Question

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P 2 R n = S n

Answer 1

Given that S be the sum, P be the product and R be the reciprocal of n term in G.P.

Let the G.P. be a,ar,a r 2 ,a r 3 ,..., a r n−1 .

According to given condition,

S n = a( r n −1 ) r−1 P= a n × r 1+2+....+n−1

The sum of first n natural numbers is n( n+1 ) 2 .

Now, substitute the value of sum of n natural numbers in above equation, we get

P= a n r n( n+1 ) 2

Now, for the value of R.

R= 1 a + 1 ar +....+ 1 a r n−1 = r n−1 + r n−2 +....+r+1 a r n−1 = 1( r n −1 ) ( r−1 ) × 1 a r n−1 = r n −1 a r n−1 ( r−1 )

Where 1,r,....,  r n−1  forms a G.P.

We have to prove that P 2 R n = S n .

Now, substitute the value of P and R in L.H.S.

P 2 R n = a 2n r n( n−1 ) × ( r n −1 ) n a n r n( n−1 ) ( r−1 ) n = a n ( r n −1 ) n ( r n −1 ) n = [ a( r n −1 ) ( r n −1 ) ] n = S n

Hence, it is proved.