Let three positive numbers a , b , c are in geometric progression, such that a , b + 8 , c are in arithmetic progression and a , b + 8 , c + 64 are in geometric progression. If the arithmetic mean of a , b , c is k , then 3 13 k is equal to

Question

Let three positive numbers a , b , c are in geometric progression, such that a , b + 8 , c are in arithmetic progression and a , b + 8 , c + 64 are in geometric progression. If the arithmetic mean of a , b , c is k , then 3 13 k is equal to,

MARKS App · Accepted Answer

Here, b 2 = a c , 2 b + 8 = a + c , b + 8 2 = a c + 64 ⇒ b 2 + 64 + 16 b = a c + 64 a ⇒ b + 4 = 4 a ⇒ b = 4 a - 4 2 b + 8 = a + c ⇒ 2 4 a + 4 = a + c ⇒ c = 7 a + 8 b 2 = a c ⇒ 4 a - 4 2 = a 7 a + 8 ⇒ 16 a 2 + 16 - 32 a = 7 a 2 + 8 a ⇒ 9 a 2 - 40 a + 16 = 0 ⇒ 9 a - 4 a - 4 = 0 ⇒ a = 4 9 , 4 But for a = 4 9 ⇒ b 0 So, a = 4 ⇒ b = 12 and c = 36 ⇒ k = a + b + c 3 = 52 3 ⇒ 3 k 13 = 3 13 × 52 3 = 4

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