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

Given, $\frac{3+5+7+\cdots +n\text{ terms }}{5+8+11+\cdots +\text{ to }10\text{ terms }}=7$

$\begin{array}{lc}\Rightarrow & \frac{\frac{n}{2}[3\times 2+(n-1\left)2\right]}{\frac{10}{2}[2\times 5+9\times 3]}=7\\ \Rightarrow & \frac{\frac{n}{2}[6+2n-2]}{5[10+27]}=7\\ \Rightarrow & \frac{n}{2}(2n+4)=7\times 5\times 37\\ \Rightarrow & n(n+2)=35\times 37\\ \Rightarrow & n(n+2)=35(35+2)\\ \therefore & n=35\end{array}$