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Sum of n Terms of Arithmetic Progression (A.P.)

Introduction

Finding the sum of the first n terms of an Arithmetic Progression (A.P.) is a common and useful pattern in aptitude tests. Many problems ask for total amounts (sums of salaries, distances, scores) where terms form an A.P. Knowing a formulaic and a shortcut approach saves time and reduces arithmetic errors.

Pattern: Sum of n Terms of Arithmetic Progression (A.P.)

Pattern

The sum of the first n terms of an A.P. is given by either formula:

Sₙ = (n/2) × [2a + (n - 1)d] or Sₙ = (n/2) × (a + l)

Where a = first term, d = common difference, n = number of terms, and l = last (nth) term.

Step-by-Step Example

Question

Find the sum of the first 20 terms of the A.P.: 3, 7, 11, 15, …

Solution

  1. Step 1: Identify a, d and n:

    First term a = 3. Common difference d = 7 - 3 = 4. Number of terms n = 20.

  2. Step 2: Find the nth (last) term l (optional):

    l = a + (n - 1)d = 3 + (20 - 1)×4 = 3 + 19×4 = 3 + 76 = 79.

  3. Step 3: Use the sum formula Sₙ = (n/2) × (a + l):

    S₂₀ = (20/2) × (3 + 79) = 10 × 82 = 820.

  4. Final Answer:

    The sum of the first 20 terms is 820.

  5. Quick Check:

    Average term = (first + last)/2 = (3 + 79)/2 = 41 → Sum = average × number of terms = 41 × 20 = 820 ✅

Quick Variations

1. Given a, d and n → use Sₙ = (n/2)[2a + (n - 1)d].

2. Given a and l (last term) and n → use Sₙ = (n/2)(a + l).

3. Given two non-consecutive term values (e.g., Tₚ and T_q) and n → first find a and d, then compute Sₙ.

4. Sum of terms in an A.P. with negative d or decreasing sequences → formulas remain the same.

Trick to Always Use

  • Step 1 → If you can easily find the last term, use Sₙ = (n/2)(a + l) - it's the fastest.
  • Step 2 → If last term is not obvious, use Sₙ = (n/2)[2a + (n - 1)d] and compute (n-1)d first to reduce mistakes.

Summary

Summary

Key takeaways for the Sum of n Terms of an A.P.:

  • Two equivalent formulas: Sₙ = (n/2)[2a + (n - 1)d] and Sₙ = (n/2)(a + l).
  • Use the (a + l) formula when the last term is known or easy to compute - it reduces arithmetic steps.
  • Quick check: average term × number of terms gives the sum.
  • Useful for word problems where totals are required (distance, money, seats, etc.).

Practice

(1/5)
1. Find the sum of the first 8 terms of the A.P.: 1, 3, 5, 7, …
easy
A. 64
B. 60
C. 68
D. 72

Solution

  1. Step 1: Identify a, d and n

    First term a = 1, common difference d = 2, number of terms n = 8.

  2. Step 2: Find the last term (l)

    Last term l = a + (n - 1)d = 1 + 7×2 = 15.

  3. Step 3: Apply the sum formula

    Use Sₙ = (n/2)(a + l) → S₈ = (8/2)(1 + 15) = 4 × 16 = 64.

  4. Final Answer:

    The sum of the first 8 terms is 64 → Option A.

  5. Quick Check:

    Average term = (1 + 15)/2 = 8 → Sum = 8 × 8 = 64 ✅

Hint: If the last term is easy to find, use Sₙ = (n/2)(a + l).
Common Mistakes: Forgetting to use (n-1) when computing the last term.
2. Find the sum of the first 12 terms of the A.P.: 4, 7, 10, 13, …
easy
A. 240
B. 246
C. 252
D. 258

Solution

  1. Step 1: Identify a, d and n

    First term a = 4, common difference d = 3, number of terms n = 12.

  2. Step 2: Compute the last term (l)

    Last term l = a + (n - 1)d = 4 + 11×3 = 4 + 33 = 37.

  3. Step 3: Apply the sum formula

    S₁₂ = (12/2)(4 + 37) = 6 × 41 = 246.

  4. Final Answer:

    The sum of the first 12 terms is 246 → Option B.

  5. Quick Check:

    Average term = (4 + 37)/2 = 20.5 → Sum = 20.5 × 12 = 246 ✅

Hint: Find l first and then use Sₙ = (n/2)(a + l) to reduce steps.
Common Mistakes: Mixing up n and (n-1) when computing the last term.
3. If the first term is 6 and the common difference is 4, find the sum of the first 10 terms of the A.P.
easy
A. 240
B. 230
C. 250
D. 260

Solution

  1. Step 1: Identify a, d and n

    Given a = 6, d = 4, n = 10.

  2. Step 2: Find the last term (l)

    l = a + (n - 1)d = 6 + 9×4 = 6 + 36 = 42.

  3. Step 3: Use the sum formula

    S₁₀ = (10/2)(6 + 42) = 5 × 48 = 240.

  4. Final Answer:

    The sum of the first 10 terms is 240 → Option A.

  5. Quick Check:

    Average term = (6 + 42)/2 = 24 → Sum = 24 × 10 = 240 ✅

Hint: Compute (n/2) first to simplify multiplication.
Common Mistakes: Using incorrect last term for large n.
4. How many terms of the A.P. 9, 17, 25, 33, … must be taken to make a sum of 231?
medium
A. 6
B. 8
C. 7
D. 9

Solution

  1. Step 1: Identify a, d and Sₙ

    First term a = 9, common difference d = 8, required sum Sₙ = 231.

  2. Step 2: Set up the sum formula

    Use Sₙ = (n/2)[2a + (n - 1)d] → 231 = (n/2)[18 + 8(n - 1)].

  3. Step 3: Multiply both sides and form quadratic

    462 = n(8n + 10) ⇒ 8n² + 10n - 462 = 0 ⇒ divide by 2 ⇒ 4n² + 5n - 231 = 0.

  4. Step 4: Solve the quadratic for n

    Discriminant Δ = 5² - 4×4×(-231) = 25 + 3696 = 3721. √Δ = 61. So n = (-5 + 61)/8 = 56/8 = 7 (positive integer).

  5. Final Answer:

    You must take 7 terms → Option C.

  6. Quick Check:

    Last term for n=7: l = 9 + 6×8 = 57 → S₇ = (7/2)(9 + 57) = 3.5 × 66 = 231 ✅

Hint: Convert Sₙ formula to a quadratic and choose the positive integer root.
Common Mistakes: Forgetting to multiply Sₙ by 2 before expanding the bracket.
5. The sum of n terms of an A.P. is given by Sₙ = 2n² + 3n. Find the 15th term of the sequence.
medium
A. 59
B. 60
C. 62
D. 61

Solution

  1. Step 1: Know the relation between Sₙ and Tₙ

    Term Tₙ = Sₙ - Sₙ₋₁. Given Sₙ = 2n² + 3n.

  2. Step 2: Compute S₁₅ and S₁₄

    S₁₅ = 2(15)² + 3(15) = 2×225 + 45 = 450 + 45 = 495.
    S₁₄ = 2(14)² + 3(14) = 2×196 + 42 = 392 + 42 = 434.

  3. Step 3: Subtract to get T₁₅

    T₁₅ = S₁₅ - S₁₄ = 495 - 434 = 61.

  4. Final Answer:

    The 15th term is 61 → Option D.

  5. Quick Check:

    You may also derive Tₙ algebraically as Sₙ - Sₙ₋₁; numeric subtraction confirms 61 ✅

Hint: When Sₙ is given as a formula, use Tₙ = Sₙ - Sₙ₋₁ to get any term quickly.
Common Mistakes: Forgetting to compute Sₙ₋₁ before subtracting.

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