Question

If $\frac{5+9+13+.....n \, terms}{7+9+11+.......(n+1)terms}=\frac{17}{16}$ ,then n is equal to

• A. 7
• B. 12
• C. 8
• D. none of these.

Answer 1

Answer: (A)

7

Answer 2

$\frac{5+9+13+.....n{\text{ terms}}}{7+9+11+......\left(n+1\right) {\text{terms}}}$ $\Rightarrow \frac{\frac{n}{2}\left[2\times5+\left(n-1\right)4\right]}{\frac{n+1}{2}\left[2\times7+\left(n+1-1\right)2\right]} = \frac{17}{16}$ $\Rightarrow \frac{n\left[4n+6\right]}{\left(n+1\right)\left(2n+14\right) } = \frac{17}{16}$ $\Rightarrow \frac{n\left(2n+3\right)}{\left(n+1\right)\left(n+7\right)} = \frac{17}{16}$ $\Rightarrow \frac{2n^{2}+3n}{n^{2}+8n+7} = \frac{17}{16}$ $\Rightarrow 32n^{2}+48n = 17n^{2}+136n+119$ $\Rightarrow 15n^{2}-88n -119 = 0$ $\Rightarrow \left(n-7\right)\left(15n+17\right) = 0$ $\Rightarrow n=7$