## NCERT Class 10 Maths Chapter 5: Arithmetic Progressions## Exercise 5.1
- The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
- The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
- The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
- The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
I. Let the taxi fare for n km be t It is given that, t t t t Since, t Hence, the given situation forms an Arithmetic Progression. II. Let t Let t t t Since, t Hence, the given situation does not form an Arithmetic Progression. III. Let the cost of digging for n metre be t It is given that, t t t t Since, t Hence, the given situation forms an Arithmetic Progression. IV. Let the amount of money for n years be t It is given that, t t t t Since, t Hence, the given situation does not form an Arithmetic Progression.
- a = 10, d = 10
- a = -2, d = 0
- a = 4, d = - 3
- a = - 1, d = 1/2
- a = - 1.25, d = - 0.25
I. a = 10 a a a The first four terms are 10, 20, 30 and 40. II. a = -2 a a a The first four terms are -2, -2, -2 and -2. III. a = 4 a a a The first four terms are 4, 1, -2 and -5. IV. a = -1 a a a The first four terms are -1, -1/2, 0 and 1/2. V. a = -1.25 a a a The first four terms are -1.25, -1.5, -1.75 and -2.
- 3, 1, - 1, - 3, . . .
- - 5, - 1, 3, 7, . . .
- 1/3, 5/3, 9/3, 13/3, . . .
- 0.6, 1.7, 2.8, 3.9, . . .
I. First term = a = 3 d = a II. First term = a = -5 d = a III. First term = a = 1/3 d = a IV. First term = a = 0.6 d = a
- 2, 4, 8, 16, . . .
- 2, 5/2, 3, 7/2, . . .
- 1.2, - 3.2, - 5.2, - 7.2, . . .
- 10, - 6, - 2, 2, . . .
- 3, 3 + √2, 3 + 2√2, 3 + 3√2, . . .
- 0.2, 0.22, 0.222, 0.2222, . . .
- 0, - 4, - 8, -12, . . .
- -1/2, -1/2, -1/2, -1/2, . . .
- 1, 3, 9, 27, . . .
- a, 2a, 3a, 4a, . . .
- a, a
^{2}, a^{3}, a^{4}, . . . - √2, √8, √18, √32, . . .
- √3, √6, √9, √12, . . .
- 1
^{2 }, 3^{2}, 5^{2}, 7^{2}, . . . - 1
^{2 }, 3^{2}, 5^{2}, 73, . . .
I. t t Since, t Hence, it is not an Arithmetic Progression. II. t t Since, t Hence, it is an Arithmetic Progression. First term = a = 2 Common difference = d = ½ Four more terms would be a a a a III. t t Since, t Hence, it is an Arithmetic Progression. First term = a = -1.2 Common difference = d = -2 Four more terms would be a a a a IV. t t Since, t Hence, it is an Arithmetic Progression. First term = a = -10 Common difference = d = 4 Four more terms would be a a a a V. t t Since, t Hence, it is an Arithmetic Progression. First term = a = 3 Common difference = d = √2 Four more terms would be a a a a VI. t t Since, t Hence, it is not an Arithmetic Progression. VII. t t Since, t Hence, it is an Arithmetic Progression. First term = a = 0 Common difference = d = -4 Four more terms would be a a a a VIII. t t Since, t Hence, it is an Arithmetic Progression. First term = a = -1/2 Common difference = d = 0 Four more terms would be a a a a IX. t t Since, t Hence, it is not an Arithmetic Progression. X. t t Since, t Hence, it is an Arithmetic Progression. First term = a Common difference = d = a Four more terms would be a a a a XI. t t Since, t Hence, it is not an Arithmetic Progression. XII. t t Since, t Hence, it is an Arithmetic Progression. First term = a = √2 Common difference = d = √2 Four more terms would be a a a a XIII. t t Since, t Hence, it is not an Arithmetic Progression. XIV. t t Since, t Hence, it is not an Arithmetic Progression. XV. t t Since, t Hence, it is an Arithmetic Progression. First term = a Common difference = d = 24 Four more terms would be a a a a ## Exercise 5.2
I. a = 7 + 7(3) = 7 + 21 = 28 II. a 0 = -18 + 9d 18 = 9d d = 2 III. a -5 = a + (17)(-3) -5 = a - 51 46 = a IV. a 3.6 = -18.9 +(n - 1)2.5 22.5 = (n - 1)2.5 9 = n - 1 n = 10 V. a Hence, the completed table is:
I. For the given AP, a = 10 and d = 7 - 10 = -3 a = 10 + 29(-3) = 10 - 87 = -77 Hence, (C) is the correct answer. II. For the given AP, a = -3 and d = -1/2 - (-3) = -1/2 + 3 = 5/2 a = -3 + 10(5/2) = -3 + 25 = 22 Hence, (B) is the correct answer.
I. a = 2 a a 24 = 2d d = 12 a II. a a 3 - a 16 = 2a a d = a = 8 - 13 = -5 a = a III. a = 5 a a 19/2 = 5 + 3d 9/2 = 3d 3/2 = d a a = 5 + 3 = 8 IV. a = -4 a a a 6 = -4 + 5d 10 = 5d d = 2 a a a a V. a a 38 = a + d … (I) -22 = a + 5d … (II) Subtract (I) from (II) -22 - 38 = a + 5d - (a + d) -60 = 4d -15 = d Using d = -15 in (I) 38 = a - 15 53 = a a = 23 a a
First term = a = 3 Common difference = d = 8 - 3 = 5 a a 78 = 3 + (n - 1)5 75 = (n - 1)5 15 = n - 1 n = 16 Hence, 78 is the 16
- 7, 13, 19, . . . , 205
- 18, , 13, . . . , - 47
I. First term = a = 7 Last term = l = 205 Common difference = d = 13 - 7 = 6 l = a + (n - 1)d 205 = 7 + (n - 1)6 198 = (n - 1)6 33 = n - 1 n = 34 Hence, the given AP has 34 terms. II. First term = a = 18 Last term = l = -47 Common difference = d = 31/2 - 18 = -5/2 l = a + (n - 1)d -47 = 18 + (n - 1)(-5/2) -65 = (1 - n)(5/2) -130 = (1 - n)5 -26 = 1 - n n = 27 Hence, the given AP has 27 terms.
Let us assume that -150 is a term of the given AP. First term = a = 11 Common Difference = 8 - 11 = -3 a a -150 = 11 + (n - 1)(-3) -161 = (n - 1)(-3) 161/3 + 1 = n 164/3 = n But this contradicts the fact that n is a natural number. The contradiction has arisen due to the wrong assumption that -150 is a term of the given AP.
a a 38 = a + 10d … (I) 73 = a + 15d … (II) Subtract (I) from (II) 73 - 38 = a + 15d - (a + 10d) 35 = 5d 7 = d Using d = 7 in (I) 38 = a + 70 a = -32 a a
a a a 106 = a + 49d … (I) 12 = a + 2d … (II) Subtract (II) from (I) 106 - 12 = a + 49d - (a + 2d) 94 = 47d 2 = d Using d = 2 in (II) 12 = a + 4 a = 8 a a
a a a -8 = a + 8d … (I) 4 = a + 2d … (II) Subtract (II) from (I) -8 - 4 = a + 8d - (a + 2d) -12 = 6d -2 = d Using d = -2 in (II) 4 = a - 4 a = 8 Let the x a 0 = 8 + (x - 1)(-2) -8 = (x - 1)(-2) 4 = x - 1 x = 5 Hence, the 5
a a a + 16d - (a + 9d) = 7 7d = 7 d = 1 Hence, the common difference for the given AP is 1.
Let the required term be a First term = a = 3 Common difference = d = 15 - 3 = 12 a a + (n - 1)d = a + 53d + 132 (n - 1)12 = 53(12) + 132 (n - 1) = 53 + 11 n - 1 = 64 n = 65 Hence, the 65
A A + 99d - (a + 99d) = 100 A - a = 100 A = A - a = 100 Hence, the difference in their 1000
The list of 3 digit numbers that are divisible by 7 is 105, 112, …, 994 a = 105 d = 7 Last term = a + (n - 1)d 994 = 105 + (n - 1)7 889 = (n - 1)7 127 = n - 1 128 = n Hence, there are 128 three digit numbers that are divisible by 7.
Multiples of 4 between 10 and 250 are 12, 16, 20, …, 248 a = 12 d = 4 Last term = a + (n - 1)d 248 = 12 + (n - 1)4 236 = (n - 1)4 59 = n - 1 n = 60 Hence, there are 60 multiples of 4 between 10 and 250.
We need to find n such that A A = 63 d a = 3 d A A + (n - 1)d 63 + (n - 1)2 = 3 + (n - 1)7 60 + 2n - 2 = 7n - 7 65 = 5n n = 13 Hence, the 13
a a a + 6d - (a + 4d) = 12 2d = 12 d = 6 a a + 2d = 16 a + 2(6) = 16 a = 16 - 12 a = 4 a Hence, the required AP is 4, 10, 16, …
Reverse the given AP: 253, 248, …, 13, 8, 3 a a = 253 d = 248 - 253 = -5 a = 253 - 95 = 158 Hence, the 20
a a a + 3d + a + 7d = 24 2a + 10d = 24 2 (a + 5d) = 24 a + 5d = 12 = a a + 5d + a a a Subtract a a a + 9d - (a + 5d) = 20 4d = 20 d = 5 Use d = 5 in a + 5d = 12 a + 5(5) = 12 a + 25 = 12 a = -13 Therefore, first term = -13 and common difference = 5. Hence, the first 3 terms will be: a a a
Subba Rao's salary each year can be represented as an AP: 5000, 5200, 5400, . . . Salary in 1995 = First term = a = 5000 Increment each year = Common difference = d = 200 a a 7000 = 5000 + (n - 1)200 2000 = (n - 1)200 10 = n - 1 n = 11 The 11 Hence, Subba Rao's salary reached Rs 7000 in the year 2005.
Ramkali's savings each week can be represented as an AP: 5, 6.75, 8.5, . . . Starting saving = First term = a = 5 Savings increment per week = Common difference = d = 1.75 a a 20.75 = 5 + (n - 1)(1.75) 15.75 = (n - 1)(1.75) 9 = n - 1 10 = n The 10 Hence, Ramkali's savings became Rs 20.75 in the 10 ## Exercise 5.3
- 2, 7, 12, . . ., to 10 terms.
- -37, -33, -29, . . ., to 12 terms
- 0.6, 1.7, 2.8, . . ., to 100 terms.
- 1/15, 1/12 , 1/10, . . ., to 11 terms.
Hence, the sum of the given AP is 245. Hence, the sum of the given AP is -180. Hence, the sum of the given AP is 5505.
**7 + + 14 + . . . + 84****34 + 32 + 30 + . . . + 10****-5 + (-8) + (-11) + . . . + (-230)**
I. = a + (n - 1)d a = 7, d = 21/2 - 7 = 7/2, a 84 = 7 + (n - 1)7/2 77 = (n - 1)7/2 22 = n - 1 n = 23 Hence, the sum of the given AP is 1046.5 II. a a = 34, d = 32 - 34 = -2, a 10 = 34 + (n - 1)(-2) -24 = (n - 1)(-2) 12 = n - 1 n = 13 Hence, the sum of the given AP is 286. III. a a = -5, d = -8 - (-5) = -3, a -230 = -5 + (n - 1)(-3) -225 = (n - 1)(-3) 75 = n - 1 n = 76 Hence, the sum of the given AP is -8930.
**given a = 5, d = 3, a**_{n}= 50, find n and S_{n}.**given a = 7, a**_{13}= 35, find d and S_{13}.**given a**_{12}= 37, d = 3, find a and S_{12}.**given a**_{3}= 15, S_{10}= 125, find d and a_{10}.**given d = 5, S**_{9}= 75, find a and a_{9}.**given a = 2, d = 8, S**_{n}= 90, find n and a_{n}.**given a = 8, a**_{n}= 62, S_{n}= 210, find n and d.**given a**_{n}= 4, d = 2, S_{n}= -14, find n and a.**given a = 3, n = 8, S = 192, find d.****given l = 28, S = 144, and there are total 9 terms. Find a.**
a 50 = 5 + (n - 1)3 45 = (n - 1)3 15 = n - 1
Hence, n = 16 and S II. a a 35 = 7 + 12d 28 = 12d
Hence, d = 7/3 and S
a a 37 = a + 33
Hence, a = 4 and S
a a 15 = a + 2d
Using d = -1 in Equation (I) 15 = a + 2(-1) 15 = a - 2 17 = a a = 17 + 9(-1) = 17 - 9 = 8
Hence, d = -1 and a
a a a
Hence, a = -35/3 and a
But n has to be a positive integer, so n = -9/2 is rejected. Therefore, a = 2 + 4(8) = 2 + 32 = 34
Hence, n = 5 and a
a a 62 = 8 + 5d 54 = 5d
Hence, n = 6 and d = 54/5.
a 4 = a + (n - 1)2 4 = a + 2n - 2 6 = a + 2n a = 6 - 2n But n has to be a positive integer, so n = -2 is rejected. Therefore, a = 6 - 2n = 6 - 2(7) = 6 - 14
Hence, n = 7 and a = -8.
Hence, d = 6.
Hence, a = 4.
First term = a = 9 Common difference = 17 - 9 = 8 But, n has to be a positive integer, so n = -53/4 is rejected. Therefore, n = 12. Hence, 12 terms of the given AP must be taken to get a sum of 636.
First term = a = 5 Last term = l = 45 l = a a a 45 = 5 + 15d 40 = 15d
Hence, the number of terms in the AP is 16 and the common difference is 8/3.
First term = a = 17 Last term = l = a Common difference = d = 9 a 350 = 17 + (n - 1)9 333 = (n - 1)9 37 = n - 1
Hence, the AP consists of 38 terms, and the sum of the AP is 6973.
a Common difference = d = 7 a a 149 = a + 21(7) 149 = a + 147 a = 2 Hence, the sum of the first 22 terms of the given AP is 1661.
Common difference = d = a d = 18 - 14 = 4 a a 14 = a + 4 a = 10
a 17 - 7 = a + 8d - a - 3d 10 = 5d d = 2 Using d = 2 in a + 3d = 7 a + 3(2) = 7 a + 6 = 7 a = 1 Hence, the sum of first n terms of the given AP is n
- a
_{n}= 3 + 4n - a
_{n}= 9 - 5n
I. a a a a a a Since, a a Hence, the sum of the first 15 terms of the given AP is 525. II. a a a a a a Since, a a Hence, the sum of the first 15 terms of the given AP is -465.
It is given that S First term = a = S S
Sum of the first two terms = S S
S 4 = 3 + a
Therefore, the second term is 1. S S 3 = 4 + a
Therefore, the third term is -1. S S S -60 = -45 + a
Therefore, the 10 Similarly, S a a a a
Therefore, the n
The list of first 40 positive integers divisible by 6 can be represented as an AP: 6 × 1, 6 × 2, 6 × 3, … … , 6 × 40 6, 12, 18, … … , 240 First term = a = 6 Common difference = d = 6 a Hence, the sum of first 40 positive integers divisible by 6 is 4920.
The list of first 15 multiples of 8 forms an AP: 8 × 1, 8 × 2, 8 × 3, … … , 8 × 15 8, 16, 24, … … , 120 First term = a = 8 Common difference = d = 8 a Hence, the sum of first 15 multiples of 8 is 4920.
The list of odd numbers between 0 and 50 form an AP: 1, 3, 5, … …, 49 First term = a = 1 Last term = l = 49 Common difference = d = 2 a 49 = 1 + (n - 1)2 48 = (n - 1)2 24 = n - 1 n = 25 Hence, the sum of the odd numbers between 0 and 50 is 625.
The penalty for the delay forms an AP: 200, 250, 300, … … First term = a = 200 Common Difference = d = 50 a n = 30 a a a a Penalty for 30 days = S Hence, the contractor has to pay Rs 27750 penalty for a delay of 30 days.
The values of each of the prizes can be represented as an AP: a, a - 20, a- 2(20), … …, a - 6(20) a, a - 20, a - 40, … …, a - 120 Sum of the given AP = S a a a a a a a Hence, the value of each of the seven prizes is Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, Rs 40.
The number of trees planted by 3 sections of each class can be represented as an AP: 3 × 1, 3 × 2, 3 × 3, … …, 3 × 12 3, 6, 9, … …, 36 First term = a = 3 Common difference = d = 6 - 3 = 3 Last term = a Total number of trees planted by the students = S Hence, the total number of trees planted by the students is 234.
Let the radii be denoted as R The radii of the semi-circles formed by the spiral are: 0.5, 1, 1.5, 2.0, … … The radii form an AP where, First term = a = 0.5 Common difference = 1 - 0.5 = 0.5 n = 13 Sum of all the radii = S Length of the perimeter of a circle = 2πR Length of the perimeter of a semi-circle = 2πR/2 = πR Length of the given spiral = L = πR L = π(R L = π(S L = (45.5)22/7 L = 143 Hence, the length of the given spiral is 143 cm.
Number of logs present in each row can be represented as an AP: 20, 19, 18, … … Logs in the first row = a = 20 Common difference = d = -1 Total number of logs = S = 200 For n = 25, a = 20 + 24(-1) = 20 - 24 = -4 But, it is not possible to have negative number of logs in the last row, so n = 25 is rejected. Therefore, n = 16. a = 20 + 15(-1) = 20 - 15 = 5 Hence, the number of logs present in the top row is 5 and the number of rows is 16.
Distance of first potato from bucket = 5m Distance of second potato from bucket = 5 + 3 = 8m Distance of third potato from bucket = 8 + 3 = 11m The distances of the potatoes from the bucket form an AP: 5, 8, 11, … … upto 10 terms First term = a = 5 Common difference = d = 3 The competitor has to run back to the bucket after picking up a potato. Therefore, the total distance covered = 2 × [Sum of distances of each potato from the bucket] = 2 × [S Total distance covered by the competitor = 2 × 185 = 370m. ## Exercise 5.4 (Optional)
First term = a = 121 Common difference = d = 117 - 121 = -4 Since, the required term is negative. Therefore, a Hence, 32
It is given that a and a a + 2d + a + 6d = 6 2a + 8d = 6 2(a + 4d) = 6 a + 4d = 3 = a a = 3 - 4d (a + 2d) (a + 6d) = 8 Use a = 3 - 4d (3 - 4d + 2d) (3 - 4d + 6d) = 8 (3 - 2d) (3 + 2d) = 8 3 9 - 4d 1 = 4d ¼ = d
Therefore, a = 3 - 4( a = 3 - 2 = 1 OR a = 3 + 2 = 5
Hence, the sum of the first 16 terms of the given AP is either 76 or 20.
Each rung is 25 cm apart and the distance between first and last rung is 5/2 m = 250 cm apart. Therefore, number of rungs = 250/25 + 1 = 10 + 1 = 11 Since, the length of each of the rungs decreases uniformly, it will form an AP where First term = a = 45 Last term = a Length of wood required to make the rungs = S Hence, the total length of wood required to make the rungs is 385 cm.
The house numbers form an AP: 1, 2, 3, … …, 49 First term = a = 1 Common difference = d = 1 Last term = a The sum of the numbers of houses that precede x = S The sum of the numbers of house that succeed x = S But, house number x has to be from 1 to 49, so x = -35 is rejected. Hence, x = 35.
The height of each step from the bottom increases by ¼ m, so it forms an AP: ¼, ¼ + ¼, ¼ + ¼ + ¼, … …, upto 15 terms ¼, ½, ¾, … … …, upto 15 terms First term = a = ¼ Common difference = d = ¼ Volume of concrete required for each step = Height × Tread × Width Tread = ½ m (fixed) Width = 50 m (fixed) Volume of concrete required to build the first step = ¼ × ½ × 50 m Volume of concrete required to build the second step = (¼ + ¼) × ½ × 50 m Therefore, total volume of concrete required for the terrace is = (a × ½ × 50) + (a = ½ × 50 × [a + a = ½ × 50 × [S Total volume of concrete required = ½ × 50 × [30] = 750 m Hence, 750 m Next TopicClass 10 Chapter 6 |