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

**Solution:**

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

According to the given conditions,

a_{5} = 4 (a_{3})

ar^{4} = 4ar ^{2}

r ^{2} = 4

r = ± 2

Also,

S_{2} = a (1 - r^{2})/(1 - r)

Case I : for r = 2

⇒ a (1 - 2^{2})/(1 - 2)

⇒ a (1 - 4)/(- 1) = - 4

⇒ 3a - 4

⇒ a = - 4/3

Case II: for r = - 2

⇒ a (1 - (- 2)^{2})/(1 - (- 2))

⇒ a (1 - 4)/(1 + 2) = - 4

⇒ - 3a/3 = - 4

⇒ a = 4

Thus, the required G.P is - 4/3, - 8/3, - 16/3, .... or 4, - 8, 16, ....

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 16

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

**Summary:**

The sum of the first two terms of the series was given as -4 and given the conditions, the G.P comes out to be 4, - 8, 16.... or - 4/3, - 8/3, - 16/3

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