Let a ,   b ,   c and d are in a geometric progression such that a < b < c < d ,   a + d = 112 and b + c = 48 . If the geometric progression is continued with a as the first term, then the sum of the first six terms is

Step-by-step solution

Let r be the common ratio a + d = 112 ⇒ a + a r 3 = 112 b + c = 48 ⇒ a r + a r 2 = 48 Dividing the first equation by the second equation we get a 1 + r 3 a 1 + r r = 112 48 = 1 + r 1 - r + r 2 r r + 1 = 7 3 ⇒ 3 r 2 - 3 r + 3 = 7 r ⇒ 3 r 2 - 10 r + 3 = 0 r = 3 or 1 3 Given they are in ascending order ⇒ r = 3 r = 3 ⇒ a = 112 1 + r 3 = 112 28 = 4 ⇒ a = 4 , b = 12 , c = 36 , d = 108 ⇒ Sum of first 6 terms = 4 ( 3 6 - 1 ) ( 3 - 1 ) = 1456