The sum of first three terms of a G.P. is \( \dfrac{39}{10}\) and their product is 1. Find the common ratio and the terms.

Asked by Pragya Singh | 1 year ago |  85

1 Answer

Solution :-

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 ago

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