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

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(A) 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 $n$ terms of an

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(10m - $8$) : (14m - $5$)

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(A) 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 $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.Given the ratio of sums: $\frac{S_{n1}}{S_{n2}} = \frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{5n - 3}{7n + 2}$.This simplifies to $\frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{5n - 3}{7n + 2}$.We need the ratio of the $m^{th}$ terms: $\frac{a_m1}{a_m2} = \frac{a_1 + (m - 1)d_1}{a_2 + (m - 1)d_2}$.Multiply the numerator and denominator of the required ratio by $2$: $\frac{2a_1 + 2(m - 1)d_1}{2a_2 + 2(m - 1)d_2}$.Comparing this with the sum ratio expression,we set $n - 1 = 2(m - 1)$,which gives $n = 2m - 1$.Substitute $n = 2m - 1$ into the ratio $\frac{5n - 3}{7n + 2}$:Ratio $= \frac{5(2m - 1) - 3}{7(2m - 1) + 2} = \frac{10m - 5 - 3}{14m - 7 + 2} = \frac{10m - 8}{14m - 5}$.Thus,the ratio of the $m^{th}$ terms is $(10m - 8) : (14m - 5)$.

The ratio of the sums of first $n$ terms of two $A.P.s$ is $(5n - 3) : (7n + 2)$. Find the ratio of their $m^{th}$ terms.

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