Question

If the ratio of the sum of n terms of two APs is(7n + 1) : (4n + 27), then the ratio of their 11th terms is.

• A. $4:3$
• B. $3:4$
• C. $2:3$
• D. $3:2$

Answer 1

The correct option is A $4:3$

Let the two series' be $T_n$ and $T_n^{\prime }$ with first terms $a$ and $a^{\prime }$ and common differences $d$ and $d^{\prime }$

The ratio of the sums of the series' $S_n$ and $S_n^{\prime }$ is given as,

$\frac{S_n}{S_n^{\prime }}=\frac{\left[n/2\right][2a+[n-1]d]}{\left[n/2\right][2a^{\prime }+[n-1]d^{\prime }]}=\frac{7n+1}{4n+27}$
Or,

$\frac{a+\left[(n-1)/2\right]d}{a^{\prime }+\left[(n-1)/2\right]d^{\prime }}=\frac{7n+1}{4n+27}$ $...\left(1\right)$

We have to find,

$\frac{T_{11}}{T_{11}^{\prime }}=\frac{a+10d}{a^{\prime }+10d^{\prime }}$

Choosing $(n-1)/2=10$ or $n=21$ in $\left(1\right)$ we get

$\frac{T_{11}}{T_{11}^{\prime }}=\frac{a+10d}{a^{\prime }+10d^{\prime }}=\frac{7\left(21\right)+1}{4\left(21\right)+27}=\frac{148}{111}=\frac{4}{3}.$
Hence, option A.