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

Solution(By Examveda Team)

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

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l} ∵ 15 \times \left( { - 119} \right) = 1785\\ - 1785 = 17 \times \left( {105} \right)\\ - 88 = 17 - 105 \end{array} \right\}\]

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