Sequences & SerieseasyPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Sum of Squares of AP Terms = 15220 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
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
A1022010220
B1286012860
C1522015220correct
D1978019780
Solution
Step 1: Tn=Sn−Sn−1T_n=S_n-S_{n-1}. Sn−1=3(n−1)2+5(n−1)=3n2−6n+3+5n−5=3n2−n−2S_{n-1}=3(n-1)^2+5(n-1)=3n^2-6n+3+5n-5=3n^2-n-2.
Tn=(3n2+5n)−(3n2−n−2)=6n+2.T_n=(3n^2+5n)-(3n^2-n-2)=6n+2.
 Step 2: (6n+2)2=36n2+24n+4(6n+2)^2=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.
 Step 3: ∑n=110n2=10⋅11⋅216=385\sum_{n=1}^{10}n^2=\dfrac{10\cdot11\cdot21}{6}=385, ∑n=110n=10⋅112=55\sum_{n=1}^{10}n=\dfrac{10\cdot11}{2}=55, ∑n=1104=40\sum_{n=1}^{10}4=40.
∴ 36(385)+24(55)+40=13860+1320+40=15220.\therefore\ 36(385)+24(55)+40=13860+1320+40=15220.
 Correct answer: (3)
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