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About Arithmetic Progressions — Class 10 Olympiad

Find nth terms and sums of APs; solve competition-level sequence problems and pattern-based challenges. This topic is part of the Olympiad Class 10 mathematics syllabus (chapter: Module 5). On this page you can practice 59 questions across three difficulty levels — 20 easy, 19 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 35-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Arithmetic Progressions — solved examples for Class 10 Olympiad

Example 1easy

The `n`th term of an arithmetic progression is given by `a_n = 3n - 5`. What is the sum of its first term and its common difference?
  1. A)1
  2. B)2
  3. C)3
  4. D)4

Step-by-step solution

  1. The first term, `a_1 = 3(1) - 5 = 3 - 5 = -2`.
  2. The second term, `a_2 = 3(2) - 5 = 6 - 5 = 1`.
  3. The common difference, `d = a_2 - a_1 = 1 - (-2) = 3`.
  4. The sum of the first term and common difference is `a_1 + d = -2 + 3 = 1`.

Answer: 1

Example 2medium

If the `p`-th term of an Arithmetic Progression is `1/q` and the `q`-th term is `1/p`, where `p ≠ q`, what is the `pq`-th term of the progression?
  1. A)0
  2. B)p+q
  3. C)1
  4. D)1/pq

Step-by-step solution

  1. Let the first term be `a` and the common difference be `d`. The `n`-th term is `a_n = a + (n-1)d`.
  2. Given `a_p = a + (p-1)d = 1/q` (Eq 1) and `a_q = a + (q-1)d = 1/p` (Eq 2).
  3. Subtracting Eq 2 from Eq 1: `(p-1 - (q-1))d = 1/q - 1/p` which simplifies to `(p-q)d = (p-q)/pq`. Since `p ≠ q`, `d = 1/pq`.
  4. Substitute `d` into Eq 1: `a + (p-1)/pq = 1/q`. This gives `a = 1/q - (p-1)/pq = (p - (p-1))/pq = 1/pq`. The `pq`-th term is `a_{pq} = a + (pq-1)d = 1/pq + (pq-1)/pq = (1 + pq - 1)/pq = pq/pq = 1`.

Answer: 1

Example 3hard

If the ratio of the sum of the first `p` terms to the sum of the first `q` terms of an arithmetic progression is `p^2/q^2`, where `p ≠ q`, what is the ratio of the `p`th term to the `q`th term (`a_p/a_q`)?
  1. A)(p+1)/(q+1)
  2. B)(2p-1)/(2q-1)
  3. C)(p-1)/(q-1)
  4. D)p/q

Step-by-step solution

  1. Given `S_p / S_q = p^2 / q^2`. Using `S_n = n/2 × [2a + (n-1)d]`, we get `(p/2 × [2a + (p-1)d]) / (q/2 × [2a + (q-1)d]) = p^2 / q^2`.
  2. Simplifying, `[2a + (p-1)d] / [2a + (q-1)d] = p/q`. Cross-multiplying gives `q[2a + (p-1)d] = p[2a + (q-1)d]`.
  3. Expanding and rearranging: `2aq + pqd - qd = 2ap + pqd - pd` which leads to `2a(q-p) = (q-p)d`. Since `p ≠ q`, `q-p ≠ 0`, so `2a = d`.
  4. Now, the ratio of the `p`th term to the `q`th term is `a_p / a_q = (a + (p-1)d) / (a + (q-1)d)`. Substitute `d = 2a`: `(a + (p-1)2a) / (a + (q-1)2a) = (a(1 + 2p - 2)) / (a(1 + 2q - 2)) = (2p-1) / (2q-1)`.

Answer: (2p-1)/(2q-1)

Practice questions on Arithmetic Progressions

  1. Q1.easy

    A debt of ₹2450 is to be paid in monthly installments, which form an arithmetic progression. If the first installment is ₹50 and the last installment is ₹300, how many installments were paid?
    1. A)12
    2. B)13
    3. C)14
    4. D)15
    Show answer

    Answer: 14

    Hint: Use the formula for the sum of an AP, `S_n = n/2 × (a_1 + a_n)`, to find the number of installments.

  2. Q2.easy

    The sum of the first `n` terms of an arithmetic progression is given by `S_n = 2n^2 + 3n`. What is the 5th term of this AP?
    1. A)20
    2. B)21
    3. C)22
    4. D)23
    Show answer

    Answer: 21

    Hint: Remember that the `n`th term `a_n` can be found using the relationship `a_n = S_n - S_{n-1}`.

  3. Q3.easy

    In an arithmetic progression, if the sum of the first `m` terms is `n`, and the sum of the first `n` terms is `m` (where `m ≠ n`), then what is the sum of the first `(m+n)` terms?
    1. A)0
    2. B)m+n
    3. C)-(m+n)
    4. D)mn
    Show answer

    Answer: -(m+n)

    Hint: Write out the sum formulas for `S_m` and `S_n`. Subtract one equation from the other to find expressions for `(m-n)d` and `(m-n)a`.

  4. Q4.medium

    Let `a_1, a_2, ..., a_n` be an Arithmetic Progression with common difference `d ≠ 0`. What is the sum `S_n = 1/(a_1 a_2) + 1/(a_2 a_3) + ... + 1/(a_n a_{n+1})`?
    1. A)n/(a_1 d)
    2. B)n/(a_1 a_{n+1})
    3. C)d/(a_1 a_{n+1})
    4. D)1/(a_1 a_{n+1})
    Show answer

    Answer: n/(a_1 a_{n+1})

    Hint: Consider expressing each term `1/(a_k a_{k+1})` as a difference of two fractions using the common difference `d`.

  5. Q5.medium

    A company's profit in the first year was ₹ 1,00,000. Each year, the profit increased by a constant amount. If the total profit over the first 5 years was ₹ 7,50,000, what was the profit in the 7th year?
    1. A)₹ 2,00,000
    2. B)₹ 2,25,000
    3. C)₹ 2,75,000
    4. D)₹ 2,50,000
    Show answer

    Answer: ₹ 2,50,000

    Hint: The annual profits form an Arithmetic Progression. Use the sum of the first 5 terms to find the common difference.

  6. Q6.medium

    The ratio of the sum of `m` terms to the sum of `n` terms of an Arithmetic Progression is `m^2/n^2`. Find the ratio of the `m`-th term to the `n`-th term.
    1. A)m/n
    2. B)mn
    3. C)(2m-1)/(2n-1)
    4. D)(m+n)/(m-n)
    Show answer

    Answer: (2m-1)/(2n-1)

    Hint: Write out the formulas for `S_m`, `S_n`, `a_m`, and `a_n` in terms of `a` (first term) and `d` (common difference), then simplify the given ratio.

  7. Q7.hard

    Consider an arithmetic progression `a_1, a_2, a_3, ...` with common difference `d=3`. If `a_1 + a_5 + a_9 + ... + a_{4k-3} = 150` for `k=5`, find the first term `a_1`.
    1. A)3
    2. B)4
    3. C)6
    4. D)7
    Show answer

    Answer: 6

    Hint: The terms `a_1, a_5, a_9, ...` also form an arithmetic progression. Determine its common difference and the number of terms.

  8. Q8.hard

    Let `p_1, p_2, p_3` be three distinct prime numbers that form an arithmetic progression. What is the value of `p_1 + p_2 + p_3`?
    1. A)11
    2. B)15
    3. C)17
    4. D)19
    Show answer

    Answer: 15

    Hint: Consider the common difference and the properties of prime numbers, especially in relation to divisibility by 3.

  9. Q9.hard

    Let `S_n` denote the sum of the first `n` terms of an arithmetic progression. If `S_m = S_n` for distinct positive integers `m` and `n` (`m ≠ n`), what is the value of `S_{m+n}`?
    1. A)0
    2. B)m+n
    3. C)m-n
    4. D)mn
    Show answer

    Answer: 0

    Hint: Set up the equations for `S_m` and `S_n` using the sum formula `S_k = k/2 × [2a + (k-1)d]` and simplify. Look for a relationship that simplifies `2a + (m+n-1)d`.

These are 9 of the 59 questions available for Arithmetic Progressions. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.