Question

If $\frac{3+5+7+....nterms}{5+8+11+....10terms}=7$,the value of $n$ is

• A. 35
• B. 36
• C. 37
• D. 40.

Answer 1

Answer: (A)

Since $\frac{3+5+7+.....n\text{terms}}{5+8+11+.....10\text{terms}}=7$
$\therefore \frac{\frac{n}{2}\left[2\times 3+\left(n-1\right)2\right]}{\frac{10}{2}\left[2\times 5+\left(10-1\right)3\right]}=7$
or $\frac{n\left[6+2n-2\right]}{\left[10\left(10+27\right)\right]}=7$
or $n\left[2n+4\right]=7\times 370=2590$
or $n\left(n+2\right)=1295$
or $n^2+2n-1295=0$
or $\left(n-35\right)\left(n+37\right)=0$
or $n=35\left[\because n=-37\text{ is not possible}\right]$