Question

The sum n terms of two A Ps are in ratio $\frac{7n+1}{4n+27}$ Find the ratio of their 11th terms

• A. $\frac{148}{111}$
• B. $\frac{213}{311}$
• C. $\frac{221}{343}$
• D. $\frac{114}{157}$

Answer 1

The correct option is A $\frac{148}{111}$

Let $a_1$ and $a_2$is the first term and $d_1$ and $d_2$ is the common difference of two A.P
$\Rightarrow \frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_1+(n-1)d_1}=\frac{7n+1}{4n+27}$
$\Rightarrow \frac{a_1+\left(\frac{n-1}{2}{d}_1\right)}{a_1+\left(\frac{n-1}{2}{d}_1\right)}=\frac{7n+1}{4n+27}$
For Ratio of 11 th term
$\Rightarrow \frac{n-1}{2}=10$
$\Rightarrow n-1=20\Rightarrow n=21$
S0 ratio of 11 th term is$\Rightarrow \frac{7\times 21+1}{4\times 21+27}$
$\Rightarrow \frac{148}{111}$