Mathematics · Sequence and Series

JEE Main 2025 — 29 January, Morning Shift — Question 57

Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800 .

Then its $11^{\text {th }}$ term is

Answer: C

$\mathrm{S}_{3}=3 \mathrm{a}+3 \mathrm{~d}=54$

$\Rightarrow \mathrm{a}+\mathrm{d}=18$

$S_{20}=10(2 a+19 d)$

$\Rightarrow 10(36+17 \mathrm{~d})$

$\Rightarrow 1600<10(36+17 \mathrm{~d})<1800$

$\Rightarrow 160<36+17 \mathrm{~d}<180$

$\Rightarrow 124<17 d<144$

$\Rightarrow 7 \frac{5}{17}<\mathrm{d}<8 \frac{8}{17}$

Common difference will be natural number $\Rightarrow \mathrm{d}=8 \Rightarrow \mathrm{a}=10$

$\Rightarrow \mathrm{a}_{11}=10+10 \times 8=90$

Answer key and solution verified before publishing.

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