Question

If the sums of n terms of two AP.'s are in the ratio (3n + 2) : (2n + 3), then find the ratio of their 12th terms.

Answer 1

Let the first terms of the two A.P.'s be a and a'; and their common difference be d and d'.

Now,

$$\begin{gathered} \frac{S_n}{S_n\text{'}}=\frac{\left(3n+2\right)}{\left(2n+3\right)} \\ \Rightarrow \frac{{\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]}{{\displaystyle \frac{n}{2}}\left[2a\text{'}+\left(n-1\right)d\text{'}\right]}=\frac{\left(3n+2\right)}{\left(2n+3\right)} \\ \Rightarrow \frac{{\displaystyle \left[2a+\left(n-1\right)d\right]}}{{\displaystyle \left[2a\text{'}+\left(n-1\right)d\text{'}\right]}}=\frac{\left(3n+2\right)}{\left(2n+3\right)} \\ \mathrm{Let}n=23 \\ \Rightarrow \frac{2a+\left(23-1\right)d}{2a\text{'}+\left(23-1\right)d\text{'}}=\frac{3\times 23+2}{2\times 23+3} \\ \Rightarrow \frac{2a+22d}{2a\text{'}+22d\text{'}}=\frac{69+2}{46+3} \\ \Rightarrow \frac{2\left(a+11d\right)}{2\left(a\text{'}+11d\text{'}\right)}=\frac{71}{49} \\ \therefore \frac{a_{12}}{a_{12}\text{'}}=\frac{71}{49} \end{gathered}$$

So, the ratio of their 12th terms is 71 : 49.