Exercise 6.3 — Sum of AP

Sum of n terms of an arithmetic progression.

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Exercise 6.3 — Sum of n Terms of an Arithmetic Progression

Exercise 6.3 introduces the second major formula of Chapter 6 — after learning to find any individual term (a_n) of an AP in Exercise 6.2, this exercise asks a different question: what is the total when you add up the first n terms of an AP? This single formula unlocks problems ranging from simple number sums to real-life situations like total savings, total distance, or total seating capacity.

Sum Formula (two forms) Sum of Natural Numbers Finding n from Sₙ Real-Life Applications

💡 The two forms of the sum formula: Sₙ = n/2 [2a + (n−1)d] — use this when you know the first term (a) and common difference (d). Sₙ = n/2 (a + l) — use this instead when you already know the last term (l), since it avoids recalculating d.

Where This Formula Comes From

The sum formula is derived by writing the series forwards and backwards, then adding the two together — every pair (first + last, second + second-last, and so on) gives the same total, 2a + (n−1)d. Since there are n such pairs, but each term has been counted twice, the actual sum is half of n × [2a + (n−1)d].

Sₙ = n/2 [2a + (n−1)d] = n/2 (a + l)

📌 Choosing the right version: If a question gives you a, d, and n — use the first form directly. If a question gives you the first term, the last term, and how many terms there are — the second form (n/2 (a+l)) is faster and avoids an extra step.

Worked Example 1 — Using a, d and n

Solution

Find the sum of the first 22 terms of the AP: 8, 3, −2, ...

a = 8, d = 3 − 8 = −5, n = 22 S₂₂ = 22/2 [2(8) + (22−1)(−5)] S₂₂ = 11 [16 + 21(−5)] = 11 [16 − 105] = 11 × (−89) S₂₂ = −979

⚠️ Common mistake: Students often forget the negative sign carries through the entire bracket. Since d = −5 here, (n−1)d is also negative — losing that sign changes the entire final answer.

Worked Example 2 — Using First Term, Last Term and n

Solution

Find the sum of the AP: 5, 8, 11, ..., 230

First, find n: a = 5, d = 3, l = 230 230 = 5 + (n−1)(3) → (n−1) = 225/3 = 75 → n = 76 Now apply Sₙ = n/2 (a + l): S₇₆ = 76/2 (5 + 230) S₇₆ = 38 × 235 = 8,930

✅ Why this method is faster: Once you know the last term, you never need to calculate (n−1)d separately — you just add the first and last term and multiply by half the number of terms.

Sum of the First n Natural Numbers

A special and extremely common case is the sum of the first n natural numbers — 1, 2, 3, ..., n. This is simply an AP with a = 1 and d = 1, so substituting into the sum formula gives a shortcut worth memorising directly:

Sum of first n natural numbers = n(n+1)/2

Solution

Find the sum of the first 50 natural numbers

Using n(n+1)/2 with n = 50: Sum = 50 × 51 / 2 = 1,275

Worked Example 3 — Finding n When the Sum Is Given

Solution

How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636?

a = 9, d = 8, Sₙ = 636 636 = n/2 [2(9) + (n−1)(8)] 1272 = n [18 + 8n − 8] = n [8n + 10] 8n² + 10n − 1272 = 0 → 4n² + 5n − 636 = 0 Solving the quadratic: n = 12 (rejecting the negative root) n = 12

📐 Why the negative root is rejected: Since n represents a count of terms, it must always be a positive whole number — any negative or fractional solution from the quadratic is discarded automatically.

Real-Life Application

Sum-of-AP questions frequently appear as word problems — total savings over months with a fixed monthly increase, total logs stacked in rows that decrease by a fixed amount, or total prize money distributed in unequal instalments that form an AP. The method is always the same: identify a, d and n (or a, l and n) from the situation, then apply the appropriate sum formula.

Solution

A person saves ₹32 in the first month and increases the saving by ₹4 every following month. Find the total savings in the first 12 months.

a = 32, d = 4, n = 12 S₁₂ = 12/2 [2(32) + (12−1)(4)] S₁₂ = 6 [64 + 44] = 6 × 108 S₁₂ = ₹648

Common Mistakes to Avoid

Confusing aₙ (nth term) with Sₙ (sum of n terms): These are two completely different formulas — aₙ gives one term, Sₙ gives a total. Read the question carefully to see which one is actually being asked for.
Forgetting to find n first: In problems where the last term is given but n is not, you must calculate n using the nth-term formula before you can use Sₙ = n/2(a+l).
Sign errors with negative common difference: As in Worked Example 1, a negative d must be carried through every bracket carefully.
Not rejecting invalid roots: When Sₙ = 636 leads to a quadratic in n, always reject negative or non-integer solutions — n must be a positive whole number.
Mixing up the two sum formula forms: Using n/2(a+l) without actually knowing the last term (l) is a frequent error — check which values the question actually gives you before choosing a formula.

⛔ High-value exam tip: Word problems that describe a real situation (savings, distance, seating) and ask for a total after a fixed number of periods are extremely common 3- and 4-mark questions in board exams. Practise translating the words into a, d and n before attempting the calculation.

Quick Reference — Key Formulas from This Exercise

Formula	When to Use
`Sₙ = n/2 [2a + (n−1)d]`	When a, d and n are known
`Sₙ = n/2 (a + l)`	When first term, last term and n are known
`Sum of first n natural numbers = n(n+1)/2`	Special case: a = 1, d = 1

What This Exercise Prepares You For

The sum-of-AP formula connects directly to Exercise 6.2 — nth Term of AP, since finding n or the last term is often the first step before applying the sum formula. It also lays the groundwork for Exercise 6.4 — Geometric Progression, where a parallel but different sum formula is introduced for progressions that grow by multiplication instead of addition.

📐 Board Exam Tip (CBSE, Telangana & AP): Questions combining the nth-term formula and the sum formula in a single multi-part question are very common — practise moving fluently between the two rather than treating them as separate topics.