Mathematics · Sequence and Series

JEE Advanced 2022 — Paper 1 — Question 10

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 $\mathrm{n} \geq 1$ .

Then, which of the following is/are TRUE?

Option A:
$T_{20}=1604$
Option B:
$\quad \sum_{k=1}^{20} T_{k}=10510$
Correct
Option C:
$T_{30}=3454$
Correct
Option D:
$\quad \sum_{k=1}^{30} T_{k}=35610$

Answer: B, C

Step-by-step solution

$\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{T}_{\mathrm{n}+1}-\mathrm{T}_{\mathrm{n}}\right)=\sum \mathrm{a}_{\mathrm{n}}$

$\Rightarrow \mathrm{T}_{\mathrm{n}+1}-\mathrm{T}_{1}=\sum \mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 7+(\mathrm{n}-1) 8]$

$\mathrm{T}_{\mathrm{n}+1}=\mathrm{n}(4 \mathrm{n}+3)+\mathrm{T}_{1}$

$\mathrm{T}_{\mathrm{n}+1}=4 \mathrm{n}^{2}+3 \mathrm{n}+3$

(A) $\mathrm{T}_{20}=4 \times 19^{2}+3 \times 9+3$

=1444+27+3=1474

(B) $\sum_{\mathrm{k}=0}^{19} \mathrm{~T}_{\mathrm{n}+1}=\sum_{\mathrm{k}=0}^{19} \mathrm{k}(4 \mathrm{k}+3)+3$ $=\sum_{\mathrm{k}=0}^{19}\left(4 \mathrm{k}^{2}+3 \mathrm{k}+3\right)=10510$

(D) $\sum_{\mathrm{k}=1}^{30} \mathrm{~T}_{\mathrm{k}}=\sum_{\mathrm{n}=0}^{29} \mathrm{~T}_{\mathrm{n}+1}=\sum_{\mathrm{n}=0}^{29} \mathrm{n}(4 \mathrm{n}+3)+3$

=4 \times\left(\frac{29 \times 30 \times 59}{6}\right)+\frac{3(29 \times 30)}{2}+90=35615

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2022
Paper: Paper 1
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression

Let l 1, l 2, ldots ., l 100 be consecutive terms of an arithmetic with common difference d 1 , and let w 1, w 2, ldots . ., w 100 be…

Answer: B, C

Step-by-step solution

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