Find the 20th term of AP: 2, 7, 12, 17, ... and sum of first 20 terms

Question

For the AP: 2, 7, 12, 17, …, find:

The 20th term ( $a_{20}$ )
The sum of the first 20 terms ( $S_{20}$ )

(NCERT Class 10, Chapter 5 — Arithmetic Progressions)

Solution — Step by Step

First term: $a = 2$

Common difference: $d = 7 - 2 = 5$

Quick check: $12 - 7 = 5$ , $17 - 12 = 5$ . Consistent. We’re good.

a_{20} = a + (20 - 1)d = 2 + 19 \times 5 = 2 + 95 = \mathbf{97}

Since we already know $a_{20} = 97$ , we use the faster version of the sum formula:

S_{20} = \frac{20}{2} \times (a + a_{20}) = 10 \times (2 + 97) = 10 \times 99 = \mathbf{990}

Why This Works

The $n$ th term formula $a_n = a + (n-1)d$ counts how many “jumps” of $d$ we make from the first term. To reach the 20th term, we make 19 jumps (not 20 — the first term is our starting point, not a jump).

The sum formula $S_n = \frac{n}{2}(a + a_n)$ comes from a clever trick: pair the first and last terms, second and second-last, and so on. Each pair adds up to the same value ( $a + a_n$ ), and there are $n/2$ such pairs.

Alternative Method — Using the other sum formula

If you don’t know $a_n$ , use $S_n = \frac{n}{2}[2a + (n-1)d]$ :

S_{20} = \frac{20}{2}[2(2) + 19(5)] = 10[4 + 95] = 10 \times 99 = 990

Both sum formulas give the same answer. Use $S_n = \frac{n}{2}(a + a_n)$ when you already know the last term. Use $S_n = \frac{n}{2}[2a + (n-1)d]$ when you don’t. In board exams, showing both forms earns no extra marks — pick whichever is faster.

Common Mistake

The classic blunder: using $n$ instead of $(n-1)$ in the $n$ th term formula. Students write $a_{20} = 2 + 20 \times 5 = 102$ — which is actually $a_{21}$ . Remember, the first term is $a_1 = a + 0 \times d$ , so the 20th term uses $(20 - 1) = 19$ as the multiplier.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the other sum formula

Common Mistake

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