The ratio of the sums of $m$ and $n$ terms of an A.P. is $m^{2} : n^{2}$ . Show that the ratio of $m^{t h}$ and $n^{t h}$ term is $(2 m - 1) (2 n - 1)$ .

Open in App

Solution

$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}} \frac{\frac{m}{2} (2 a_{1} + (m - 1) d_{1})}{\frac{n}{2} (2 a_{2} + (n - 1) d_{2})} = \frac{m^{2}}{n^{2}} \frac{2 a_{1} + (m - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{m}{n} \frac{a_{1} + \frac{(m - 1) d_{1}}{2}}{a_{2} + \frac{(n - 1) d_{2}}{2}} = \frac{m}{n}$
Put $m = 2 m - 1$ and $n = 2 n - 1$
$\frac{a_{1} + (m - 1) d_{1}}{a_{2} + (n - 1) d_{2}} = \frac{2 m - 1}{2 n - 1} \frac{a_{m}}{a_{n}} = \frac{2 m - 1}{2 n - 1}$
Hence proved

Suggest Corrections

Similar questions

The ratio of the sums of m and n terms of an A.P. is $m^{2}$ : $n^{2}$ . Show that the ratio of the $m^{t h}$ and $n^{t h}$ term is $(2 m - 1) : (2 n - 1)$ . [4 MARKS]

The ratio of the sum of the m and n terms of on A.P. is $m^{2}$ : $n^{2}$ . Show that the ratio of $m^{t h}$ and $n^{t h}$ term is (2m - 1) : (2n - 1).