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 A.P. Find the numbers

Solution:

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 ....(1)

Now, (a - 1), (ar - 7), (ar² - 21) forms an A.P.

Therefore,

(ar - 7) - (a - 1) = (ar² - 21) - (ar - 7)

⇒ ar - a - 6 = ar² - ar - 14

⇒ ar² - ar - ar + a = 8

⇒ ar² - 2ar + a = 8

⇒ a (r ² + 1- 2r ) = 8

⇒ 7a (r ² - 2r + 1) = 56 ....(2)

From (1) and (2) , we get

⇒ 7 (r² - 2r + 1) = (1 + r + r ²)

⇒ 7r² - 14r + 7 - 1 - r - r ² = 0

⇒ 6r² - 15r + 6 = 0

⇒ 2r² - 5r + 2 = 0

⇒ 2r² - 4r - r + 2 = 0

⇒ 2r (r - 2) - 1(r - 2) = 0

⇒ (2r - 1)(r - 2) = 0

⇒ r = 1/2, 2

Case I:

Substituting r = 2 in (1) , we obtain

a (1 + 2 + 2²) = 56

⇒ 7a = 56

⇒ a = 8

Hence, the three numbers are in G.P are 8, 16 and 32.

Case II:

Substituting r = 1/2 in (1) , we obtain

a (1 + 1/2 + (1/2)²) = 56

⇒ 7/4 a = 56

⇒ a = 32

Hence, the three numbers are in G.P are 32, 16 and 8.

Thus, the three required numbers are 8, 16 and 32

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NCERT Solutions Class 11 Maths Chapter 9 Exercise ME 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 A.P. Find the numbers.

Summary:

If the sum of three numbers in G.P is 56 and if we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P, then the numbers are 8, 16 and 32