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

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Answer 1 · 2018-04-29T05:15:46

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.