The sums of n terms of two arithmatic series | vedclass

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Question

(a) We have $\frac{{{S_{{n_1}}}}}{{{S_{{n_2}}}}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{\frac{n}{2}[2{a_1} + (n - 1){d_1}]}}{{\frac{n}{2}[2{a_

Vedclass · Accepted Answer

$53 : 155$

Vedclass · Answer

(a) We have $\frac{{{S_{{n_1}}}}}{{{S_{{n_2}}}}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{\frac{n}{2}[2{a_1} + (n - 1){d_1}]}}{{\frac{n}{2}[2{a_2} + (n - 1){d_2}]}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{2\left[ {{a_1} + \left( {\frac{{n - 1}}{2}} \right)\,{d_1}} \right]}}{{2\left[ {{a_2} + \left( {\frac{{n - 1}}{2}} \right)\,{d_2}} \right]}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{{a_1} + \left( {\frac{{n - 1}}{2}} \right)\,{d_1}}}{{{a_2} + \left( {\frac{{n - 1}}{2}} \right)\,{d_2}}} = \frac{{2n + 3}}{{6n + 5}}$

Put $n = 25$ then $\frac{{{a_1} + 12{d_1}}}{{{a_2} + 12{d_2}}} = \frac{{2(25) + 3}}{{6(25) + 3}}$

==> $\frac{{{T_{{{13}_1}}}}}{{{T_{{{13}_2}}}}} = \frac{{53}}{{155}}$.

The sums of $n$ terms of two arithmatic series are in the ratio $2n + 3:6n + 5$, then the ratio of their ${13^{th}}$ terms is

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