Question

If $\frac{2+5+8+.....nterms}{7+11+15+........nterms}=\frac{23}{35}$, then n value is ___________?

• A. $17$
• B. $15$
• C. $18$
• D. $23$

Answer 1

Answer: (B)

$15$

Sum of $n$ terms of an Arithmetic Progression with first term $a$ and common difference $d$ is $\frac{n}{2}\times [2a+(n-1\left)d\right]$

$\frac{2+5+8+...\text{n terms}}{7+11+15+...\text{n terms}}=\frac{\frac{n}{2}\times [2\times 2+(n-1)\times 3]}{\frac{n}{2}\times [2\times 7+(n-1)\times 4]}$

$\Rightarrow \frac{4+(n-1)\times 3}{14+(n-1)\times 4}=\frac{23}{35}$

$\Rightarrow 140+105n-105=322+92n-92$

$\Rightarrow n=15$

So, Option (B)