Question

The ratio of the sum of n terms of two A.P.’s is $(7n+1):(4n+27)$. Find the ratio of their $m^{th}$ terms.

Answer 1

Let $a_1$, $a_2$ be the first terms and $d_1$, $d_2$ the common differences of the two given A.P.'s .

Then the sums of their n terms are given by $S_n$ = ($\frac{n}{2}$) [2$a_1$ + (n —1) $d_1$], and

$S_n$ = ($\frac{n}{2}$) [2$a_2$ + (n —1) $d_2$]

$\frac{S_n}{S_n{}^{\prime }}$ = $\frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_2+(n-1\left)d_2\right]}$

= $\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}$

It is given that

$\frac{S_n}{S_n{}^{\prime }}$ = $\frac{7n+1}{4n+27}$ ... (i)

To find the ratio of the $m^{th}$ terms of the two given AP ' s, we replace n by (2m - 1) in equation (i).

Then we get

$\frac{2a_1+(2m-2)d_1}{2a_2+(2m-2)d_2}$ = $\frac{14m-6}{8m+23}$

Hence the ration of the $m^{th}$ terms of the two A.P 's is (14 m - 6) : (8 m + 23)