Sum-of-three-numbers-in-GP-be-14-If-one-is-added-to-first-and-second-and-1-is-subtracted-from-the-third-the-new-numbers-are-in-AP-The-smallest-of-them-is-

Question Number 12827 by 786786AM last updated on 03/May/17

Sum of three numbers in GP be 14 . If one is

added to first and second and 1 is subtracted

from the third, the new numbers are in AP .

The smallest of them is

Answered by mrW1 last updated on 04/May/17

a_{1} = a

a_{2} = a q

a_{3} = {a q}^{2}

a_{1} + a_{2} + a_{3} = 14

a (1 + q + q^{2}) = 14

(a_{2} + 1) - (a_{1} + 1) = (a_{3} - 1) - (a_{2} + 1)

2 (a_{2} + 1) = (a_{1} + 1) + (a_{3} - 1)

2 (a q + 1) = a + 1 + {a q}^{2} - 1

{a q}^{2} - 2 a q + a - 2 = 0

{a q}^{2} + a q + a - 3 a q - 2 = 0

14 - 3 a q - 2 = 0

a q = 4

a + a q + {a q}^{2} = a + 4 + \frac{16}{a} = 14

a + \frac{16}{a} = 10

a^{2} - 10 a + 16 = 0

a = \frac{10 \pm \sqrt{100 - 4 \times 16}}{2} = \frac{10 \pm 6}{2} = 8 o r 2

q = \frac{1}{2} o r 2

t h e n u m b e r s a r e 2, 4, 8 o r 8, 4, 2

\Rightarrow t h e s m a l l e s t o f t h e m i s 2 .

Answered by sandy_suhendra last updated on 04/May/17

the other way, we can start from AP

AP : (a - b), a, (a + b)

GP : (a - b - 1), (a - 1), (a + b + 1)

(a - b - 1) + (a - 1) + (a + b + 1) = 14

3 a - 1 = 14

3 a = 15 \Rightarrow a = 5

GP : (4 - b), 4, (6 + b)

(4 - b) (6 + b) = 4^{2}

24 - 2 b - b^{2} = 16

b^{2} + 2 b - 8 = 0

(b + 4) (b - 2) = 0

b = - 4 \Rightarrow 8, 4, 2

b = 2 \Rightarrow 2, 4, 8