Question

The ratio of sum to $n$ terms of two $A.P.s$ is $\frac{8n+1}{7n+3}$ for every $n\in N$. Find the ratio of their $7th$ terms and $mth$ terms.

Answer 1

As we know that sum of $n$ terms in an A.P. is given as-

$S_n=\frac{n}{2}(2a+(n-1)d]$

Therefore,

$\frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{8n+1}{7n+3}$

$\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{8n+1}{7n+3}$

$\frac{a_1+\left(\frac{n-1}{2}\right)d_1}{a_2+\left(\frac{n-1}{2}\right)d_2}=\frac{8n+1}{7n+3}.....\left(1\right)$

Now,

For $7^{th}$ term-

$\frac{n-1}{2}=6$

$\Rightarrow n=6\times 2+1=13$

Substituting the value of $n$ in equation $\left(1\right)$, we have

$\frac{a_1+6d_1}{a_2+6d_2}=\frac{8\times 13+1}{7\times 13+3}=\frac{104+1}{91+3}=\frac{105}{94}$

For $m^{th}$ term-

$\frac{n-1}{2}=(m-1)$

$\Rightarrow n=2m-2+1=2m-1$

Substituting the value of $n$ in equation $\left(1\right)$, we have

$\frac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\frac{8\times (2m-1)+1}{7\times (2m-1)+3}=\frac{16m-8+1}{14m-7+3}=\frac{16m-7}{14m-4}$