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

$53 : 155$

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(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 sums of $n$ terms: $\frac{S_{n_1}}{S_{n_2}} = \frac{2n + 3}{6n + 5}$.Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$,we have:$\frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{2n + 3}{6n + 5}$$\frac{a_1 + \frac{n - 1}{2}d_1}{a_2 + \frac{n - 1}{2}d_2} = \frac{2n + 3}{6n + 5}$.To find the ratio of the $13^{th}$ terms,we need the expression to be in the form $\frac{a_1 + 12d_1}{a_2 + 12d_2}$.Setting $\frac{n - 1}{2} = 12$,we get $n - 1 = 24$,so $n = 25$.Substituting $n = 25$ into 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 $(2n + 3) : (6n + 5)$. Then the ratio of their $13^{th}$ terms is:

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