CBSE Important Questions Class 10 Maths Chapter 5 – Arithmetic Progression

Class 10 Maths Chapter 5 Arithmetic Progression is part of the CBSE Class 10 Maths syllabus. It explains what an AP is: a list of numbers in which each term is obtained by adding a fixed number, called the common difference, to the term before it. Students learn the formula for the nth term, aₙ = a + (n − 1)d, and the formula for the sum of the first n terms.

These CBSE important questions for Class 10 Maths Chapter 5 cover counting two-digit numbers divisible by 7, identifying which pattern is an AP, finding the common difference when two terms are given, finding d when the nth term or the sum of n terms is given as a formula, and finding the sum of a given number of terms. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. How many two-digit numbers are divisible by 7?

Answer: The first two-digit number divisible by 7 is 14, and the last one is 98.
So the list is 14, 21, 28, ..., 98, which is an AP with a = 14 and d = 7.
Using aₙ = a + (n − 1)d:
98 = 14 + (n − 1) × 7
84 = (n − 1) × 7
n − 1 = 12
n = 13
So 13 two-digit numbers are divisible by 7.

Q2. Consider a number pattern with 'a' as its first term. If every number except the first term is 'd' more than its preceding term, what will be the 4th term of the pattern?
(a) a + 3d
(b) 3a + d
(c) a + 4d
(d) a + d

Answer: (a) a + 3d
This pattern is an AP with first term a and common difference d.
Using the formula aₙ = a + (n − 1)d with n = 4:
a₄ = a + (4 − 1)d = a + 3d

Q3. Which of the following number patterns has an equal difference between its two consecutive terms?
(a) 1, 1, 1, 4, 4, 4, 7, 7, 7, ...
(b) 2, 4, 6, 8, 12, 16, 20, ...
(c) 4, 6, 8, 10, 12, ...
(d) 6, 9, 10, 13, 14, 17, 18, ...

Answer: (c) 4, 6, 8, 10, 12, ...
In this list the difference between every pair of consecutive terms is the same: 6 − 4 = 2, 8 − 6 = 2, 10 − 8 = 2, and so on. So it is an AP with d = 2.
In the other options the differences keep changing, so they are not APs.

Q4. The 3rd term of an AP is −40 and the 13th term is 0. Find the common difference.

Answer: Using aₙ = a + (n − 1)d:
a₃ = a + 2d = −40 ...(i)
a₁₃ = a + 12d = 0 ...(ii)
Subtracting (i) from (ii):
10d = 0 − (−40) = 40
d = 4
(Putting d = 4 in (i): a + 8 = −40, so a = −48.)

Q5. If the sum of n terms of an AP is given by n² + n, then tₙ equals
(a) n
(b) 2n
(c) 3n
(d) 4n

Answer: (b) 2n
We use the fact that tₙ = Sₙ − Sₙ₋₁.
Sₙ = n² + n
Sₙ₋₁ = (n − 1)² + (n − 1) = n² − 2n + 1 + n − 1 = n² − n
tₙ = (n² + n) − (n² − n) = 2n

Q6. The nth term of an AP is 7 − 4n. Find its common difference.

Answer: The common difference is d = tₙ − tₙ₋₁.
tₙ = 7 − 4n
tₙ₋₁ = 7 − 4(n − 1) = 7 − 4n + 4 = 11 − 4n
d = (7 − 4n) − (11 − 4n) = 7 − 4n − 11 + 4n = −4
So the common difference is −4, which means the AP is decreasing.

Q7. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Answer: Given a₁₇ − a₁₀ = 7.
Using the formula:
[a + 16d] − [a + 9d] = 7
a + 16d − a − 9d = 7
7d = 7
d = 1

Q8. The 3rd term of an AP is −40 and the 13th term is 0. Then the common difference equals
(a) −40
(b) −4
(c) 4
(d) 40

Answer: (c) 4
From a + 2d = −40 and a + 12d = 0, subtracting gives 10d = 40, so d = 4.

Q9. If the 3rd term and the 5th term of an AP are 5 and 9 respectively, the common difference of the AP is
(a) 3
(b) 2
(c) 4
(d) 8

Answer: (b) 2
a₃ = a + 2d = 5 ...(i)
a₅ = a + 4d = 9 ...(ii)
Subtracting (i) from (ii): 2d = 4, so d = 2.

Q10. Write the formula for the sum of the first n terms of an AP. Find the sum of the first 20 terms of the AP 2, 7, 12, ...

Answer: Formula: Sₙ = n/2 [2a + (n − 1)d], where a is the first term and d the common difference.
If the last term l is known, we can also use Sₙ = n/2 (a + l).
Here a = 2, d = 7 − 2 = 5, n = 20.
S₂₀ = 20/2 [2(2) + (20 − 1)(5)]
= 10 [4 + 95]
= 10 × 99
= 990

Q11. Which term of the AP 3, 8, 13, 18, ... is 78?

Answer: Here a = 3 and d = 8 − 3 = 5.
Let the nth term be 78.
aₙ = a + (n − 1)d
78 = 3 + (n − 1)(5)
75 = (n − 1)(5)
n − 1 = 15
n = 16
So 78 is the 16th term of this AP.

Q12. Find the sum of the first 15 multiples of 8.

Answer: The multiples of 8 are 8, 16, 24, ... which form an AP with a = 8 and d = 8.
Here n = 15.
Sₙ = n/2 [2a + (n − 1)d]
S₁₅ = 15/2 [2(8) + (15 − 1)(8)]
= 15/2 [16 + 112]
= 15/2 × 128
= 15 × 64
= 960

Related Study Material on Extramarks

• CBSE Important Questions Class 10 Maths Chapter 1
• CBSE Important Questions Class 10 Maths Chapter 2
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