The 20 th term from the end of the progression 20,19 1/4, 18 1/2, 17 3/4, ldots,-129 1/4 is :-

Mathematics · Sequence and Series

JEE Main 2024 — 27 January, Shift 2 — Question 5

The $20^{\text {th }}$ term from the end of the progression $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots,-129 \frac{1}{4}$ is :-

$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}$

This is A.P. with common difference $d_{1}=-1+\frac{1}{4}=-\frac{3}{4}$

$-129 \frac{1}{4}, \ldots \ldots \ldots \ldots . ., 19 \frac{1}{4}, 20$

This is also A.P. $\mathrm{a}=-129 \frac{1}{4}$ and $\mathrm{d}=\frac{3}{4}$

Required term $=$

$-129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right)$

$=-129-\frac{1}{4}+15-\frac{3}{4}=-115$

Answer key and solution verified before publishing.

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