If 5 + 9 + 13 + ...\, to\,n\, terms7 + 9 + 11 + ...\, to\, ( n + 1 )\, terms = 1716, then n = ?

Answer: Option B

Sum of 5 + 9 + 13 + . . . . to n terms
$$\begin{array}{rl}& =\frac{n}{2}\left[2a+\left(n-1\right)d\right]\\ & \text{Here}a=5,d=9-5=4\\ & \therefore \text{Sum}=\frac{n}{2}\left[2\times 5+\left(n-1\right)\times 4\right]\\ & =\frac{n}{2}\left[10+4n-4\right]\\ & =\frac{n}{2}\left[6+4n\right]\\ & =n\left(3+2n\right)\end{array}$$
and sum of 7 + 9 + 11 + . . . . to (n + 1) terms
$$\begin{array}{rl}& =\frac{n+1}{2}\left[2\times 7+\left(n+1-1\right)2\right]\\ & =\frac{n+1}{2}\left[14+2n\right]\\ & =\left(n+1\right)\left(7+n\right)\\ & \therefore \frac{5+9+13+...\text{to}n\text{terms}}{7+9+11+...\text{to}\left(n+1\right)\text{terms}}=\frac{17}{16}\\ & \Rightarrow \frac{n\left(3+2n\right)}{\left(n+1\right)\left(7+n\right)}=\frac{17}{16}\\ & \Rightarrow 16n\left(3+2n\right)=17\left(n+1\right)\left(7+n\right)\\ & \Rightarrow 48n+32n^2=17\left(n^2+8n+7\right)\\ & \Rightarrow 48n+32n^2=17n^2+136n+119\\ & \Rightarrow 48n+32n^2-17n^2-136n-119=0\\ & \Rightarrow 15n^2-88n-119=0\\ & \Rightarrow 15n^2-105n+17n-119=0\end{array}$$

$$\left\{\begin{array}{l}\because 15\times \left(-119\right)=1785\\ -1785=17\times \left(105\right)\\ -88=17-105\end{array}\right\}$$

$$\begin{array}{rl}& \Rightarrow 15n\left(n-7\right)+17\left(n-7\right)=0\\ & \Rightarrow \left(n-7\right)\left(15n+17\right)=0\end{array}$$
Either $$n-7=0,$$   then $$n=7$$
or $$15n+13=0,$$    then $$n=\frac{-13}{15}$$   which is not possible being fraction
∴ n = 7