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If $\frac{5+9+13+\ldots \text { to } n \text { terms }}{7+9+11+\ldots \text { to }(n+1) \text { terms }}=\frac{17}{16}$, then n = ?

A. 8 B. 7 C. 10 D. 11 Answer: Option B
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Solution(By Apex Team)

$\begin{aligned}&\text{ Sum of }5+9+13+\ldots\text{ to }n\text{ terms }\\ &=\frac{n}{2}[2a+(n-1)d]\\ &\text{ Here }a=5,\ d=9-5=4\\ &\therefore\text{ Sum }=\frac{n}{2}[2\times5+(n-1)\times4]\\ &=\frac{n}{2}[10+4n-4]\\ &=\frac{n}{2}[6+4n]\\ &=n(3+2n)\end{aligned}$ and sum of 7 + 9 + 11 + . . . . to (n + 1) terms $\begin{aligned}&=\frac{n+1}{2}[2\times7+(n+1-1)2]\\ &=\frac{n+1}{2}[14+2n]\\ &=(n+1)(7+n)\\ &\therefore\frac{5+9+13+\ldots\text{ to }n\text{ terms }}{7+9+11+\ldots\text{ to }(n+1)\text{ terms }}=\frac{17}{16}\\ &\Rightarrow\frac{n(3+2n)}{(n+1)(7+n)}=\frac{17}{16}\\ &\Rightarrow16n(3+2n)=17(n+1)(7+n)\\ &\Rightarrow48n+32n^2=17\left(n^2+8n+7\right)\\ &\Rightarrow48n+32n^2=17n^2+136n+119\\ &\Rightarrow48n+32n^2-17n^2-136n-119=0\\ &\Rightarrow15n^2-88n-119=0\\ &\Rightarrow15n^2-105n+17n-119=0\end{aligned}$
$\begin{array}{l} \left\{\begin{array}{l} \because 15 \times(-119)=1785 \\ -1785=17 \times(105) \\ -88=17-105 \end{array}\right\} \\ \\ \Rightarrow 15 n(n-7)+17(n-7)=0 \\ \Rightarrow(n-7)(15 n+17)=0 \\ \text { Either } n-7=0, \text { then } n=7 \end{array}$ $\text { or } 15 n+13=0, \text { then } n=$ $\Large\frac{-13}{15}$ which is not possible being fraction ∴ n = 7

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