JEE Main 2026 — Mathematics Algebra

Given the sum of the first $n$n terms of the A.P. is $S_n = 3n^2 + 5n$Sn​=3n2+5n.

The $n$n-th term of the A.P. is given by:

$$T_n = S_n - S_{n-1}$$Tn​=Sn​−Sn−1​

$$T_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)]$$Tn​=(3n2+5n)−[3(n−1)2+5(n−1)]

$$T_n = 3n^2 + 5n - [3(n^2 - 2n + 1) + 5n - 5]$$Tn​=3n2+5n−[3(n2−2n+1)+5n−5]

$$T_n = 3n^2 + 5n - (3n^2 - 6n + 3 + 5n - 5)$$Tn​=3n2+5n−(3n2−6n+3+5n−5)

$$T_n = 6n + 2$$Tn​=6n+2

We need to find the sum of the squares of the first $10$10 terms:

$$\sum_{n=1}^{10} T_n^2 = \sum_{n=1}^{10} (6n + 2)^2$$n=1∑10​Tn2​=n=1∑10​(6n+2)2

$$\sum_{n=1}^{10} T_n^2 = \sum_{n=1}^{10} (36n^2 + 24n + 4)$$n=1∑10​Tn2​=n=1∑10​(36n2+24n+4)

$$\sum_{n=1}^{10} T_n^2 = 36 \sum_{n=1}^{10} n^2 + 24 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 4$$n=1∑10​Tn2​=36n=1∑10​n2+24n=1∑10​n+n=1∑10​4

Using the formulas for the sum of squares and sum of first $n$n natural numbers:

$$\sum_{n=1}^{10} T_n^2 = 36 \left( \dfrac{10 \times 11 \times 21}{6} \right) + 24 \left( \dfrac{10 \times 11}{2} \right) + 4 \times 10$$n=1∑10​Tn2​=36(610×11×21​)+24(210×11​)+4×10

$$\sum_{n=1}^{10} T_n^2 = 36(385) + 24(55) + 40$$n=1∑10​Tn2​=36(385)+24(55)+40

$$\sum_{n=1}^{10} T_n^2 = 13860 + 1320 + 40 = 15220$$n=1∑10​Tn2​=13860+1320+40=15220

Answer: $15220$15220