The sum of n terms of two A.P.’s are in the ratio 5n + 4 : 9n + 6. Then, the ratio of their 18^th term is

Solution(By Examveda Team)

Let a₁, d₂ be the first terms of two ratios S and S' and d₁, d₂ be their common difference respectively
Then,
$$\eqalign{ & {S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]\,{\text{and}} \cr & S{'_n} = \frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right] \cr & {\text{Now,}} \cr & \,\frac{{{S_n}}}{{S{'_n}}} = \frac{{\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} \cr & = \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} \cr & {\text{But}}\,\frac{{{S_n}}}{{S{'_n}}} = \frac{{5n + 4}}{{9n + 6}} \cr & \therefore \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \frac{{5n + 4}}{{9n + 6}} \cr} $$
Now we have to find the ratios in 18^th term
Here n = 18
$$\eqalign{ & \therefore \frac{{2{a_1} + \left( {18 - 1} \right){d_1}}}{{2{a_2} + \left( {18 - 1} \right){d_2}}} = \frac{{5\left( {2n - 1} \right) + 4}}{{9\left( {2n - 1} \right) + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5\left( {2 \times 18 - 1} \right) + 4}}{{9\left( {2 \times 18 - 1} \right) + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5 \times 35 + 4}}{{9 \times 35 + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{175 + 4}}{{315 + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{179}}{{321}} \cr} $$