Question

The ratio between the sum of n terms of two A.P.'s is $3n+8:7n+15$ .Find the ratio between their 12th terms.

• A. $\frac{7}{12}.$
• B. $\frac{7}{16}.$
• C. $\frac{7}{20}.$
• D. $\frac{7}{18}.$

Answer 1

The correct option is B $\frac{7}{16}.$

$\frac{S_n}{S_n^{\prime }}=\frac{\left[n/2\right][2a+[n-1]d]}{\left[n/2\right][2a^{\prime }+[n-1]d^{\prime }]}=\frac{3n+8}{7n+15}$ or

$\frac{a+\left[(n-1)/2\right]d}{a^{\prime }+\left[(n-1)/2\right]d^{\prime }}=\frac{3n+8}{7n+15}$ $...\left(1\right)$

We have to find $\frac{T_{12}}{T_{12}^{\prime }}=\frac{a+11d}{a^{\prime }+11d^{\prime }}$

Choosing $(n-1)/2=11$ or $n=23$ in $\left(1\right)$, we get

$\frac{T_{12}}{T_{12}^{\prime }}=\frac{a+11d}{a^{\prime }+11d^{\prime }}=\frac{3\left(23\right)+8}{7\left(23\right)+15}=\frac{77}{176}=\frac{7}{16}$

Hence, option B.