Worksheet Solutions: Arithmetic Progression | Mathematics Class 10 ICSE PDF Download

Section A – Very Short Answer Questions 

Q1. Write the common difference of the AP: 12, 7, 2, −3, …
Solution:
d = 7 − 12 = −5

Q2. Find the 5th term of the AP: 4, 9, 14, …
Solution:
a = 4, d = 9 − 4 = 5
t5 = a + (5 − 1)d = 4 + 4×5 = 24

Q3. If the 3rd term of an AP is 10 and 5th term is 18, find the common difference.
Solution:
t3 = a + 2d = 10
t5 = a + 4d = 18
Subtracting → (a + 4d) − (a + 2d) = 18 − 10 → 2d = 8 → d = 4

Q4. The 1st term of an AP is 7 and common difference is 0. Write the 4th term.
Solution:
t4 = a + (4 − 1)d = 7 + 3×0 = 7

Q5. Find the arithmetic mean between 8 and 20.
Solution:
AM = (a + b)/2 = (8 + 20)/2 = 14

Section B – Short Answer Questions (2–3 marks each)

Q6. Find the 12th term of the AP: 5, 11, 17, …
Solution:
a = 5, d = 11 − 5 = 6
t12 = a + (12 − 1)d = 5 + 11×6 = 5 + 66 = 71

Q7. Which term of the AP: 15, 12, 9, … will be −6?
Solution:
a = 15, d = −3, tn = −6
tn = a + (n − 1)d
−6 = 15 + (n − 1)(−3)
−6 = 15 − 3n + 3 → −6 = 18 − 3n
−3n = −24 → n = 8
So, −6 is the 8th term.

Q8. Find the sum of the first 25 terms of the AP: 7, 10, 13, …
Solution:
a = 7, d = 3, n = 25
Sn = n/2 [2a + (n − 1)d]
S25 = 25/2 [2×7 + 24×3] = 25/2 [14 + 72] = 25/2 × 86 = 1075

Q9.The sum of first 15 terms of an AP is 630 and its first term is 7. Find the common difference.
Solution:
Sn = n/2 [2a + (n − 1)d]
630 = 15/2 [2×7 + 14d]
630 = 15/2 [14 + 14d]
630 = (15/2)(14)(1 + d)
630 = 105(1 + d)
1 + d = 6 → d = 5

Q10. Insert 4 arithmetic means between 3 and 23.
Solution:
We need AP: 3, A1, A2, A3, A4, 23 → 6 terms in total.
a = 3, l = 23, n = 6
d = (l − a)/(n − 1) = (23 − 3)/5 = 20/5 = 4
Means are: 3+4=7, 11, 15, 19
So, AMs = 7, 11, 15, 19

Section C – Long Answer Questions (4–5 marks each)

Q11. Find the sum of the first 30 multiples of 9.
Solution:
AP: 9, 18, 27, …
a = 9, d = 9, n = 30
Sn = n/2 [2a + (n − 1)d]
S30 = 30/2 [2×9 + 29×9]
= 15 [18 + 261]
= 15 × 279 = 4185

Q12. The 7th term of an AP is 32 and 13th term is 62. Find the first term, common difference, and sum of first 20 terms.
Solution:
t7 = a + 6d = 32 … (i)
t13 = a + 12d = 62 … (ii)

Subtracting (ii) − (i):
(a + 12d) − (a + 6d) = 62 − 32
6d = 30 → d = 5

Put in (i): a + 6×5 = 32 → a + 30 = 32 → a = 2

Now, S20 = 20/2 [2a + (20 − 1)d]
= 10 [2×2 + 19×5]
= 10 [4 + 95] = 10 × 99 = 990

Q13. A farmer arranges trees in rows. First row has 20 trees, second 18, third 16, and so on. If the arrangement continues until a row has 0 trees, find:
(i) The number of rows
(ii) The total number of trees

Stepwise solution:
Given AP: a = 20, d = −2
Last non-negative term is 0.
0 = a + (n − 1)d
0 = 20 + (n − 1)(−2)
−20 = −2(n − 1) → n − 1 = 10 → n = 11
S11 = n/2 [2a + (n − 1)d]
S11 = 11/2 [40 + 10(−2)] = 11/2 [40 − 20] = 11/2 × 20 = 110
Answer: (i) 11 rows (ii) 110 trees