Question

If $\frac {3+5+7+....n \,terms} {5+8+11+....10\, terms} = 7$,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\times3+\left(n-1\right)2\right]}{\frac{10}{2} \left[ 2\times5+\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\times370=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 $n= 35 \quad \left[\because n=-37{\text{ is not possible}}\right]$