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. $$\frac{{5n - 1}}{{3n - 1}}$$
Answer: Option B
Solution(By Examveda Team)
In first A.P. let its first term be a1 and common difference d1and 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)
$$\eqalign{ & \Rightarrow \frac{{{a_{1n}}}}{{{a_{2n}}}} = \frac{{6n + 2}}{{10n + 2}} \cr & \Rightarrow \frac{{{a_{1n}}}}{{{a_{2n}}}} = \frac{{3n + 1}}{{5n + 1}} \cr} $$
Related Questions on Progressions
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A. 5
B. 6
C. 4
D. 3
E. 7
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A. 600
B. 765
C. 640
D. 680
E. 690
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