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 arithmetic progression. Find the numbers.

Asked by Abhisek | 1 year ago |  64

##### Solution :-

Let’s consider the three numbers in G.P. to be as a, ar, and ar2.

Then from the question, we have

a + ar + ar2 = 56

a (1 + r + r2) = 56

$$a=\dfrac{56}{1+r+r^2}$$.........(1)

Also, given

a – 1, ar – 7, ar2 – 21 forms an A.P.

So, (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)

ar – a – 6 = ar– ar – 14

ar– 2ar + a = 8

ar– ar – ar + a = 8

a(r+ 1 – 2r) = 8

a (r – 1)2 = 8 … (2)

$$\dfrac{56}{1+r+r^2}(r-1)^2=8$$

7(r2 – 2r + 1) = 1 + r + r2

7r2 – 14 r + 7 – 1 – r – r2 = 0

6r2 – 15r + 6 = 0

6r2 – 12r – 3r + 6 = 0

6r (r – 2) – 3 (r – 2) = 0

(6r – 3) (r – 2) = 0

r = 2, $$\dfrac{1}{2}$$

When r = 2, a = 8

When r = $$\dfrac{1}{2}$$, a = 32

Thus,

When r = 2, the three numbers in G.P. are 8, 16, and 32.

When r = $$\dfrac{1}{2}$$, the three numbers in G.P. are 32, 16, and 8.

Therefore in either case, the required three numbers are 8, 16, and 32.

Answered by Pragya Singh | 1 year ago

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