Let’s consider the three numbers in G.P. to be as a, ar, and ar^{2}.

Then from the question, we have

a + ar + ar^{2} = 56

a (1 + r + r^{2}) = 56

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

Also, given

a – 1, ar – 7, ar^{2} – 21 forms an A.P.

So, (ar – 7) – (a – 1) = (ar^{2} – 21) – (ar – 7)

ar – a – 6 = ar^{2 }– ar – 14

ar^{2 }– 2ar + a = 8

ar^{2 }– ar – ar + a = 8

a(r^{2 }+ 1 – 2r) = 8

a (r – 1)^{2} = 8 … (2)

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

7(r^{2} – 2r + 1) = 1 + r + r^{2}

7r^{2} – 14 r + 7 – 1 – r – r^{2} = 0

6r^{2} – 15r + 6 = 0

6r^{2} – 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 agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).