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
a3 = 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 + 10r2 = 39r
10r2 – 29r + 10 = 0
10r2 – 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) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).