Question

The sum of n terms of A.P. are in the ratio of $\frac{7n+1}{4n+27}$ or $7n+1:4n+27$ then find the ratio of
${11}^{th}$ term
$r^{th}$ term
$n^{th}$ term (last term)

Answer 1

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

Suppose $S{n}_1,S{n}_2$ denote the sum of $n$ terms of two A.P s.

$\frac{S{n}_1}{S{n}_2}=\frac{7n+1}{4n+27}$ (given)

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

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

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

Put $\frac{n-1}{2}=m-1$, we get

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

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

$\Rightarrow n=2m-1$

$\Rightarrow \frac{a_{m_1}}{a_{m_2}}=\frac{7(2m-1)+1}{4(2m-1)+27}$

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

$\Rightarrow a_{m_1}:a_{m_2}=14m-6:8m+23$

Now Ratio of $11$ terms are

$=14\times 11-6:8\times 11+23$

$=148:111$

Ratio of $r^{th}$ terms $=14r-6:8r+23$

Ratio of $n^{th}$ term $=14n-6:18n+23$