## NCERT Solutions Class 11th Maths Chapter 9: Sequences and Series## Exercise 9.1
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a Hence, the first five terms of the given sequence are 3, 8, 15, 24, 35.
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a Hence, the first five terms of the given sequence are 1/2, 2/3, 3/4, 4/5 and 5/6. 3. a
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a Hence, the first five terms of the given sequence are 2, 4, 8, 16, 32.
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a Hence, the first five terms of the given sequence are -1/6, 1/6, 1/2, 5/6 and 7/6.
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a Hence, the first five terms of the given sequence are 25, -125, 625, -3125, and 15625.
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a Hence, the first five terms of the given sequence are 3/2, 9/2, 21/2, 21 and 75/2.
n 17 24
n 7
n 9
n 20 = 20(18)/23 = 360/23
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a 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 + …
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a 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) + …
n First term = a Second term = a Third term = a Fourth term = a Fifth term = a 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) + …
a n Third term = a Fourth term = a Fifth term = a Sixth term = a Vales of a n = 1 ⇒ a n = 2 ⇒ a n = 3 ⇒ a n = 4 ⇒ a n = 5 ⇒ a ## Exercise 9.2
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 = S = 1001/2 × [2(1) + (1000)2] = 1001/2 × [2 + 2000] = 1001/2 × 2002 = 1001 = 1002001 Hence, the sum of odd numbers from 1 to 2001 is 1002001.
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 = S = 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.
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 = S = 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
20 = 2 - 114 = -112 Hence, proved that the 20
Required sum = S First term = a = -6 Common difference = -11/2 - (-6) = -11/2 + 6 = 1/2 S -25 = n/2 × [2(-6) + (n - 1)/2] -50 = n [-12 + (n - 1)/2] -50 = -12n + n(n - 1)/2 -50 = [-24n + n -100 = n n n 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.
General term of an AP = a p q a 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
Now, a 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
Now, sum of first pq terms: S = 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).
First term = a = 25 Common difference = d = 22 - 25 = -3 Let the number of terms of the AP whose sum is 116 be n. S 116 = n/2 × [2(25) + (n - 1)(-3)] 232 = n × [50 - 3n + 3] 232 = 53n - 3n 3n 3n 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.
Therefore, the sum 116 is obtained with 8 terms of the AP. Last term = a = 25 + 7(-3) = 25 - 21 = 4 Hence, the last term is 4.
k 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 = S = n/2 × [2(6) + (n - 1)5] = n/2 × [12 + 5n - 5] = n/2 × [5n + 7] = (5n + 7)/2
Sum of n terms of the AP = S (pn + qn 2qn 2qn 2qn Compare the coefficients of n 2q = d Hence, common difference of the given AP is 2q.
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 a Hence, the ratio of the 18
Sum of the first p terms of the AP = S Sum of the first q terms of the AP = S It is given that the sum of first p terms of the AP is equal to the sum of the first q terms. Therefore, S 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(p 2a (p - q) + d(p 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 S Hence, the sum of first p + q terms of the AP is 0.
Let the first term of the AP = A and the common difference = d Sum of first p terms of the AP = S a = p/2 × [2A + (p - 1)d]
Sum of first q terms of the AP = S b = q/2 × [2A + (q - 1)d]
Sum of first r terms of the AP = S c = r/2 × [2A + (r - 1)d]
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)
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)
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.
Let the first term of the AP = a and the common difference = d Sum of m terms = S Sum of n terms = S It is given that the ratio of the sum of m and terms of the AP is m S m 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) = a a Hence, proved.
Let the first term of the AP = a and the common difference = d Sum of n terms of the AP = S 3n 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:
2a - d = 10 2a - 6 = 10 2a = 16
m 164 = 8 + (m - 1)6 156 = (m - 1)6 26 = m - 1
Hence, the value of m is 27.
Let the five numbers between 8 and 26 be A The AP will be 8, A Firs term = a = 8 a a + 6d = 26 8 + 6d = 26 6d = 18 d = 3 A A A A A Therefore, five numbers to between 8 and 26 in order to make the sequence an AP are 11, 14, 17, 20, and 23.
Arithmetic Mean between a and b = (a + b)/2 It is given that that Arithmetic Mean between a and b is (a (a + b)/2 = (a (a + b) (a a a ab b b (a/b) (a/b) Compare the exponents n - 1 = 0
Hence, value of n is 1.
Let the m numbers between 1 and 31 be A The AP will be 1, A First term = a = 1 a a + (n - 1)d = 31 1 + (m + 2 - 1)d = 31 (m + 1)d = 30
A A … A A It is given that the ratio of the 7 a [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
Hence, m = 14 numbers were inserted between 1 and 31.
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 30 = a + 29d = 100 + 29(5) = 100 + 145 = 245 Hence, the man will pay Rs 245 in the 30
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 = S 180 (n - 2) = n/2 × [2(120) + (n - 1)5] 360 (n - 2) = n × [240 + 5n - 5] 360n - 720 = n × [5n + 235] 360n - 720 = 5n 5n 5(n n n 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 9.3
First term = a = 5/2 Common ratio = r = (5/4)/(5/2) = 1/2 20 = 5/2 n = 5/2
Common ratio of the GP = r = 2 First term = a 8 192 = a(2) 192 = 128a
12 = 3/2 × (2) = 3 × 2 Hence, the 12
Let the first term of the GP = a and the common ratio = r 5 p = ar 8 q = ar 11 s = ar Now, a
AND a
Therefore, q/p = s/q
Hence, proved.
Let the first term of the GP = a = -3 and the common ratio = r It is given that 4 a ar ar -3r -3r = 9 r = -3 7 = (-3) (-3) = (-3) Hence, the 7
Common ratio = r = 2√2/2 = √2 a ar 2 (√2) (√2) (√2) (√2) Compare the exponents n + 1 = 14 n = 13 Hence, 128 is the 13
Common ratio = r = 3/√3 = √3 a ar √3 (√3) (√3) (√3) Compare the exponents n = 12 Hence, 729 is the 12
Common ratio = r = 1/9 × 3/1 = 1/3 a ar 1/3 × (1/3) 1/(3) (3) Compare the exponents n = 9 Hence, 1/19683 is the 9
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 x x = ± 1 Hence, the given numbers will be in GP when x = ± 1.
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 = S Sum of 20 terms of the given GP = S = 0.15(1 - (0.1) = 1/6 × {1 - (0.1)
First term = a = √7 Common ratio = r = √21/√7 = √3 Sum of n terms of the GP = S = √7(1 - (√3) Rationalise √7(1 - (√3) = √7(1 - (√3) = √7(1 - (√3) = √7((√3)
First term = 1 Common ratio = r = -a/1 = -a Sum of n terms of the GP = S = 1(1 - (-a) = (1 - (-a)
First term = a = x Common ratio = r = x Sum of n terms of the GP = S = x
= 2 (11) = 22
The expansion of First term = a = 3 Common ratio = r = 3 We know that sum of n terms of a GP = S Sum of 11 terms of the given GP = S = 3(1 - 3 = 3/2 × (3 Hence,
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 ⇒ a ⇒ Therefore, a/r + a + ar = 39/10 1/r + 1 + r = 39/10 (1 + r + r 10 (1 + r + r 10r 10r 10r 5r (2r - 5) - 2 (2r - 5) = 0 (2r - 5) (5r - 2) = 0 (2r - 5) = 0 ⇒ 2r = 5 ⇒ OR (5r - 2) = 0 ⇒ 5r = 2 ⇒ Hence, the three terms of the GP are 5/2, 1 and 2/5 OR 2/5, 1 and 5/2.
First term = a = 3 Common ratio = r = 3 S 120 = 3 (1 - 3 40 = (1 - 3 -80 = 1 - 3 -81 = -3 3 3 Compare the exponents
Hence, 4 terms of the given GP are needed in order to get the sum 120.
Let the GP be a, ar, ar First term = a Common ratio = r It is given that the sum of first three terms of the GP is 16. Therefore, a + ar + ar a(1 + r + r
It is also given that the sum of next three terms of the GP is 128. Therefore, ar ar Substitute a = 16/(1 + r + r 16/(1 + r + r 16r r r Compare the exponents
Now, a = 16/(1 + r + r a = 16/(1 + 2 + 2
Sum of n terms of the GP = S = 16(1 - 2 = -16(1 - 2
Hence, first term of the GP = 16/7, common ratio = 2 and sum of n terms = 16/7 × (2
First term = a = 729 7 ar 729r r r Compare the exponents
S = 729(1 - (2/3) = 729(1 - 128/2187)/(1/3) = 2187 (2187 - 128)/2187 = 2059 Hence, S
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, S -4 = a (1 -4 = a (1 - r)(1 + r)/(1 - r)
It is also given that the fifth term is 4 times the third term. Therefore, 4a 4ar 4r 4 = r
Now, -4 = a(1 + r) For r = 2, -4 = a(1 + 2) -4 = 3a
For r = -2, -4 = a(1 - 2) -4 = -a
Hence, the GP will be either 4, -8, 16, … OR -4/3, -8/3, -16/3, …
Let the first term of the GP = a and the common ratio = r. 4 10 16 Now, a y/x = r a z/y = r Therefore, y/x = z/y. Hence, x, y and z are in GP with the common ratio r
S This can be rewritten as S = 8/9 × [(10 - 1) + (100 - 1) + (1000 - 1) + (10000 - 1) + … n terms] = 8/9 × [(10 - 1) + (10 = 8/9 × [10 + 10 = 8/9 × [10(1 - 10 = 8/9 × [10(1 - 10 = 8/9 × [10(10 = 8/9 × [10(10 = Hence, the sum of give sequence is 80(10
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 = S S = 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.
Sequence formed by the products of corresponding terms of the given two sequences: aA, arAR, ar Second term/First term = arAR/aA = rR Third term/Second term = ar Hence, the obtained sequence forms a GP with common ratio rR.
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, a ar ar a(r
It is also given that the second term is greater than the 4 ar ar = ar -18 = ar -18 = ar(r
Now, -18/r(r -18/r(r -18/r = 9 -18 = 9r
Thus, a = 9/(r ⇒ The four terms will be 3, 3(-2), 3(-2) 3, -6, 12, -24 Hence, the first four terms of the given GP are 3, -6, 12, -24.
Let the first term of the GP = A and the common ratio = R. p q r Now, a = A = A = A = 1 × 1 = 1 Hence, proved that a
First term = a Let common ratio = r n Product of n terms = P = a × ar × ar = a 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 = a Square both sides P P P P Hence, proved.
Let the first term of the GP = a and the common ratio = r. Sum of the first n terms of the GP = a(1 - r From the (n + 1) Therefore, sum of the terms from (n + 1) Now, a Ratio of the sum of first n terms to the sum of terms from (n + 1) = [a(1 - r = a(1 - r = 1/r Hence, proved.
It is given that a, b, c and d form a GP. Therefore, b/a = c/b = d/c
On the RHS, we have (ab + bc + cd) = (ab + d(a + c)) = a = a = a = a = a = a = a = a = (b = (a Hence, proved. 26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
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 4 81 = ar 81 = 3r 27 = r
2 = 3(3) = 9 3 = 3(3) 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.
We know that the geometric mean between a and b is given by √ab. Therefore, (a Square both sides (a (a (a a a a a a a (a/b) Compare the exponents 2n + 1 = 0 2n = -1
Hence, n is -1/2.
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,
Square both sides (a + b) a Subtract 4ab from both sides a a (a - b) a - b = √(32ab)
Add Equation (I) and (II) a + b + (a - b) = 6√ab + 4√2ab 2a = (√ab) (6 + 4√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
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).
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
AND √ab = G Square both sides
We know that (x - y) (a - b) (a - b) (a - b) (a - b) (a - b) Take square root on both sides (a - b) = √4(A + G) (A - G) a - b = 2√(A + G) (A - G)
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
Thus, a = 2√(A + G) (A - G) + b a = 2√(A + G) (A - G) + A - √(A + G)(A - G)
Hence, proved that the two numbers are A ± √(A + G)(A - G).
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 n 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 2 = 30 × 2 Number of bacteria at the end of the 4 = 30 × 2 Number of bacteria at the end of the n = 30 × 2
Initial deposited amount = Rs 500 Interest rate after each year = 10% Amount at the end of the 1 Amount at the end of the 2 Amount at the end of the 3 and so on until the 10 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 10 Hence, the Rs 500 will amount to Rs 500 × (1.1)
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
AND √ab = 5 Square both sides
We can form the quadratic equation with the following format: x = x = x Hence, the required quadratic equation is x ## Exercise 9.4
The series can be rewritten as 1 × (1 + 1) + 2 × (2 + 1) + 3 × (3 + 1) + 4 × (4 + 1) + … Thus, the n Sum of the given series = = = = 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
The series can be rewritten as 1 × (1 + 1) × (1 + 2) + 2 × (2 + 1) × (2 + 2) + 3 × (3 + 1) × (3 + 2) + … Thus, the n = (n = n = n Sum of the given series = = = [n(n + 1)/2] = [n(n + 1)/2] = n(n + 1)/2 × [n(n + 1)/2 + (2n + 1) + 2] = n(n + 1)/2 × [n = n(n + 1)/4 × [n = n(n + 1)/4 × [n = 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
The series can be rewritten as (2(1) + 1) × 1 Thus, the n Sum of the given series = = 2 = 2 × [n(n + 1)/2] = n = n(n + 1)/2 × [n(n + 1) + (2n + 1)/3] = n(n + 1)/2 × [(3n = n(n + 1)/2 × (3n = n(n + 1)(3n
The series can be rewritten as 1/(1 × (1 + 1)) + 1/(2 × (2 + 1)) + 1/(3 × (3 + 1)) + … Thus, the n Sum of the given series = = = [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)
The series can be rewritten as (1 + 4) Thus, the n Sum of the given series = = = n(n + 1)(2n + 1)/6 + 8 × n(n + 1)/2 + 16n Now, the last term of the series = 20 So, n = 16 Sum of the given series = S = 16(17)(33)/6 + 4 × 16(17) + 256 = 8(17)(11) + 1088 + 256 = 1496 + 1344 = 2840
The series can be rewritten as 1 × 3 × (1 × 3 + 5) + 2 × 3 × (2 × 3 + 5) + 3 × 3 × (3 × 3 + 5) + … Thus, the n Sum of the given series = = 9 = 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)
First term of the series = 1 Second term of the series = 1 Third term of the series = 1 Thus, the n = n(n + 1)(2n + 1)/6 = n(2n = n(2n = (2n = n Sum of the given series = = 1/3 × = 1/3 × [n(n + 1)/2] = n = 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)
n = n(n = n Sum of the series = S = = [n(n + 1)/2] = n = 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 × [3n = n(n + 1)/2 × [3n = n(n + 1)(3n
n Sum of the series = S = Now,
The terms of 2 First term of the GP = a = 2 Common ratio = r = 2 Sum of the n terms of the GP = a(1 - r = 2(1 - 2 = 2(2 Hence, sum of the given series = = n(n + 1)(2n + 1)/6 + 2(2
n Sum of the series = S = 4 = 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(2n = n × [2(2n = n × [4n = n/3 × (4n = n/3 × [(2n) = n/3 × (2n + 1)(2n - 1) = n(2n + 1)(2n - 1)/3 ## Miscellaneous Exercise
Let there be an AP with first term = a and common difference = d. (m + n) (m - n) m Sum of (m + 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 × a Hence, proved that the sum of (m + n)
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
AND (a - d) × a × (a + d) = 440 a (a 8 (64 - d 64 - d d
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.
Let there be an AP with first term = a and common difference = d. Sum of the first n terms of the AP = S Sum of the first 2n terms of the AP = S Sum of the first 3n terms of the AP = S Now, S S S S S S Multiply both sides by 3 3(S
Hence, proved.
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 = a 399 = 203 + (n - 1)7 196 = (n - 1)7 28 = n - 1
Therefore, there are 29 numbers between 200 and 400 that are divisible by 7. Sum of the AP = S S = 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.
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 = a 100 = 2 + (n - 1)2 98 = (n - 1)2 49 = n - 1
Therefore, there are 50 numbers from 1 to 100 that are divisible by 2. Sum of the AP = S S = 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 = a 100 = 5 + (n - 1)5 95 = (n - 1)5 19 = n - 1
Therefore, there are 20 numbers from 1 to 100 that are divisible by 5. Sum of the AP = S S = 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 = a 100 = 10 + (n - 1)10 90 = (n - 1)10 9 = n - 1
Therefore, there are 10 numbers from 1 to 100 that are divisible by 2 and 5. Sum of the AP = S S = 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.
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 = a 97 = 13 + (n - 1)4 84 = (n - 1)4 21 = n - 1
Therefore, there are 22 two digit numbers that yield 1as the remainder when divided by 4. Sum of the AP = S S = 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.
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 = S It is given that 120 = 3(1 - 3 40 = (1 - 3 80 = 3 3 3 Compare the exponents
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 - r 315 = 5(1 - 2 63 = (1 - 2 63 = 2 2 2 Compare the exponents
Last term of the GP = a Hence, the number of terms is 6 and the last term of the GP is 160.
First term = a = 1 Let the common ratio of the GP be r. Third term of the GP = a Fifth term of the GP = a It is given that the sum of the fifth and the third terms of the GP is 90. Therefore, a r r Using the quadratic formula r = [-1 ± √361]2 = (-1 ± 19)/2 When r r r
When r r r The roots are not real. Therefore, r Hence, common ratio of the GP is ± 3.
Let the three numbers in GP be a, ar, and ar It is given that the sum of these numbers is 56. Therefore, a + ar + ar a(1 + r + r
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, ar Common difference of the AP = ar - 7 - (a - 1) = ar ar - 7 - a + 1 = ar ar - a - 6 = ar ar ar a(r 56/(1 + r + r 7 (r 7r 6r 3(2r 2r 2r 2r(r - 2) - 1(r - 2) = 0 (r - 2) (2r - 1) = 0 (r - 2) = 0 ⇒ (2r - 1) = 0 ⇒ 2r = 1 ⇒ Therefore, When r = 2: a = 56/(1 + 2 + 2 = 56/7
The three numbers are 8, 8(2), 8(2) When r = 1/2: a = 56/(1 + 1/2 + 1/4) = 56/[(4 + 2 + 1)/4] = 56 (4)/7
The three numbers are 32, 32(1/2), 32(1/2) Hence, the required three numbers are 8, 16 and 32 OR 32, 16, and 8.
Let the terms of the GP be T 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, T [T [T [T [T ar(r
Hence, the common ratio of the GP is 4.
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 = S 56 = 2 × [2a + 3d] 28 = 2a + 3d 3d = 28 - 2a d = (28 - 2(11))/3 d = (28 - 22)/3
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
Hence, the number of terms in the AP is 11.
(a + bx)/(a - bx) = (b + cx)/(b - cx) (a + bx) (b - cx) = (b + cx) (a - bx) ab + b 2b 2b b b
(b + cx)/(b - cx) = (c + dx)/(c - dx) (b + cx) (c - dx) = (c + dx) (b - cx) bc + c 2c 2c c
Therefore, b/a = c/b = d/c Hence, a, b, c, and d are in GP.
Let the terms in the GP be a, ar, ar First term = a Common ratio = r Sum of the n terms of the GP = S S = a(1 - r Product of the n terms of the GP = a × ar × ar P = a = a Sum of the reciprocals of n terms of the GP = 1/a + 1/ar + 1/ar R = 1/a × [1 + 1/r + 1/r = 1/a × (r The terms in r r Thus, R = 1/ar = (1 - r Now, P = a = a = [a(1 - r = S Hence, proved.
Let the first term of the AP be A and its common difference be D. p a = A + (p - 1)D q b = A + (q - 1)D r c = A + (r - 1)D A b - c = A + (q - 1)D - A - (r - 1)D b - c = D(q - 1 - r + 1) b - c = D(q - r)
A a - b = A + (p - 1)D - A - (q - 1)D a - b = D(p - 1 - q + 1) a - b = D(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.
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 ab ab ab(b - a) + 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.
Given, a, b, c, and d are in GP b c ad = bc … (iii) Required to prove (an + bn), (bn + cn), (cn + dn) are in GP i.e., (bn + cn) Taking LHS (bn + cn) = (b = (ac)n = 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.
Given, a and b are the roots of x So, we have a + b = 3 and ab = p … (i) Also, c and d are the roots of x 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 = xr From (i) and (ii), we get x + xr = 3 x (1 + r) = 3 And, xr xr On dividing, we get xr r 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 = x cd = x (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 = x cd = x (q + p)/(q - p) = (-288 - 18)/(-288 + 18) = -306/-270 = 17/15 (q + p) : (q - p) = 17 : 15 Hence, proved.
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) (a + b)
We know that (a - b) (a - b) (a - b) (a - b) Take square root on both sides
Add (i) and (ii) a + b + a - b = 2m√ab/n + 2√ab × √(m 2a = [2m√ab + 2√ab√(m 2an = 2√ab (m + √(m
Substitute the value of a in (i) a + b = 2m√ab/n b = 2m√ab/n - a b = 2m√ab/n - √ab [m + √(m b = √ab × [2m - m - √(m
Now, a/b = [√ab [m + √(m a/b = [m + √(m
Hence, proved.
It is given that a, b, c are in AP. Therefore, Common difference = b - a = c - b 2b = a + c
It is also given that b, c, d are in GP. Therefore, Common ratio = c/b = d/c
Also, 1/c, 1/d, 1/e are in AP. Therefore, Common difference = 1/d - 1/c = 1/e - 1/d
2/(c 2b/c 2(a + c)/2c (a + c)/c (a + c)/c = (e + c)/e (a + c)e = (e + c)c ae + ce = ec + c
Hence, a, c, and e are in GP.
S = 5/9 × [9 + 99 + 999 + … upto n terms] = 5/9 × [(10 - 1) + (100 - 1) + (1000 - 1) + … upto n terms] = 5/9 × [(10 = 5/9 × [(10 The terms in 10 Sum of n terms of the GP = 10 = 10(1 - 10 Therefore, S = 5/9 × [10(1 - 10 = 5/9 × [10(1 - 10 = 5/9 × [10(10 = 50(10
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 - 1/10 = 6/9 × [(1 + 1 + 1 + … upto n terms) - (1/10 The terms in 1/10 Sum of n terms of the GP = 1/10 = 1/10 × (1 - 1/10 Therefore, S = 6/9 × [n × 1 - 1/10 × (1 - 1/10 = 6/9 × [n - 1/10 × (1 - 1/10 = 6/9 × [n - (1 - 1/10 = 2n/3 - 2(1 - 1/10
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 n 20 = 4(400) + 80 = 1600 + 80 = 1680 Hence, the 20
Sum of the given series = S = 3 + 7 + 13 + 21 + 31 + … + a S = 3 + 7 + 13 + 21 + … + a Subtract both equations S - S = [3 + 7 + 13 + 21 + 31 + … + a 0 = [3 + (7 + 13 + 21 + 31 + … + a 0 = 3 + [(7 - 3) + (13 - 7) + (21 - 13) + (31 - 21) + … + (a a The terms in 4 + 6 + 8 + 10 + … + upto (n - 1) terms form an AP. Therefore, a a a a a a Now, Sum of the series = S = = = n(n + 1)(2n + 1)/6 + n(n + 1)/2 + n = n × [(n + 1)(2n + 1)/6 + (n + 1)/2 + 1] = n × [(2n = n/6 × [2n = n/6 × [2n = n/6 × 2[n
According to the question, it is given that: S S S We need to prove that 9 S Taking LHS = 9 S = 9 × [n(n + 1)(2n + 1)/6] = 9 × [n(n + 1)(2n + 1)] = [n(n + 1)(2n + 1)] = [n(n + 1)(2n + 1)/2] Taking RHS = S = [n(n + 1)/2] = [n(n + 1)/2] = [n(n + 1)/2] = [n(n + 1)/2] = [n(n + 1)/2] = [n(n + 1)(2n + 1)/2] Since, LHS = RHS. Hence, proved that 9 S
The n = [n(n + 1)/2] 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) = n So, a = n = (n + 1) = (n Sum of the given series = S = = 1/4 = 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 × [2n = n/24 × [2n = n(2n
n = n n LHS = [1 × 2 = = = [ Numerator = = [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 × [3n = n(n + 1)/12 × [3n = n(n + 1)/12 × [3n = 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(n + 1)/2] = 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 × [3n = n(n + 1)/12 × [3n = n(n + 1)/12 × [3n = 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 × 2 = [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.
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 = a 6000 = 500 + (n - 1)500 5500 = (n - 1)500 11 = n - 1
Sum of the AP = S = 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.
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 = a 18000 = 1000 + (n - 1)1000 17000 = (n - 1)1000 17 = n - 1
Sum of the AP = S = 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.
Number of letters mailed by the person originally = 4 Number of letters mailed by the person's friends = 4 × 4 = 4 Number of letters mailed by the next recipients = 4 × 4 … and so on. The number of letters mailed in the chain at every step form a GP 4, 4 First term = a = 4 Common ratio = r = 4 Letters sent until the 8 = 4(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 8
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 15 = 10000 + 14(500) = 10000 + 7000 = Rs 17000 Total amount after 20 years = a = 10000 + 20(500) = 10000 + 10000 = 20000 Hence, the amount in 15
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 × 4 = 5 = 5 × 1024 = Rs 5120 Hence, the value of the machine after the end of 5
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 - x x x x x(x + 32) - 17(x + 32) = 0 (x + 32) (x - 17) = 0 (x + 32) = 0 ⇒ OR (x - 17) = 0 ⇒ 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. |