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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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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.

Sum
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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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Chapter 8: Sequences and Series - EXERCISE 8.2 [Page 146]

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NCERT Mathematics [English] Class 11
Chapter 8 Sequences and Series
EXERCISE 8.2 | Q 16. | Page 146

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