Maths Q2 – JEE-MAIN 2026

Home Maths Sequences and SeriesQuestion #395

Let the sum of the first

n

terms of an A.P. be

3 n^{2} + 5 n

. Then the sum of squares of the first 10 terms of the A.P. is:

Clue 1 of 3: Arithmetic Progression Properties

Recall the relationship between the sum of $n$ terms ( $S_{n}$ ) and the $n$ -th term ( $a_{n}$ ) of an A.P.

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Step 1: Find the general term of the A.P.✦ Active

The sum of the first $n$ terms is given by $S_{n} = 3 n^{2} + 5 n$ . The $n$ -th term $a_{n}$ is given by $a_{n} = S_{n} - S_{n - 1}$ .

S_{n - 1} = 3 (n - 1)^{2} + 5 (n - 1) = 3 (n^{2} - 2 n + 1) + 5 n - 5 = 3 n^{2} - 6 n + 3 + 5 n - 5 = 3 n^{2} - n - 2

Therefore, the general term is:

a_{n} = (3 n^{2} + 5 n) - (3 n^{2} - n - 2) = 6 n + 2

Step 2: Calculate the square of the general term○ Expand

The square of the $n$ -th term is:

a_{n}^{2} = (6 n + 2)^{2} = 36 n^{2} + 24 n + 4

Step 3: Sum the squares of the first 10 terms○ Expand

We need to calculate $\sum_{n = 1}^{10} a_{n}^{2} = \sum_{n = 1}^{10} (36 n^{2} + 24 n + 4)$ . Using the sum formulas $\sum_{n = 1}^{N} k = \frac{N (N + 1)}{2}$ and $\sum_{n = 1}^{N} k^{2} = \frac{N (N + 1) (2 N + 1)}{6}$ for $N = 10$ :

\sum_{n = 1}^{10} n = \frac{10 (11)}{2} = 55

\sum_{n = 1}^{10} n^{2} = \frac{10 (11) (21)}{6} = 385

The sum of squares is:

36 \sum_{n = 1}^{10} n^{2} + 24 \sum_{n = 1}^{10} n + \sum_{n = 1}^{10} 4 = 36 (385) + 24 (55) + 4 (10)

= 13860 + 1320 + 40 = 15220

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Maths Question 2 – JEE-MAIN 2026

Clue 1 of 3: Arithmetic Progression Properties

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