Question

If $\frac{3+5+7+......+n\left(terms\right)}{5+8+11+.....+10\left(terms\right)}=7$, then the value of $n$ is

• A. $35$
• B. $36$
• C. $37$
• D. $40$

Answer 1

The correct option is A $35$

$S_n=$ Sum of n terms of an A.P.

$=\frac{n}{2}[2a+(n-1\left)d\right]$

where $a=$ first term

$d=$ common difference

$\therefore \frac{3+5+7+......+nterms}{5+8+11+......+10terms}=7$

$\Rightarrow \frac{\frac{n}{2}[2\times 3+(n-1)\times 2]}{\frac{10}{2}[2\times 5+(10-1)\times 3]}=7$

$\Rightarrow \frac{n(2n+4)}{370}=7$

$\Rightarrow 2n^2+4n-2590=0$

$\Rightarrow n^2+2n-1295=0$

$\Rightarrow n^2+37n-35n-1295=0$

$\Rightarrow n(n+37)-35(n+37)=0$

$\Rightarrow (n-35)(n+37)=0$

$\Rightarrow n=35$