Mathematics - Sequence & Series Question with Solution | TestHub

Given $\mathrm{n}=20;{\mathrm{S}}_{20}=$ ? Series (1) $\to 3,7,\underline{11},15,19,23,27,\underline{31},35$, $39,43,47$, $\underline{51},55,59\dots$ Series (2) $\to 1,6,\underline{11},16,21,26,\underline{31},36,41$, $46,\underline{51},56$, $61,66,71$. The common terms between both the series are $11,31,51,71\dots$ Above series forms an Arithmetic progression (A.P). Therefore, first term (a) $=11$ and common difference $(\mathrm{d})=20$ Now, ${\mathrm{S}}_{\mathrm{n}}=\frac{n}{2}[2a+(n-1)d]$ $$\begin{array}{rl}& {\mathrm{S}}_{20}=\frac{20}{2}[2\times 11+(20-1)20]\\ & {\mathrm{S}}_{20}=10[22+19\times 20]\\ & {\mathrm{S}}_{20}=10\times 402=4020\\ & \therefore {\mathrm{S}}_{20}=4020\end{array}$$