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

##### 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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