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
$\mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790$
$2 a+19 d=79$
$S_{10}=\frac{10}{2}[2 a+9 d]=145$
$2 a+9 d=29$
From (1) and (2) $\mathrm{a}=-8, d=5$
$$ \begin{aligned} & S_{15}-S_{5}=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \ & =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \ & =405-10 \ & =395 \end{aligned} $$
