The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from t

1.	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.
Answer» Let the three numbers be a, ar, ar² According to the question a + ar + ar² = 56 … (1) Let us subtract 1,7,21 we get, (a – 1), (ar – 7), (ar² – 21) The above numbers are in AP. If three numbers are in AP, by the idea of the arithmetic mean, we can write 2b = a + c 2 (ar – 7) = a – 1 + ar² – 21 = (ar² + a) – 22 2ar – 14 = (56 – ar) – 22 2ar – 14 = 34 – ar 3ar = 48 ar = 48/3 ar = 16 a = 16/r …. (2) Now, substitute the value of a in equation (1) we get, (16 + 16r + 16r²)/r = 56 16 + 16r + 16r² = 56r 16r² – 40r + 16 = 0 2r² – 5r + 2 = 0 2r² – 4r – r + 2 = 0 2r(r – 2) – 1(r – 2) = 0 (r – 2) (2r – 1) = 0 r = 2 or 1/2 Substitute the value of r in equation (2) we get, a = 16/r = 16/2 or 16/(1/2) = 8 or 32 ∴ The three numbers are (a, ar, ar²) is (8, 16, 32)

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.

Answer»

Let the three numbers be a, ar, ar²

According to the question

a + ar + ar² = 56 … (1)

Let us subtract 1,7,21 we get,

(a – 1), (ar – 7), (ar² – 21)
The above numbers are in AP.

If three numbers are in AP, by the idea of the arithmetic mean, we can write 2b = a + c

2 (ar – 7) = a – 1 + ar² – 21

= (ar² + a) – 22

2ar – 14 = (56 – ar) – 22
2ar – 14 = 34 – ar

3ar = 48

ar = 48/3

ar = 16

a = 16/r …. (2)

Now, substitute the value of a in equation (1) we get,

(16 + 16r + 16r²)/r = 56

16 + 16r + 16r² = 56r

16r² – 40r + 16 = 0

2r² – 5r + 2 = 0

2r² – 4r – r + 2 = 0

2r(r – 2) – 1(r – 2) = 0

(r – 2) (2r – 1) = 0

r = 2 or 1/2

Substitute the value of r in equation (2) we get,

a = 16/r

= 16/2 or 16/(1/2)

= 8 or 32

∴ The three numbers are (a, ar, ar²) is (8, 16, 32)