The sum of the first 20 terms common between the series $3+7+11+15+$ and $1+6+11+$ $16+\ldots .$. is
The sum of the first 20 terms common between the series $3+7+11+15+$ and $1+6+11+$ $16+\ldots .$. is
4000
4020
4200
4220
Solution
Given $\mathrm{n}=20 ; \mathrm{S}_{20}=$ ?
Series (1) $\rightarrow 3,7, \underline{11}, 15,19,23,27, \underline{31}, 35$, $39,43,47$,
$\underline{51}, 55,59 \ldots$
Series (2) $\rightarrow 1,6, \underline{11}, 16,21,26, \underline{31}, 36,41$, $46, \underline{51}, 56$, $61,66,71$.
The common terms between both the series are $11,31,51,71 \ldots$
Above series forms an Arithmetic progression (A.P).
Therefore, first term (a) $=11$ and common difference $(\mathrm{d})=20$
Now, $\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 a+(n-1) d]$
$
\begin{aligned}
&\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{aligned}
$