JEE Advanced 2022 · Paper 1 · Mathematics Q10

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MathematicsMultiple correct+4 / -2Question 10 of 54

Let $a_1, a_2, a_3, \ldots$ be an arithmetic progression with $a_1 = 7$ and common difference 8. Let $T_1, T_2, T_3, \ldots$ be such that $T_1 = 3$ and $T_{n+1} - T_n = a_n$ for $n \ge 1$ . Then, which of the following is/are TRUE?

T_{20} = 1604

\sum_{k=1}^{20} T_k = 10510

T_{30} = 3454

\sum_{k=1}^{30} T_k = 35610

Answer and worked solution

Official answer: B, C

Closed form. $a_n=8n-1$ , so
$T_n=3+\sum_{k=1}^{n-1}(8k-1)=3+4n(n-1)-(n-1)=4n^2-5n+4.$

Single terms.

$T_{20}=1600-100+4=1504$ , so (A) is false.
$T_{30}=3600-150+4=3454$ , so (C) is true.

Sums. $\sum_{k=1}^nT_k=4\sum k^2-5\sum k+4n$ .

$n=20$ : $4(2870)-5(210)+80=10510$ , so (B) is true.
$n=30$ : $4(9455)-5(465)+120=35615\ne35610$ , so (D) is false.

Topics:Special series sums Arithmetic progression

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JEE Advanced 2022 · Paper 1, Mathematics question 10