# 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 ar^{2} .

From the given condition,

a + ar + ar^{2} = 56

⇒ a (1 + r + r^{2}) = 56 ....(1)

Now, (a - 1), (ar - 7), (ar^{2} - 21) forms an A.P.

Therefore,

(ar - 7) - (a - 1) = (ar^{2} - 21) - (ar - 7)

⇒ ar - a - 6 = ar^{2} - ar - 14

⇒ ar^{2} - ar - ar + a = 8

⇒ ar^{2} - 2ar + a = 8

⇒ a (r ^{2} + 1- 2r ) = 8

⇒ 7a (r ^{2} - 2r + 1) = 56 ....(2)

From (1) and (2) , we get

⇒ 7 (r^{2} - 2r + 1) = (1 + r + r ^{2})

⇒ 7r^{2} - 14r + 7 - 1 - r - r ^{2} = 0

⇒ 6r^{2} - 15r + 6 = 0

⇒ 2r^{2} - 5r + 2 = 0

⇒ 2r^{2} - 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 + 2^{2}) = 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

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