Class 11 Mathematics - Chapter Sequence and Series NCERT Solutions | The sum of three numbers in G.P. is 56.

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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 obtain an arithmetic progression. Find the numbers.

Answer

Let the three numbers in G.P. be aar, and ar2.

From the given condition, a + ar + ar2 = 56

⇒ a (1 + r + r2) = 56

rightwards double arrow a space equals fraction numerator 56 over denominator 1 plus r plus r squared end fraction space space space space... space left parenthesis 1 right parenthesis

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

⇒ (r – 1)2 = 8 … (2)

rightwards double arrow fraction numerator 56 over denominator 1 plus r plus r squared end fraction open parentheses r minus 1 close parentheses squared equals 8 space space space space space space space space space space open square brackets U sin g space left parenthesis 1 right parenthesis close square brackets

⇒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

therefore space r space equals space 2 comma 1 half

When r = 2, a = 8

When r space equals space 1 half space comma space a space equals space 32

Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.

When r space equals space 1 half, 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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