Mathematics · Sequence and Series

JEE Main 2024 — 5 April, Shift 1 — Question 27

Let $a_{1}, a_{2}, a_{3}, \ldots$ be in an arithmetic progression of positive terms.

Let $A_{k}=a_{1}{ }^{2}-a_{2}{ }^{2}+a_{3}{ }^{2}-a_{4}{ }^{2}+\ldots+a_{2 k-1}{ }^{2}-a_{2 k}{ }^{2}$ . If $\mathrm{A}_{3}=-153, \mathrm{~A}_{5}=-435$ and

$\mathrm{a}_{1}{ }^{2}+\mathrm{a}_{2}{ }^{2}+\mathrm{a}_{3}{ }^{2}=66$ , then $\mathrm{a}_{17}-\mathrm{A}_{7}$ is equal to $\qquad$

Answer: 910

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{d} \rightarrow$ common diff.

$\mathrm{A}_{\mathrm{k}}=-\mathrm{kd}[2 \mathrm{a}+(2 \mathrm{k}-1) \mathrm{d}]$

$\mathrm{A}_{3}=-153$

$\Rightarrow 153=13 \mathrm{~d}[2 \mathrm{a}+5 \mathrm{~d}]$

$51=\mathrm{d}[2 \mathrm{a}+5 \mathrm{~d}]$

$\mathrm{A}_{5}=-435$

$435=5 \mathrm{~d}[2 \mathrm{a}+9 \mathrm{~d}]$

$87=\mathrm{d}[2 \mathrm{a}+9 \mathrm{~d}]$

$(2)-(1)$

$36=4 d^{2}$

$\mathrm{d}=3, \mathrm{a}=1$

$\mathrm{a}_{17}-\mathrm{A}_{7}=49-[-7.3[2+39]]=910$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression

If frac1√(1)+√(2)+frac1√(2)+√(3)+ldots+frac1√(99)+√(100)=m and 1/1 × 2+1/2 × 3+ldots+1/99 × 100=n , then the point (m, n) lies on the line

Answer: 910

Step-by-step solution

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