Question

The ratio of the sum of first $n$ terms of two A.Ps is $(7n+1):(4n+27),$ the ratio of their $10th$ term is $a:b$, find $a+b$.

Answer 1

Given: The ratio of sum of $n$ terms of two AP's $=(7n+1):(4n+27)$

Let the sum of n terms of first A.P and second A.P be $S_{n1}$ and $S_{n2}$ respectively.

So, $\frac{S_{n1}}{S_{n2}}=\frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_2+(n-1\left)d_2\right]}$

$\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+27}$

$\frac{a_1+\frac{n-1}{2}{d}_1}{a_2+\frac{n-1}{2}{d}_2}=\frac{7n+1}{4n+27}$---- (1)

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

$n=2m-1$

From (1), we get

$\frac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\frac{7(2m-1)+1}{4(2m-1)+27}$

$\frac{a_{m1}}{a_{m2}}=\frac{14m-6}{8m+23}$

Thus the ratio of $m^{th}$ terms of two AP’s is $(14m–6):(8m+23).$

Thus the ratio of ${10}^{th}$ terms of two AP’s is $134:103$