The ratio of the sums of first n terms of two | vedclass

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(C) Let the first terms of the two $A.P.s$ be $a_1$ and $a_2$,and their common differences be $d_1$ and $d_2$ respectively.The sum of the first $n$ te

Vedclass · Accepted Answer

$\frac{119}{138}$

Vedclass · Answer

(C) Let the first terms of the two $A.P.s$ be $a_1$ and $a_2$,and their common differences be $d_1$ and $d_2$ respectively.The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given the ratio of the sums of $n$ terms:$\frac{S_{n,1}}{S_{n,2}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{4n+3}{5n-7}$$\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{4n+3}{5n-7}$We need to find the ratio of the $15^{th}$ terms,which is $\frac{a_1 + 14d_1}{a_2 + 14d_2}$.To get $14d$ in the expression,we set $\frac{n-1}{2} = 14$,which gives $n-1 = 28$,or $n = 29$.Substituting $n = 29$ in the ratio:$\frac{2a_1 + 28d_1}{2a_2 + 28d_2} = \frac{4(29)+3}{5(29)-7}$$\frac{2(a_1 + 14d_1)}{2(a_2 + 14d_2)} = \frac{116+3}{145-7}$$\frac{a_1 + 14d_1}{a_2 + 14d_2} = \frac{119}{138}$Thus,the ratio of the $15^{th}$ terms is $\frac{119}{138}$.

The ratio of the sums of first $n$ terms of two $A.P.s$ is $\frac{4n+3}{5n-7}$. Find the ratio of $15^{th}$ terms of the $A.P.s$.

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