Sequences And Series Question 7
Question 7 - 2024 (27 Jan Shift 2)
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 :-
(1) -118
(2) -110
(3) -115
(4) -100
Show Answer
Answer (3)
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. $\mathrm{a}=-129 \frac{1}{4}$ and $\mathrm{d}=\frac{3}{4}$
Required term $=$
$-129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right)$
$=-129-\frac{1}{4}+15-\frac{3}{4}=-115$
