If the sum of n, 2n and infinite terms of G.P. are `S_(1),S_(2)` and `S` respectively, then prove that `S_(1)(S_(1)-S)=S(S_(1)-S_(2)).`

Answer

N/a

Explanation

Let the first term and common ratio of G.P. be 'a' and 'r' respectively.
`therefore" "S_(1)=" sum of n terms"=(a(1-r^(n)))/(1-r)`
`S_(1)=" sum of 2n terms"=(a(1-r^(2n)))/(1-r)`
`S=" sum of infinite terms"=(a)/(1-r)`
`"Now, "S_(1)-S=(a(1-r^(n)))/(1-r)-(a)/(1-r)`
`=(a(1-r^(n)-1))/(1-r)-(ar^(n))/(1-r)" ...(1)"`
`S_(1)-S_(2)=(a(1-r^(n)))/(1-r)-(a(1-r^(2n)))/(1-r)`
`=(a(1-r^(n)-1+r^(2n)))/(1-r)`
`=(-ar^(n)(1-r^(n)))/(1-r)" ...(2)"`
`"L.H.S. "=S_(1)cdot(S_(1)-S)=(a(1-r^(n)))/(1-r){(-ar^(n))/(1-r)}" [From eq.(1)]"`
`=(a)/(1-r){(-ar^(n)(1-r^(n))}/(1-r)}`
`=Scdot(S_(1)-S_(2))" [From eq.(2)"`
`"= R. H. S."`