Jee Main Jan 30 Shift 1 Mathematics - Question1
Question 1
In an arithmetic progression if sum of 20 terms is 790 and sum of 10 terms is 145 , then $S _{15}-S _5$ is (when $S _n$ denotes sum of $n$ terms)
(1) 400
(2) 395
(3) 385
(4) 405
Show Answer
Answer: (2)
Solution:
$S _{20}=\frac{20}{2}[2 a+19 d]=790$
$2 a+19 d=79$
$S _{10}=\frac{10}{2}[2 a+9 d]=145$
$2 a+9 d=29$
from (1) and (2) $a=-8, \quad d=5$
$S _{15}-S _5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d]$
$=\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20]$
$=405-10$
$=395$
BETA
