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Question
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
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Solution
Let a be the first term and r be the common ratio of the G.P. According to the given conditions,
S2 = `-4 = ("a"(1 - "r"^2))/(1 - "r")` ........(i)
a5 = 4 × a3
⇒ ar4 = 4ar2 ⇒ r2 = 4
∴ r = ± 2
From (i) we obtain
-4 = `("a"[1 - (2)^2])/(1 - 2)` for r = 2
⇒ `-4 = ("a"(1 - 4))/-1`
⇒ −4 = a(3)
⇒ a = `(-4)/3`
Also, −4 = `("a"[1 - (-2)^2])/(1 - (-2))` for r = −2
⇒ `-4 = ("a"(1 - 4))/(1 + 2)`
⇒ `-4 = ("a"(-3))/3`
⇒ a = 4
Thus, the required G. P. is `(-4)/3, (-8)/3, (-16)/3` ,.... or 4, −8, 16, −32 ........
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