Let a,ar,ar2,ar3,ar4,ar5 be six terms of a G.P. where ' a ' is first term and r is common ratio. According to given conditions, we have ar3−a=5⇒a(r3−1)=52 . . . (i) and a+ar+ar2=26 ⇒a(1+r+r2)=26 . . . (ii) To find: a(1+r+r2+r3+r4+r5) Consider a[1+r+r2+r3+r4+r5] =a[1+r+r2+r3(1+r+r2)] =a[1+r+r2][1+r3] . . . (iii) Divide (i) by (ii), we get
r3−1
1+r+r2
=2 we know r3−1=(r−1)(1+r+r2) ∴r−1=2⇒r=3 and a=2 ∴a(1+r+r2+r3+r4+r5)=a(1+r+r2)(1+r3) =2(1+3+9)(1+27)=26×28=728