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JEE Main 20265 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 5 April · Morning Shift · Mathematics · Sequences and Series

Let the sum of the first n terms of an A.P. be 3n^2 + 5n . Then the...

Let the sum of the first nn terms of an A.P. be 3n2+5n3n^2 + 5n. Then the sum of squares of the first 1010 terms of the A.P. is:

Options

  1. A1022010220
  2. B1286012860
  3. C1522015220
  4. D1978019780

Correct answer

C. 1522015220

Step-by-step solution

Given the sum of the first nn terms of the A.P. is Sn=3n2+5nS_n = 3n^2 + 5n.

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

Tn=Sn−Sn−1T_n = S_n - S_{n-1}

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

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

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

Tn=6n+2T_n = 6n + 2

We need to find the sum of the squares of the first 1010 terms:

∑n=110Tn2=∑n=110(6n+2)2\sum_{n=1}^{10} T_n^2 = \sum_{n=1}^{10} (6n + 2)^2

∑n=110Tn2=∑n=110(36n2+24n+4)\sum_{n=1}^{10} T_n^2 = \sum_{n=1}^{10} (36n^2 + 24n + 4)

∑n=110Tn2=36∑n=110n2+24∑n=110n+∑n=1104\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

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

∑n=110Tn2=36(10×11×216)+24(10×112)+4×10\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=110Tn2=36(385)+24(55)+40\sum_{n=1}^{10} T_n^2 = 36(385) + 24(55) + 40

∑n=110Tn2=13860+1320+40=15220\sum_{n=1}^{10} T_n^2 = 13860 + 1320 + 40 = 15220

Answer: 1522015220
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