20^{text {th }} term from end of progression 20,19 frac{1}{4}, 18 frac{1}{2}, 17 frac{3}{4}, ldots .

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8. Sequences and Series

hard

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 :-

A

$-118$

B

$-110$

C

$-115$

D

$-100$

(JEE MAIN-2024)

Solution

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

This is $A.P$. with common difference

$ d_1=-1+\frac{1}{4}=-\frac{3}{4} $

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

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

Required term $=$

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

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

Standard 11

Mathematics

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