Question

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

• A. $7$
• B. $6$
• C. $9$
• D. None of these

Answer 1

The correct option is A

$7$

Step 1. Find the value of $n$:

Given, $\frac{(5+9+13+...n)}{\left[7+9+11+...(n+1)\right]}=\frac{17}{16}$

Take $(5+9+13+\dots n)$

Here, $a=5$ and $d=4$

$\begin{array}{rcl}& \Rightarrow & S_1=\frac{n}{2}\left[10+(n-1)4\right]\\ & =& \frac{n}{2}\left(6+4n\right)\end{array}$

Now take, $\left[7+9+11+\dots (n+1)\right]$

Here, $a=7$ and $d=2$

$\Rightarrow$$S_2=\frac{(n+1)}{2}(14+2n)$

Step 2. Divide $S_1$ by $S_2$, we get

$\frac{(5+9+13+...n)}{\left[7+9+11+...(n+1)\right]}=\frac{17}{16}$

$\Rightarrow$ $\frac{S_1}{S_2}=\frac{17}{16}$

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

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

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

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

$\Rightarrow$ $n=7,-\frac{17}{15}$

$n$ cannot be negative.

$\mathbf{\therefore }\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}$

Hence, option (A) is correct.