then find the value of n

Question:

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

Solution:

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

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

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

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

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

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

$\Rightarrow n=35, n=-37$

Rejecting the negative value, we get:

$\Rightarrow n=35$

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