Arithmetic Progressions for Math Olympiad: Complete Guide

Why This Matters

Arithmetic progressions are sequences with beautiful patterns, and Olympiad papers test your ability to find and exploit these patterns. From finding specific terms to calculating sums, AP problems appear in every competition.

For Class 10 students, Olympiad AP problems go beyond direct formula application. They test pattern recognition, reverse engineering, and combining AP concepts with algebra and geometry.

Best Strategy

Common Pitfalls

Mistakes:

• Common difference sign — $d$ can be negative (decreasing AP). Do not assume $d > 0$.
• nth term vs sum — $a_n$ is a single term; $S_n$ is the sum of first $n$ terms. $a_n = S_n - S_{n-1}$.
• Starting index — The first term is $a_1 = a$ (when $n = 1$), not $a_0$.
• Sum of negative APs — The sum can decrease. Maximum sum occurs just before terms become negative.

Practice Questions

Try these!

Question 1

Find the sum of the first 20 terms of AP: 5, 8, 11, 14, ...

Solution: $a = 5$, $d = 3$, $n = 20$.
$S_{20} = \frac{20}{2}[2(5) + 19(3)] = 10[10 + 57] = 10 \times 67 = 670$.

Question 2

The 7th term of an AP is 32 and 13th term is 62. Find the AP.

Solution: $a + 6d = 32$ and $a + 12d = 62$.
Subtracting: $6d = 30$, $d = 5$. Then $a = 32 - 30 = 2$.
AP: 2, 7, 12, 17, 22, ...

Question 3

How many terms of AP 3, 7, 11, ... must be taken so that sum equals 1710?

Solution: $\frac{n}{2}[6 + (n-1)4] = 1710$
$n(4n + 2) = 3420$
$4n^2 + 2n - 3420 = 0$
$2n^2 + n - 1710 = 0$
$n = \frac{-1 + \sqrt{1 + 13680}}{4} = \frac{-1 + 117}{4} = 29$.

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