If `S_1,S_2a n dS_3` be respectively the sum of n, 2n and 3n terms of a G.P., prove that `S_1(S_3-S_2)=S_1(S_2-S_1)^2`

Answer

N/a

Explanation

Let the first term and common ratio of G.P. be 'a' and 'r' respectively.
`S_(1)= "sum of n terms"=(a(1-r^(n)))/(1-r)" ...(1)"`
`S_(2)= "sum of 2n terms"=(a(1-r^(2n)))/(1-r)" ...(2)"`
`S_(3)= "sum of 3n terms"=(a(1-r^(3n)))/(1-r)" ...(3)"`
`"Now, "S_(2)-S_(1)=(a(1-r^(2n)))/(1-r)-(a(1-r^(n)))/(1-r)" [From eqs. (1) and (2)]"`
`=(a[1-r^(2n)-1+r^(n)])/(1-r)`
`=(ar^n(1-r^(n)))/(1-r)`
`"and L.H.S."=(S_(2)-S_(1))^(2)=(a^(2)cdotr^(2n)(1-r^(n))^(2))/(1-r)^(2)" ...(4)"`
`"Again "S_(3)-S_(2)=(a(1-r^(3n)))/(1-r)-(a(1-r^(2n)))/(1-r)" [From eqs. (2) and (3)]"`
`=(a(1-r^(3n)-1+r^(2n)))/(1-r)`
`=(acdotr^(2n)(1-r^(n)))/(1-r)`
`"and R.H.S."=S_(1)cdot(S_(3)-S_(2))`
`=(a(1-r^(n)))/(1-r).(ar^(2n)(1-r^(n)))/(1-r)`
`=(a^(2)cdotr^(2n)cdot(1-r^(n))^(2))/((1-r)^(2))`
`therefore" L.H.S. = R.H.S."`