# If m,n,s,t are in G.P , then 1/m,1/n,1/s,1/t , are in ?

Apr 29, 2018

Let the common ratio of the GP in Question be $r$.

Then, $n \div m = r , s \div n = r , t \div s = r \ldots \ldots \ldots \ldots \left(\ast\right)$.

Now, $\frac{1}{n} \div \frac{1}{m} = \frac{m}{n} = \frac{1}{r} \ldots \ldots \ldots \ldots \left[\because , \left(\ast\right)\right] \ldots \ldots \left({\ast}^{1}\right)$.

$\frac{1}{s} \div \frac{1}{n} = \frac{n}{s} = \frac{1}{r} \mathmr{and} \frac{1}{t} \div \frac{1}{s} = \frac{s}{t} = \frac{1}{r} \ldots \ldots \left({\ast}^{2}\right)$.

From $\left({\ast}^{1}\right) \mathmr{and} \left({\ast}^{2}\right)$, we find,

$\frac{1}{n} \div \frac{1}{m} = \frac{1}{s} \div \frac{1}{n} = \frac{1}{t} \div \frac{1}{s} = \frac{1}{r}$.

This proves the assertion.