Arithmetic Progressions — Class 10 MCQs with Answers

Arithmetic Progressions — important questions & MCQs with answers (Class 10 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Common difference of the AP 5, 8, 11, 14, … is:

   A2B−3C5D3
   Show answerHide answer

   Answer: 3

   Explanation: Common difference d = term − previous term: d = 8 − 5 = 3, and the same gap of 3 repeats through 11, 14, confirming a genuine AP.

2. Q2 of 6Easy

   In an AP, the common difference d is:

   Aa₂ − a₁Ba₁ + a₂Ca₁/a₂Da₁ × a₂
   Show answerHide answer

   Answer: a₂ − a₁

   Explanation: By definition, the common difference is any term minus the one right before it, d = a₂ − a₁ — not a sum, ratio, or product of terms.

3. Q3 of 6Medium

   Sum of first n terms of AP 9, 17, 25... is 636. Then n is:

   A12B11C10D13
   Show answerHide answer

   Answer: 12

   Explanation: Sₙ = (n/2)[2a + (n − 1)d] with a = 9, d = 8 gives n(4n + 5) = 636, a quadratic that factors as (n − 12)(4n + 53) = 0, so n = 12.

4. Q4 of 6Medium

   Assertion (A): The 20th term of the AP 6, 11, 16, … is 101. Reason (R): The nth term of an AP with first term a and common difference d is aₙ = a + (n – 1)d.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
   Show answerHide answer

   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: A is true: a₂₀ = 6 + (20 – 1) × 5 = 6 + 95 = 101. R is true, and A is found directly from R.

5. Q5 of 6Hard

   If the 5th term of an AP is 20 and the 9th term is 36, then the first term is:

   A0B4C2D8
   Show answerHide answer

   Answer: 4

   Explanation: Write a + 4d = 20 and a + 8d = 36; subtracting gives 4d = 16, so d = 4, and a = 20 − 4×4 = 4.

6. Q6 of 6Hard

   Assertion (A): For the AP 30, 27, 24, …, the sum of the first 5 terms is equal to the sum of the first 16 terms. Reason (R): In this AP, the terms from the 6th to the 16th add up to zero.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
   Show answerHide answer

   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: A is true: S₅ = 5/2 × (60 – 12) = 120 and S₁₆ = 8 × (60 – 45) = 120. R is true (a₆ = 15 and a₁₆ = –15, so these 11 terms add to 0), and it is the reason the two sums are equal.

Arithmetic Progressions notes: key concepts (Class 10 Mathematics)

An arithmetic progression is a list in which each term differs from the previous one by a fixed amount. This chapter finds any term directly, sums the first n terms, and applies both to problems about instalments, savings and patterns.

Arithmetic progression (AP)

A sequence in which the difference between any term and the one before it is constant throughout.

Common difference (d)

The fixed gap between consecutive terms, found as aₙ − aₙ₋₁. A negative d makes the sequence decrease.

First term (a)

The starting value of the AP, from which every other term is generated by repeatedly adding d.

General (nth) term

aₙ = a + (n − 1)d gives any term directly, without listing all the terms before it.

Testing for an AP

Compute several consecutive differences; the sequence is an AP only if every one of them is the same.

Finding the number of terms

Set aₙ equal to the last term in aₙ = a + (n − 1)d and solve for n; a non-integer n means the value is not in the AP.

Sum of the first n terms

Sₙ = n/2[2a + (n − 1)d], used when the last term is unknown.

Sum using the last term

When the last term l is known, Sₙ = n/2(a + l) is far quicker — it is the average term times the number of terms.

nth term from the sum

The nth term can be recovered as aₙ = Sₙ − Sₙ₋₁, which is useful when only a sum formula is given.

Middle-term and symmetry tricks

Taking terms as a − d, a, a + d keeps the algebra symmetric and simplifies problems that fix the sum of three terms.

Applications

Situations changing by a constant amount — fixed annual increments, equal monthly savings, evenly spaced rows in a stack.