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

Question

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

Professor PC Thomas Classes · Answer

$ \begin{aligned} & \frac{\frac{\mathrm{n}}{2}[10+(\mathrm{n}-1) 4]}{\frac{\mathrm{n}+1}{2}[14+\mathrm{n} \cdot 2]}=\frac{17}{16} \Rightarrow \frac{\mathrm{n}(2 \mathrm{n}+3)}{(\mathrm{n}+1)(\mathrm{n}+7)}=\frac{17}{16} \ & 32 \mathrm{n}^2+48 \mathrm{n}=17 \mathrm{n}^2+136 \mathrm{n}+119 \ & 15 \mathrm{n}^2-88 \mathrm{n}-119=0(\mathrm{n}-7)(15 \mathrm{n}+17)=0 \ & \Rightarrow \mathrm{n}=7 \end{aligned} $