Chapter 6.4 — Exercise 6.3 — Sum of AP

Deriving the Sum Formula

Write the sum Sₙ = a + (a+d) + (a+2d) + … + (a+(n−1)d) forwards, then write the identical sum backwards, term for term. Adding the two versions together, every pair of matching terms sums to exactly the same value — (first + last) — repeated n times:

The second form is often faster when the last term l is already known, since a + l replaces the whole 2a+(n−1)d expression directly. A useful companion fact connects the two exercises: aₙ = Sₙ − Sₙ₋₁ — any single term can be recovered as the difference between two consecutive sums, since everything the sums share up to the (n−1)th term cancels out, leaving only the one term that's new to Sₙ.

Four Sums, Directly From the Formula

The fourth row is worth a second look: spotting that d = 1/60 from two fractions with unlike denominators takes more care than the whole-number rows, but the formula itself doesn't change at all once d is correctly found — fractions substitute into n/2[2a+(n−1)d] exactly the same way whole numbers do.

Three Sums With a Known Last Term

When a problem states the AP's final term outright rather than a term count, the n/2(a+aₙ) form of the formula is the faster route: find n first from the last term, then plug both a and the last term straight in, without ever needing the longer 2a+(n−1)d expression at all.

• 7, 10½, 14, …, 84 — a=7, d=3.5, giving n=23 terms, so S = 23/2(7+84) = 1046.5.
• 34 + 32 + 30 + … + 10 — a=34, d=−2, n=13, so S = 13/2(34+10) = 286.
• −5 + (−8) + (−11) + … + (−230) — a=−5, d=−3, n=76, so S = 38(−5−230) = −8930.

Solving for Whatever's Missing

Between the nth-term formula and the sum formula, any two known quantities from {a, d, n, aₙ, Sₙ} unlock the rest:

Two of the seven cases in this part of the exercise don't resolve into a simple substitution — they require solving a genuine quadratic. Given a=2, d=8, Sₙ=90, substituting into the sum formula produces 2n²−n−45=0, which factors to give n=5 (the negative root is rejected, since a term count can't be negative). A second case, aₙ=4, d=2, Sₙ=−14, produces n²−5n−14=0, giving n=7, a=−8. Both are a direct reminder that the sum formula is genuinely quadratic in n — solving for n specifically, rather than a, d, or aₙ, means expecting a quadratic equation and checking which root actually makes sense. This is the same factoring and root-rejection work covered at length in Exercise 5.2, showing up again here inside a word problem rather than as a quadratic equation presented on its own.

A Clean Result: the Sum of Odd Numbers

Given S₇=49 and S₁₇=289 for some AP, solving the two resulting equations gives a=1 and d=2 — meaning the AP itself is 1, 3, 5, 7, …, the odd numbers. Substituting a=1, d=2 into the general sum formula simplifies remarkably:

The sum of the first n odd numbers is always a perfect square — 1+3=4, 1+3+5=9, 1+3+5+7=16, and so on. This isn't a coincidence specific to this problem; it's a genuine identity that falls directly out of the sum formula whenever a=1 and d=2, and it's worth remembering as a shortcut in its own right the next time a problem specifically asks for the sum of consecutive odd numbers starting from 1.

Five Real-World Sums

Every word problem here follows the same two-step shape: recognise the quantities involved as an AP, then apply the sum formula once the AP's a, d, and n are identified. The formula itself works whether the AP has ten terms or ten thousand, and these five problems show just how differently it can be dressed up — counting problems, stacking problems, even a relay race — while the underlying calculation never actually changes.

• Cash prizes: ₹700 split into 7 prizes, each ₹20 less than the one before. With d=−20, n=7, and total 700, solving for the largest prize gives ₹160, 140, 120, 100, 80, 60, 40.
• Tree planting: three sections each of Classes I through XII plant trees equal to their class number — 3, 6, 9, …, 36 (a=3, d=3, n=12). Total trees: S₁₂ = 234.
• A spiral of semicircles: thirteen semicircles with radii 0.5, 1.0, 1.5, …, up to 6.5 cm, alternating centres. Each semicircle's length is π×radius, so the total length is π × 0.5 × (1+2+…+13) = (22/7) × 0.5 × 91 = 143 cm.
• Stacked logs: 200 logs, 20 in the bottom row, decreasing by 1 each row up. Solving n²−41n+400=0 gives two candidate values, 25 and 16 — but 25 rows would require a negative log count partway up, which is impossible, so the valid answer is 16 rows, with 5 logs in the top row.
• A bucket-and-ball relay: ten balls spaced 3 m apart, the nearest 5 m from the bucket. Each round trip (to a ball and back) forms its own AP — 10, 16, 22, …, 64 m (a=10, d=6, n=10) — and the total distance run is S₁₀ = 370 m.

The log-stacking problem is worth a second look specifically for its rejected root: both 25 and 16 satisfy the algebra perfectly, but only one describes a physically buildable stack. Checking a quadratic's roots against the real situation, not just against the equation, matters here exactly as much as it did throughout the previous chapter. The spiral-of-semicircles and bucket-relay problems are worth pausing on too, since neither one looks like a sum-of-an-AP question until the underlying pattern is spotted: the semicircle radii and the round-trip distances both increase by a fixed amount from one to the next, which is precisely what turns a geometry problem and a running problem into the same kind of calculation as the cash-prize and tree-planting problems beside them.

Adding by Multiplying Instead

Every sum in this exercise came from an AP — a sequence built by repeated addition. Exercise 6.4 introduces a second, structurally different kind of progression, built by repeated multiplication instead — the same recognising-the-pattern skill from Exercise 6.1, applied to a genuinely different kind of fixed step. For the nth-term formula this exercise's sum formula is built directly on top of, see Exercise 6.2, or revisit the AP definition in the chapter introduction.