Question:
If the sum of the first 40 terms of the series, $3+4+8+9+$ $13+14+18+19+\ldots$ is $(102) \mathrm{m}$, then $\mathrm{m}$ is equal to:
Correct Option: 1
Solution:
$S=\underbrace{3+4}+\underbrace{8+9}+\underbrace{13+14}+\underbrace{18+19} \ldots . .40$ terms
$S=7+17+27+37+47+\ldots .20$ terms
$S_{40}=\frac{20}{2}[2 \times 7+(19) 10]=10[14+190]$
$=10[2040]=(102)(20)$
$\Rightarrow \quad m=20$