Question

If the sums of n terms of two A.P,'s are in the ratio (3n + 2) : (2n + 3), 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,

$\frac{S_n}{S_n^{{}^{\prime }}}=\frac{(3n+2)}{(2n+3)}$

$\Rightarrow \frac{\frac{n}{2}[2a+(n-1\left)d\right]}{\frac{n}{2}[2a^{{}^{\prime }}+(n-1\left)d^{{}^{\prime }}\right]}=\frac{(3n+2)}{(2n+3)}$

$\Rightarrow \frac{[2a+(n-1\left)d\right]}{[2a^{{}^{\prime }}+(n-1\left)d^{{}^{\prime }}\right]}=\frac{(3n+2)}{(2n+3)}$

Let n = 23

$\Rightarrow \frac{2a+(23-1)d}{2a^{{}^{\prime }}+(23-1)d^{{}^{\prime }}}=\frac{3\times 23+2}{2\times 23+3}$

$\Rightarrow \frac{2a+22d}{2a^{{}^{\prime }}+22d^{{}^{\prime }}}=\frac{69+2}{46+3}$

$\Rightarrow \frac{2(a+11d)}{2(a^{{}^{\prime }}+11d^{{}^{\prime }})}=\frac{71}{49}$

$\therefore \frac{a_{12}}{a_{{12}^{{}^{\prime }}}}=\frac{71}{49}$

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