Question

If the sums of n terms of two arithmetic progressions are in the ratio $2n+5:3n+4$, then write the ratio of their $m^{th}$ terms.

Answer 1

It is given that

$\frac{S_n}{S_n^1}=\frac{2n+5}{3n+4}$

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

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

$\therefore \frac{S_n}{S_n^{{}^{\prime }}}=\frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_2+(n-1\left)d_2\right]}=\frac{2n+5}{3n+4}$

$\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{2n+5}{3n+4}$

$\frac{a_1+\left(\frac{n-1}{2}\right)d_1}{a_2+\left(\frac{n-1}{2}\right)d_2}=\frac{2n+5}{3n+4}$

Let, $\frac{n-1}{2}=m$

$\Rightarrow n-1=2m$

$\Rightarrow n=2m-1$

Replacing $\frac{n-1}{2}$ by m on both side of equation (ii), we get

$\frac{a_1+m{d}_1}{a_2+m{d}_2}=\frac{2(2m-1)+5}{3(2m-1)+4}$

$=\frac{4m-2+5}{6m-3+4}$

$=\frac{4m+3}{6m+1}$

$\therefore$ Ratio of the mth terms $=\frac{4m+3}{6m+1}$