The difference between the fourth term and the first term of a Geometrical Progresssion is 52. If the sum of its first three terms is 26 , then the sum of the first six terms of the progression is

Step-by-step solution

Let a, a r, a r^2, a r^3, a r^4, a r^5 be six terms of a G.P. where ' a ' is first term and r is common ratio. According to given conditions, we have a r^3-a=5 a (r^3-1 )=52 and a+a r+a r^2=26 a (1+r+r^2 )=26 To find: a (1+r+r^2+r^3+r^4+r^5 ) Consider aligned & a [1+r+r^2+r^3+r^4+r^5 ] & =a [1+r+r^2+r^3 (1+r+r^2 ) ] & =a [1+r+r^2 ] [1+r^3 ] aligned Divide (1) by (2), we get r^3-1 1+r+r^2 =2 we know r^3-1=(r-1) (1+r+r^2 ) aligned & r-1=2 r=3 and a=2 & a (1+r+r^2+r^3+r^4+r^5 ) & =a (1+r+r^2 ) (1+r^3 ) & =2(1+3+9)(1+2