The sums of n terms of two arithmetic series | vedclass

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Question

(A) Let the first terms of the two arithmetic series be $a_1$ and $a_2$,and their common differences be $d_1$ and $d_2$ respectively.Given the ratio o

Vedclass · Accepted Answer

$53 : 155$

Vedclass · Answer

(A) Let the first terms of the two arithmetic series be $a_1$ and $a_2$,and their common differences be $d_1$ and $d_2$ respectively.Given the ratio of the sum of $n$ terms: $\frac{S_{n_1}}{S_{n_2}} = \frac{2n + 3}{6n + 5}$.The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.So,$\frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{2n + 3}{6n + 5}$.Simplifying,we get $\frac{a_1 + \frac{n-1}{2}d_1}{a_2 + \frac{n-1}{2}d_2} = \frac{2n + 3}{6n + 5}$.We want the ratio of the $13^{th}$ terms,which is $\frac{T_{13_1}}{T_{13_2}} = \frac{a_1 + 12d_1}{a_2 + 12d_2}$.Comparing $\frac{n-1}{2} = 12$,we get $n - 1 = 24$,so $n = 25$.Substituting $n = 25$ in the ratio: $\frac{T_{13_1}}{T_{13_2}} = \frac{2(25) + 3}{6(25) + 5} = \frac{50 + 3}{150 + 5} = \frac{53}{155}$.

The sums of $n$ terms of two arithmetic series are in the ratio $(2n + 3) : (6n + 5)$. Find the ratio of their $13^{th}$ terms.

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