Sequences And Series Question 12
Question 12 - 2024 (30 Jan Shift 1)
Let $S _a$ denote the sum of first $n$ terms an arithmetic progression. If $S _{20}=790$ and $S _{10}=145$, then $S _{15}-S _5$ is
(1) 395
(2) 390
(3) 405
(4) 410
Show Answer
Answer (1)
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) $a=-8, d=5$
$$ \begin{aligned} & S _{15}-S _5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\ & =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\ & =405-10 \\ & =395 \end{aligned} $$
