Question

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

Answer 1

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

Then, `n-:m=r, s-:n=r, t-:s=r............(ast)`.

Now, `1/n-:1/m=m/n=1/r............[because, (ast)]......(ast^1)`.

`1/s-:1/n=n/s=1/r and 1/t-:1/s=s/t=1/r......(ast^2)`.

From `(ast^1) and (ast^2)`, we find,

`1/n-:1/m=1/s-:1/n=1/t-:1/s=1/r`.

This proves the assertion.