The sum of three numbers in G.P. is 14. If the first two terms are eac

1.	The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.
Answer» Let the three numbers be \(\frac{a}{r}, a, ar.\) ⇒ \(\frac{a}{r}\) + a + ar = 14 ⇒ a + ar + ar² = 14r…(1) First two terms are increased by 1, and third decreased by 1 \(\therefore\) \(\frac{a}{r}\) + 1, a + 1, ar -1 The above sequence is in AP. We know in AP. 2b = a + c \(\therefore\) 2(a + 1) = ar - 1 + \(\frac{a}{r} + 1\) ⇒ 2a + 2 = \(\frac{ar^2 + a}{r}\) ⇒ 2ar + 2r = ar² + a ⇒ ar² – 2ar + a = 2r …(2) Dividing 1 by 2 we get ⇒ \(\frac{a + ar + ar^2}{ar^2 - 2ar + a}\) = \(\frac{14r}{2r}\) ⇒ \(\frac{1+r+r^2}{r^2 - 2r + 1} = 7\) ⇒ 1 + r + r² = 7r²– 14r + 7 ⇒ 6r² – 15r – 6 = 0 ⇒ 6r² – 12r – 3r – 6 = 0 ⇒ 6r(r – 2) – 3(r – 2) = 0 ⇒ (6r – 3) (r – 2) = 0 ⇒ r = 2 or r = 1/2. Substituting r = 2 in 2 we get ⇒ a(2)² – 2a(2) + a = 2(2) ⇒ 4a – 4a + a = 4 ⇒ a = 4 Substituting r = 1/2 in 2 we get ⇒ a(1/2)² – 2a(1/2) + a = 2(1/2) ⇒ a = 4 ∴ substituting a and r we get the numbers as 2,4,8.

The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.

Answer»

Let the three numbers be \(\frac{a}{r}, a, ar.\)

⇒ \(\frac{a}{r}\) + a + ar = 14

⇒ a + ar + ar² = 14r…(1)

First two terms are increased by 1, and third decreased by 1

\(\therefore\) \(\frac{a}{r}\) + 1, a + 1, ar -1

The above sequence is in AP.

We know in AP.

2b = a + c

\(\therefore\) 2(a + 1) = ar - 1 + \(\frac{a}{r} + 1\)

⇒ 2a + 2 = \(\frac{ar^2 + a}{r}\)

⇒ 2ar + 2r = ar² + a

⇒ ar² – 2ar + a = 2r …(2)

Dividing 1 by 2 we get

⇒ \(\frac{a + ar + ar^2}{ar^2 - 2ar + a}\) = \(\frac{14r}{2r}\)

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

⇒ 1 + r + r² = 7r²– 14r + 7

⇒ 6r² – 15r – 6 = 0

⇒ 6r² – 12r – 3r – 6 = 0

⇒ 6r(r – 2) – 3(r – 2) = 0

⇒ (6r – 3) (r – 2) = 0

⇒ r = 2 or r = 1/2.

Substituting r = 2 in 2 we get

⇒ a(2)² – 2a(2) + a = 2(2)

⇒ 4a – 4a + a = 4

⇒ a = 4

Substituting r = 1/2 in 2 we get

⇒ a(1/2)² – 2a(1/2) + a = 2(1/2)

⇒ a = 4

∴ substituting a and r we get the numbers as 2,4,8.