Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.

Asked by Pragya Singh | 1 year ago |  119

##### Solution :-

Consider a to be the first term and r to be the common ratio of the G.P.

Given, S2 = -4

Then, from the question we have

$$s_2=-4=\dfrac{a(1-r^n)}{1-r}$$......(1)

And,

a5 = 4 x a3

ar4 = 4ar2

r2 = 4

r = ± 2

Using the value of r in (1), we have

$$s_2=-4=\dfrac{a(1-(2)^2}{1-2}$$

$$-4=\dfrac{a(1-4)}{-1}$$

-4 = a(3)

$$a=\dfrac{-4}{3}$$

for r = -2

$$s_2=-4=\dfrac{a(1-(2)^2}{1-(-2)}$$

$$-4=\dfrac{a(1-4)}{1+2}$$

$$-4=\dfrac{a(-3)}{3}$$

= 4

Therefore, the required G.P is -4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……

Answered by Abhisek | 1 year ago

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