + 8 = 25 days. |