Question

If $\frac{3+5+7+\dots \dots +\text{upto n terms}}{5+8+11+\dots \dots \text{upto 10 terms}}=7$, then find the value of n.

Answer 1

We have,

$\frac{3+5+7+\dots \dots +\text{upto n terms}}{5+8+11+\dots \dots \text{upto 10 terms}}=7$

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

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

$\Rightarrow \frac{n[6+2n-2]}{10[10+27]}=7$

$\Rightarrow \frac{n[4+2n]}{10\times 37}=7$

$\Rightarrow 4n+2n^2=7\times 30$

$\Rightarrow 2n+n^2=7\times 185$

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

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

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

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

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

$\Rightarrow n-35=0$ $[\because n+37\ne 0]$

$\Rightarrow n=35$