Let \( \dfrac{a}{r}\), a, ar be the first three terms of the G.P.

\( \dfrac{a}{r}\)+ a + ar = \( \dfrac{39}{10}\) …… (1)

(\( \dfrac{a}{r}\)) (a) (ar) = 1 …….. (2)

From (2), we have

a^{3} = 1

Hence, a = 1 [Considering real roots only]

Substituting the value of a in (1), we get

\( \dfrac{1}{r}\) + 1 + r = \( \dfrac{39}{10}\)

\( \dfrac{(1 + r + r^2)}{r}\) = \( \dfrac{39}{10}\)

10 + 10r + 10r^{2} = 39r

10r^{2} – 29r + 10 = 0

10r^{2} – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(5r – 2) (2r – 5) = 0

Thus,

r = \( \dfrac{2}{5}\) or \( \dfrac{5}{2}\)

Therefore, the three terms of the G.P. are \( \dfrac{5}{2}\), 1 and \( \dfrac{2}{5}\).

Answered by Abhisek | 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 \( \dfrac{81}{2}\).