Question

If the sum of the first $40$ terms of the series, $3+4+8+9+13+14+18+19+\dots ..$ is $\left(102\right)m$, then $m$ is equal to:

• A. $10$
• B. $25$
• C. $5$
• D. $20$

Answer 1

The correct option is D

$20$

Explanation for the correct option.

The given series $3+4+8+9+13+14+18+19+\dots ..40terms$ is neither in AP nor in G.P., but if we write it like

$\left(3+4\right)+\left(8+9\right)+\left(13+14\right)+\left(18+19\right)+\dots ..20terms=7+17+27+37+.....20terms$, it will be in AP where, $a=7,\text{and}d=10$.

Sum of $n$ terms of AP, ${\mathit{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{[}\mathbf{2}\mathbf{a}\mathbf{+}\left(n-1\right)\mathbf{d}\mathbf{]}$

$\begin{array}{rcl}\Rightarrow S_{20}& =& \frac{20}{2}[2\left(7\right)+\left(20-1\right)10]\\ & \Rightarrow & 102m=10\left(14+190\right)\\ & \Rightarrow & 102m=2040\\ \Rightarrow m& =& 20\end{array}$

Hence, option D is correct.