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.

Asked by Sakshi | 1 year ago |  30

##### Solution :-

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 = $$\dfrac{48}{3}$$

ar = 16

a = $$\dfrac{16}{r}$$ …. (2)

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

$$\dfrac{ (16 + 16r + 16r^2)}{r}$$ = 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 $$\dfrac{1}{2}$$

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

a = $$\dfrac{16}{r}$$

= $$\dfrac{16}{2}$$

= 8 or 32

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

Answered by Aaryan | 1 year ago

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