The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Let the three numbers in G.P. be a, ar, and ar2.
From the given condition, a + ar + ar2 = 56
⇒ a (1 + r + r2) = 56
a – 1, ar – 7, ar2 – 21 forms an A.P.
∴(ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)
⇒ ar – a – 6 = ar2 – ar – 14
⇒ar2 – 2ar + a = 8
⇒ar2 – ar – ar + a = 8
⇒a(r2 + 1 – 2r) = 8
⇒ a (r – 1)2 = 8 … (2)
⇒7(r2 – 2r + 1) = 1 + r + r2
⇒7r2 – 14 r + 7 – 1 – r – r2 = 0
⇒ 6r2 – 15r + 6 = 0
⇒ 6r2 – 12r – 3r + 6 = 0
⇒ 6r (r – 2) – 3 (r – 2) = 0
⇒ (6r – 3) (r – 2) = 0
When r = 2, a = 8
When 
Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.
When 
, the three numbers in G.P. are 32, 16, and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.
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Welcome to the NCERT Solutions for Class 11 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 5 , Question 10: The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we....
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