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

Introduction

किसी Arithmetic Progression (A.P.) के पहले n terms का sum निकालना aptitude tests में बहुत common और useful pattern है। कई problems में salaries, distances, scores आदि का total पूछा जाता है जहाँ terms A.P. बनाते हैं। Formula और shortcut दोनों का use calculation को तेज और error-free बनाता है।

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

Pattern

पहले n terms का sum इन formulas से निकाला जाता है:

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

जहाँ a = first term, d = common difference, n = number of terms, और l = nth (last) term है।

Step-by-Step Example

Question

निम्न A.P. के पहले 20 terms का sum निकालें: 3, 7, 11, 15, …

Solution

  1. Step 1: a, d और n पहचानें

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

  2. Step 2: nth (last) term l निकालें (optional)

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

  3. Step 3: Sum formula Sₙ = (n/2)(a + l) का use करें

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

  4. Final Answer:

    पहले 20 terms का sum = 820

  5. Quick Check:

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

Quick Variations

1. a, d और n दिए हों → Sₙ = (n/2)[2a + (n - 1)d] use करें।

2. a और l (last term) दिए हों → Sₙ = (n/2)(a + l) सबसे आसान है।

3. Tₚ और T_q जैसी दो terms से a और d निकालकर फिर Sₙ निकाल सकते हैं।

4. d negative या decreasing sequence हो तब भी formulas same रहते हैं।

Trick to Always Use

  • Step 1 → यदि last term आसानी से मिल जाए तो Sₙ = (n/2)(a + l) use करें-यह fastest method है।
  • Step 2 → यदि last term न मिले तो formula Sₙ = (n/2)[2a + (n - 1)d] use करें और पहले (n-1)d compute करें।

Summary

Summary

A.P. में sum निकालने के key points:

  • दो formula: Sₙ = (n/2)[2a + (n - 1)d] और Sₙ = (n/2)(a + l)
  • Last term पता हो तो (a + l) वाला formula सबसे simple होता है।
  • Quick check: Average term × number of terms = total sum।
  • Totals वाले word problems (distance, money, seats आदि) में बहुत useful।

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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