Question

$\frac{5+9+13+.......upton.terms}{If7+9+11+.....upto(n+1)terms}=\frac{17}{16}$ then n is equal to

• A. 7
• B. 6
• C. 9
• D. none of these

Answer 1

The correct option is A 7

$\frac{5+9+13+.....nterms}{7+9+.....to(n+1)terms}=\frac{17}{16}$

$\frac{\frac{n}{2}[10+(n-1\left)4\right]}{\frac{n+1}{2}[14+n2]}=\frac{17}{16}$

$\frac{n[4n+6]}{(n+1)(n+7)}=\frac{17}{16}$

$\frac{n[4n+6]}{(n+1)(2n+14)}=\frac{17}{16}$

$\frac{n(2n+3)}{(n+1)(n+7)}=\frac{17}{16}$

$32n^2+48n=17(n^2+8n+7)$

$15n^2+48n-136n-119=0$

$15n^2-88n-119=0$

$15n^2-105n+17n-119=0$

$15n(n-7)+17(n-7)=0$

$n=7$