JEE Advanced 2022 Paper 1 Q10 Mathematics Algebra Sequences & Series Medium

JEE Advanced 2022 Paper 1 · Q10 · Sequences & Series

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?

  1. A. $T_{20}=1604$
  2. B. $\displaystyle\sum_{k=1}^{20}T_{k}=10510$
  3. C. $T_{30}=3454$
  4. D. $\displaystyle\sum_{k=1}^{30}T_{k}=35610$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C

Solution

$a_{n}=7+8(n-1)=8n-1$. $T_{n+1}-T_{n}=8n-1$, so $T_{n}=T_{1}+\sum_{k=1}^{n-1}(8k-1)=3+8\cdot\dfrac{(n-1)n}{2}-(n-1)=3+4n(n-1)-(n-1)=3+(n-1)(4n-1)=4n^{2}-5n+4$. Check $T_{20}=4\cdot 400-100+4=1504$ (not 1604), so (A) false. $T_{30}=4\cdot 900-150+4=3454$, so (C) TRUE. $\sum_{k=1}^{N}T_{k}=4\sum k^{2}-5\sum k+4N=4\cdot\dfrac{N(N+1)(2N+1)}{6}-5\cdot\dfrac{N(N+1)}{2}+4N$. For $N=20$: $4\cdot\dfrac{20\cdot 21\cdot 41}{6}-5\cdot 210+80=11480-1050+80=10510$. So (B) TRUE. For $N=30$: $4\cdot\dfrac{30\cdot 31\cdot 61}{6}-5\cdot 465+120=37820-2325+120=35615\ne 35610$, so (D) false.

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