If $\frac{2+4+6+8------- ext { upto } n ext { terms }}{1+3+5+7------- ext { upto } n ext { terms }}=\frac{37}{36}$, then $n=$

Question

If $\frac{2+4+6+8-------	ext { upto } n 	ext { terms }}{1+3+5+7-------	ext { upto } n 	ext { terms }}=\frac{37}{36}$, then $n=$

Aicharya · Accepted Answer

sum of $n$ even natural numbers $=n(n+1)$ sum of $n$ odd natural numbers $=n^{\wedge} 2$ $n(n+1) / n^{\wedge} 2=37 / 36$ $n+1 / n=37 / 36$ cross multiplication $36(n+1)=37 n$ $36 n+36=37 n$ $37 n-36 n=36$ $n=36$

If $\frac{2+4+6+8-------\text { upto } n \text { terms }}{1+3+5+7-------\text { upto } n \text { terms…

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