NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (All Exercises)

Class 10 Maths Chapter 5: Arithmetic Progressions - Complete Solutions

Welcome! Below are detailed, step-by-step solutions for all exercises in Chapter 5 (Arithmetic Progressions) of the NCERT Class 10 Mathematics textbook for MP Board and CBSE students.

EXERCISE 5.1

Question 1

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

Solution:
Taxi fare for 1st km = Rs 15
Fare for 2 km = 15 + 8 = Rs 23
Fare for 3 km = 23 + 8 = Rs 31
Fare for 4 km = 31 + 8 = Rs 39
The list is 15, 23, 31, 39, ...
Since the difference between consecutive terms is constant (d = 8), it forms an AP.

(ii) 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.

Solution:
Let initial air volume = V
Air removed 1st time = V / 4, remaining air = V - V/4 = 3V/4
Air removed 2nd time = (1/4) × (3V/4) = 3V/16, remaining air = 3V/4 - 3V/16 = 9V/16
The list is V, 3V/4, 9V/16, ...
Common difference = 3V/4 - V = -V/4; 9V/16 - 3V/4 = -3V/16 (not equal).
Since common difference is not constant, it does NOT form an AP.

(iii) 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.

Solution:
Cost for 1st metre = Rs 150
Cost for 2 metres = 150 + 50 = Rs 200
Cost for 3 metres = 200 + 50 = Rs 250
The list is 150, 200, 250, ...
Since common difference d = 50 is constant, it forms an AP.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Solution:
Amount after year 1 = 10000(1 + 8/100)¹ = 10800
Amount after year 2 = 10000(1 + 8/100)² = 11664
Amount after year 3 = 10000(1 + 8/100)³ = 12597.12
The differences between terms are 800, 864, 933.12, which are not constant.
Hence, it does NOT form an AP.

Question 2

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10: 10, 20, 30, 40

(ii) a = -2, d = 0: -2, -2, -2, -2

(iii) a = 4, d = -3: 4, 1, -2, -5

(iv) a = -1, d = 1/2: -1, -1/2, 0, 1/2

(v) a = -1.25, d = -0.25: -1.25, -1.50, -1.75, -2.00

Question 3

For the following APs, write the first term and the common difference:

(i) 3, 1, -1, -3, ...: a = 3, d = 1 - 3 = -2

(ii) -5, -1, 3, 7, ...: a = -5, d = -1 - (-5) = 4

(iii) 1/3, 5/3, 9/3, 13/3, ...: a = 1/3, d = 5/3 - 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9, ...: a = 0.6, d = 1.7 - 0.6 = 1.1

Question 4

Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16, ...: Not an AP (4-2=2, 8-4=4).

(ii) 2, 5/2, 3, 7/2, ...: Is an AP. d = 1/2. Next 3 terms: 4, 9/2, 5.

(iii) -1.2, -3.2, -5.2, -7.2, ...: Is an AP. d = -2. Next 3 terms: -9.2, -11.2, -13.2.

(iv) -10, -6, -2, 2, ...: Is an AP. d = 4. Next 3 terms: 6, 10, 14.

(v) 3, 3+√2, 3+2√2, 3+3√2, ...: Is an AP. d = √2. Next 3 terms: 3+4√2, 3+5√2, 3+6√2.

(vi) 0.2, 0.22, 0.222, 0.2222, ...: Not an AP.

(vii) 0, -4, -8, -12, ...: Is an AP. d = -4. Next 3 terms: -16, -20, -24.

(viii) -1/2, -1/2, -1/2, -1/2, ...: Is an AP. d = 0. Next 3 terms: -1/2, -1/2, -1/2.

(ix) 1, 3, 9, 27, ...: Not an AP.

(x) a, 2a, 3a, 4a, ...: Is an AP. d = a. Next 3 terms: 5a, 6a, 7a.

(xi) a, a², a³, a⁴, ...: Not an AP.

(xii) √2, √8, √18, √32, ...: Simplified: √2, 2√2, 3√2, 4√2... Is an AP. d = √2. Next 3 terms: √50, √72, √98.

(xiii) √3, √6, √9, √12, ...: Not an AP.

(xiv) 1², 3², 5², 7², ... (1, 9, 25, 49): Not an AP.

(xv) 1², 5², 7², 73, ... (1, 25, 49, 73, ...): Is an AP. d = 24. Next 3 terms: 97, 121, 145.

EXERCISE 5.2

Question 1: Fill in the missing values

• (i) a = 7, d = 3, n = 8 ⇒ a₈ = 7 + (8-1)3 = 28
• (ii) a = -18, n = 10, a_n = 0 ⇒ 0 = -18 + 9d ⇒ d = 2
• (iii) d = -3, n = 18, a_n = -5 ⇒ -5 = a + 17(-3) ⇒ a = 46
• (iv) a = -18.9, d = 2.5, a_n = 3.6 ⇒ 3.6 = -18.9 + (n-1)2.5 ⇒ 22.5 = 2.5(n-1) ⇒ n = 10
• (v) a = 3.5, d = 0, n = 105 ⇒ a₁₀₅ = 3.5

Question 2: Choose the correct choice

(i) 30th term of AP: 10, 7, 4, ...
a = 10, d = -3 ⇒ a₃₀ = 10 + 29(-3) = -77. Correct choice: (C) -77

(ii) 11th term of AP: -3, -1/2, 2, ...
a = -3, d = -1/2 - (-3) = 5/2 ⇒ a₁₁ = -3 + 10(5/2) = 22. Correct choice: (B) 22

Question 3: Find missing terms in boxes

(i) 2, [ ], 26: b = (2 + 26)/2 = 14

(ii) [ ], 13, [ ], 3: d = (3 - 13)/2 = -5 ⇒ terms are 18, 13, 8, 3

(iii) 5, [ ], [ ], 9 1/2: a = 5, a₄ = 19/2 ⇒ 3d = 9/2 ⇒ d = 3/2. Missing terms: 6 1/2, 8

(iv) -4, [ ], [ ], [ ], [ ], 6: a = -4, a₆ = 6 ⇒ 5d = 10 ⇒ d = 2. Missing terms: -2, 0, 2, 4

(v) [ ], 38, [ ], [ ], [ ], -22: a₂ = 38, a₆ = -22 ⇒ 4d = -60 ⇒ d = -15. Terms: 53, 38, 23, 8, -7, -22

Question 4

Which term of the AP: 3, 8, 13, 18, ... is 78?
Solution: a = 3, d = 5, a_n = 78.
78 = 3 + (n - 1)5 ⇒ 75 = 5(n - 1) ⇒ n - 1 = 15 ⇒ n = 16. (16th term)

Question 5

Find the number of terms in each AP:
(i) 7, 13, 19, ..., 205: 205 = 7 + (n - 1)6 ⇒ 198 = 6(n - 1) ⇒ n = 34
(ii) 18, 15 1/2, 13, ..., -47: -47 = 18 + (n - 1)(-2.5) ⇒ -65 = -2.5(n - 1) ⇒ n = 27

Question 6

Check whether -150 is a term of the AP: 11, 8, 5, 2, ...
Solution: -150 = 11 + (n - 1)(-3) ⇒ -161 = -3(n - 1) ⇒ n = 164/3, which is not an integer. Hence, -150 is NOT a term.

Question 7

Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution: a + 10d = 38, a + 15d = 73 ⇒ 5d = 35 ⇒ d = 7, a = -32.
a₃₁ = -32 + 30(7) = 178.

Question 8

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution: a + 2d = 12, a + 49d = 106 ⇒ 47d = 94 ⇒ d = 2, a = 8.
a₂₉ = 8 + 28(2) = 64.

Question 9

If the 3rd and 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution: a + 2d = 4, a + 8d = -8 ⇒ 6d = -12 ⇒ d = -2, a = 8.
0 = 8 + (n - 1)(-2) ⇒ 2(n - 1) = 8 ⇒ n = 5. (5th term)

Question 10

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution: a₁₇ - a₁₀ = 7 ⇒ (a + 16d) - (a + 9d) = 7 ⇒ 7d = 7 ⇒ d = 1.

Question 11

Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
Solution: a = 3, d = 12. a₅₄ = 3 + 53(12) = 639.
Target term = 639 + 132 = 771.
771 = 3 + (n - 1)12 ⇒ 768 = 12(n - 1) ⇒ n - 1 = 64 ⇒ n = 65. (65th term)

Question 12

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution: Let first terms be a₁ and a₂. Difference = a₁ - a₂ = 100. Since common difference d is same, difference between 1000th terms is also (a₁ + 999d) - (a₂ + 999d) = a₁ - a₂ = 100.

Question 13

How many three-digit numbers are divisible by 7?
Solution: AP is 105, 112, 119, ..., 994.
994 = 105 + (n - 1)7 ⇒ 889 = 7(n - 1) ⇒ n - 1 = 127 ⇒ n = 128.

Question 14

How many multiples of 4 lie between 10 and 250?
Solution: AP is 12, 16, 20, ..., 248.
248 = 12 + (n - 1)4 ⇒ 236 = 4(n - 1) ⇒ n - 1 = 59 ⇒ n = 60.

Question 15

For what value of n, are the n-th terms of two APs: 63, 65, 67, ... and 3, 10, 17, ... equal?
Solution: 63 + (n - 1)2 = 3 + (n - 1)7 ⇒ 60 = 5(n - 1) ⇒ n - 1 = 12 ⇒ n = 13.

Question 16

Determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
Solution: a₇ - a₅ = 12 ⇒ 2d = 12 ⇒ d = 6.
a₃ = 16 ⇒ a + 2(6) = 16 ⇒ a = 4.
The AP is 4, 10, 16, 22, ...

Question 17

Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.
Solution: Reversing AP: 253, 248, ..., 3. Here a = 253, d = -5.
a₂₀ = 253 + 19(-5) = 253 - 95 = 158.

Question 18

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution: (a+3d)+(a+7d) = 24 ⇒ 2a+10d = 24 ⇒ a+5d = 12.
(a+5d)+(a+9d) = 44 ⇒ 2a+14d = 44 ⇒ a+7d = 22.
Subtracting: 2d = 10 ⇒ d = 5, a = -13.
First three terms are -13, -8, -3.

Question 19

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Solution: 7000 = 5000 + (n - 1)200 ⇒ 2000 = 200(n - 1) ⇒ n = 11.
Year = 1995 + 10 = 2005 (11th year).

Question 20

Ramkali saved Rs 5 in the first week and increased weekly savings by Rs 1.75. If in the n-th week her savings become Rs 20.75, find n.
Solution: 20.75 = 5 + (n - 1)1.75 ⇒ 15.75 = 1.75(n - 1) ⇒ n - 1 = 9 ⇒ n = 10.

EXERCISE 5.3

Question 1: Find the sum of APs

(i) 2, 7, 12, ... to 10 terms: S₁₀ = (10/2)[2(2) + 9(5)] = 5[4 + 45] = 245

(ii) -37, -33, -29, ... to 12 terms: S₁₂ = (12/2)[2(-37) + 11(4)] = 6[-74 + 44] = -180

(iii) 0.6, 1.7, 2.8, ... to 100 terms: S₁₀₀ = (100/2)[2(0.6) + 99(1.1)] = 50[1.2 + 108.9] = 5505

(iv) 1/15, 1/12, 1/10, ... to 11 terms: d = 1/60. S₁₁ = (11/2)[2/15 + 10/60] = (11/2)[3/10] = 33/20

Question 2: Find the sums

(i) 7 + 10 1/2 + 14 + ... + 84: a = 7, d = 3.5, l = 84 ⇒ 84 = 7 + (n-1)3.5 ⇒ n = 23.
S₂₃ = (23/2)(7 + 84) = 2093/2 = 1046 1/2

(ii) 34 + 32 + 30 + ... + 10: a = 34, d = -2, l = 10 ⇒ 10 = 34 + (n-1)(-2) ⇒ n = 13.
S₁₃ = (13/2)(34 + 10) = 286

(iii) (-5) + (-8) + (-11) + ... + (-230): a = -5, d = -3, l = -230 ⇒ -230 = -5 + (n-1)(-3) ⇒ n = 76.
S₇₆ = (76/2)(-5 - 230) = 38(-235) = -8930

Question 3: Solve sub-parts

• (i) a = 5, d = 3, a_n = 50 ⇒ n = 16, S₁₆ = (16/2)(5 + 50) = 440
• (ii) a = 7, a₁₃ = 35 ⇒ 35 = 7 + 12d ⇒ d = 7/3, S₁₃ = (13/2)(7 + 35) = 273
• (iii) a₁₂ = 37, d = 3 ⇒ 37 = a + 33 ⇒ a = 4, S₁₂ = (12/2)(4 + 37) = 246
• (iv) a₃ = 15, S₁₀ = 125 ⇒ a + 2d = 15, 5(2a + 9d) = 125 ⇒ 2a + 9d = 25 ⇒ d = -1, a₁₀ = 8
• (v) d = 5, S₉ = 75 ⇒ 75 = (9/2)(2a + 40) ⇒ a = -35/3, a₉ = 85/3
• (vi) a = 2, d = 8, S_n = 90 ⇒ 90 = (n/2)(4 + (n-1)8) ⇒ 4n² - 2n - 90 = 0 ⇒ 2n² - n - 45 = 0 ⇒ n = 5, a₅ = 34
• (vii) a = 8, a_n = 62, S_n = 210 ⇒ 210 = (n/2)(8 + 62) ⇒ n = 6, 62 = 8 + 5d ⇒ d = 54/5
• (viii) a_n = 4, d = 2, S_n = -14 ⇒ 4 = a + 2n - 2 ⇒ a = 6 - 2n. -14 = (n/2)(6 - 2n + 4) ⇒ n² - 5n - 14 = 0 ⇒ n = 7, a = -8
• (ix) a = 3, n = 8, S = 192 ⇒ 192 = 4(6 + 7d) ⇒ 48 = 6 + 7d ⇒ d = 6
• (x) l = 28, S = 144, n = 9 ⇒ 144 = (9/2)(a + 28) ⇒ 32 = a + 28 ⇒ a = 4

Question 4

How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
Solution: 636 = (n/2)[18 + (n-1)8] ⇒ 636 = n(4n + 5) ⇒ 4n² + 5n - 636 = 0.
Solving quadratic: (n - 12)(4n + 53) = 0 ⇒ n = 12.

Question 5

First term is 5, last term is 45 and sum is 400. Find n and d.
Solution: 400 = (n/2)(5 + 45) ⇒ n = 16.
45 = 5 + 15d ⇒ d = 8/3.

Question 6

First and last terms are 17 and 350, d = 9. How many terms and what is their sum?
Solution: 350 = 17 + (n-1)9 ⇒ n = 38.
S₃₈ = (38/2)(17 + 350) = 19(367) = 6973.

Question 7

Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution: 149 = a + 21(7) ⇒ a = 2.
S₂₂ = (22/2)(2 + 149) = 11(151) = 1661.

Question 8

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18.
Solution: d = 18 - 14 = 4, a = 14 - 4 = 10.
S₅₁ = (51/2)[20 + 50(4)] = (51/2)(220) = 5610.

Question 9

If sum of first 7 terms is 49 and that of 17 terms is 289, find sum of first n terms.
Solution: S₇ = 49 ⇒ (7/2)(2a+6d) = 49 ⇒ a+3d = 7.
S₁₇ = 289 ⇒ (17/2)(2a+16d) = 289 ⇒ a+8d = 17.
Solving: d = 2, a = 1.
S_n = (n/2)[2(1) + (n-1)2] = n².

Question 10

Show that a_n forms an AP and find S₁₅:
(i) a_n = 3 + 4n: a₁ = 7, d = 4. S₁₅ = (15/2)[2(7) + 14(4)] = 525
(ii) a_n = 9 - 5n: a₁ = 4, d = -5. S₁₅ = (15/2)[2(4) + 14(-5)] = -465

Question 11

If S_n = 4n - n², find S₁, S₂, a₂, a₃, a₁₀, a_n.
Solution:
S₁ = 4(1) - 1² = 3 (First term a₁ = 3)
S₂ = 4(2) - 2² = 4
a₂ = S₂ - S₁ = 4 - 3 = 1
Common difference d = 1 - 3 = -2
a₃ = 3 + 2(-2) = -1
a₁₀ = 3 + 9(-2) = -15
a_n = 3 + (n-1)(-2) = 5 - 2n

Question 12

Find the sum of first 40 positive integers divisible by 6.
Solution: 6, 12, 18, ..., 240.
S₄₀ = (40/2)(6 + 240) = 20(246) = 4920.

Question 13

Find the sum of first 15 multiples of 8.
Solution: 8, 16, 24, ..., 120.
S₁₅ = (15/2)(8 + 120) = (15/2)(128) = 960.

Question 14

Find the sum of odd numbers between 0 and 50.
Solution: 1, 3, 5, ..., 49 (n = 25 terms).
S₂₅ = (25/2)(1 + 49) = 625.

Question 15

Construction delay penalty: Rs 200 for 1st day, Rs 250 for 2nd day, etc. Delayed by 30 days.
Solution: AP: a = 200, d = 50, n = 30.
S₃₀ = (30/2)[2(200) + 29(50)] = 15[400 + 1450] = Rs 27,750.

Question 16

Total 7 prizes = Rs 700. Each prize is Rs 20 less than preceding.
Solution: S₇ = 700, n = 7, d = -20.
700 = (7/2)[2a + 6(-20)] ⇒ 200 = 2a - 120 ⇒ a = 160.
Prizes are: Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, Rs 40.

Question 17

Classes I to XII, 3 sections each, plant trees equal to class number.
Solution: Trees planted by 1 section = 1 + 2 + ... + 12 = (12 × 13)/2 = 78.
For 3 sections: Total trees = 3 × 78 = 234 trees.

Question 18

Spiral of 13 semicircles with radii 0.5 cm, 1.0 cm, 1.5 cm, ... (π = 22/7)
Solution: Total length = π(r₁ + r₂ + ... + r₁₃) = π × (13/2)[2(0.5) + 12(0.5)] = (22/7) × (13/2) × 7 = 143 cm.

Question 19

200 logs stacked: 20 in bottom, 19 in next, 18 in next.
Solution: 200 = (n/2)[2(20) + (n-1)(-1)] ⇒ 400 = n(41 - n) ⇒ n² - 41n + 400 = 0.
(n - 16)(n - 25) = 0. If n = 25, a₂₅ = 20 + 24(-1) = -4 (impossible). Thus, n = 16 rows.
Logs in top row = a₁₆ = 20 + 15(-1) = 5 logs.

Question 20

Potato race: 10 potatoes, 1st at 5m, others 3m apart.
Solution: Distance for 1st potato = 2 × 5 = 10m.
Distances form AP: 10, 16, 22, ... for 10 terms.
Total distance = (10/2)[2(10) + 9(6)] = 5[20 + 54] = 370 metres.

EXERCISE 5.4 (Optional)

Question 1

Which term of AP: 121, 117, 113, ... is its first negative term?
Solution: a_n < 0 ⇒ 121 + (n - 1)(-4) < 0 ⇒ 125 < 4n ⇒ n > 31.25.
Smallest integer n is 32 (32nd term).

Question 2

Sum of 3rd and 7th terms of AP is 6, product is 8. Find S₁₆.
Solution: a₃ + a₇ = 6 ⇒ a + 4d = 3.
(a + 2d)(a + 6d) = 8 ⇒ (3 - 2d)(3 + 2d) = 8 ⇒ 9 - 4d² = 8 ⇒ d = ±1/2.
Case 1: d = 1/2, a = 1 ⇒ S₁₆ = (16/2)[2 + 15(1/2)] = 76.
Case 2: d = -1/2, a = 5 ⇒ S₁₆ = (16/2)[10 + 15(-1/2)] = 20.

Question 3

Ladder rungs decrease from 45 cm to 25 cm. Distance between top and bottom rungs is 2 1/2 m = 250 cm. Rungs are 25 cm apart.
Solution: Number of rungs n = (250 / 25) + 1 = 11.
Total length of wood = (11 / 2)(45 + 25) = (11 / 2)(70) = 385 cm.

Question 4

Houses numbered 1 to 49. S_{x - 1} = S₄₉ - S_x.
Solution: [ (x - 1) x ] / 2 = [ (49 × 50) / 2 ] - [ x(x + 1) / 2 ]
⇒ (x² - x) + (x² + x) = 2450 ⇒ 2x² = 2450 ⇒ x² = 1225 ⇒ x = 35.

Question 5

Terrace with 15 steps, each 50 m long, 1/4 m rise, 1/2 m tread.
Solution: Volume of 1st step = 50 × (1/4) × (1/2) = 25/4 m³.
Step volumes form AP: 25/4, 50/4, 75/4, ... up to 15 terms.
Total Volume = (15 / 2)[ (25/4) + (375/4) ] = (15 / 2)(100) = 750 m³.