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# If the sums of n terms of two arithmetic progressions are in the ration $$\frac{{3n + 5}}{{5n + 7}},$$   then their nth terms are in the ration

A. $$\frac{{3n - 1}}{{5n - 1}}$$

B. $$\frac{{3n + 1}}{{5n + 1}}$$

C. $$\frac{{5n + 1}}{{3n + 1}}$$

D. $$\frac{{5n - 1}}{{3n - 1}}$$

### Solution(By Examveda Team)

In first A.P. let its first term be a1 and common difference d1
and in second A.P., first term be a2 and common difference d2, then
\eqalign{ & \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}
$$\therefore \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \frac{{3n + 5}}{{5n + 7}}$$
$${\text{Substituting n = 2n - 1, then}}$$
$$\frac{{2{a_1} + \left( {2n - 2} \right){d_1}}}{{2{a_2} + \left( {2n - 2} \right){d_2}}} =$$     $$\frac{{3\left( {2n - 1} \right) + 5}}{{5\left( {2n - 1} \right) + 7}}$$
$$\Rightarrow \frac{{{a_1} + \left( {n - 1} \right){d_1}}}{{{a_2} + \left( {n - 1} \right){d_2}}} =$$     $$\frac{{6n - 3 + 5}}{{10n - 5 + 7}}$$    (Dividing by 2)
\eqalign{ & \Rightarrow \frac{{{a_{1n}}}}{{{a_{2n}}}} = \frac{{6n + 2}}{{10n + 2}} \cr & \Rightarrow \frac{{{a_{1n}}}}{{{a_{2n}}}} = \frac{{3n + 1}}{{5n + 1}} \cr}
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