If a , b , c are positive integers forming the first three terms of an increasing geometric progression with common ratio r , where b − a is a perfect cube and lo g 6 a + lo g 6 b + lo g 6 c = 6

Question

If a , b , c are positive integers forming the first three terms of an increasing geometric progression with common ratio r , where b − a is a perfect cube and lo g 6 ​ a + lo g 6 ​ b + lo g 6 ​ c = 6 . Let S denote the sum of 10 terms of another G.P. whose first term is a and common ratio is 2 5 r ​ . Then

TestHub · Accepted Answer

Let a = r α ​ , b = α , c = α r . Then ab c = α 3 = 6 6 ⇒ α = 36 , so b = 36 , a = r 36 ​ , c = 36 r . b − a = 36 ⋅ r r − 1 ​ must be a perfect cube, with a , c positive integers and r > 1 . Checking the possible ratios, only r = 4 works ( b − a = 27 ) . So a = 9 , b = 36 , c = 144 and a + b + c = 189 . S has first term 9 and ratio 2 5 ⋅ 4 ​ = 10 : S = 9 ⋅ 9 1 0 10 − 1 ​ = 1 0 10 − 1 , which is ten 9s, so the digit sum is 90.

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