TS ICET 23 May 2019 Shift 2 Paper

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Question : 128 of 200

Marks: +1, -0

The sum of first three terms of a geometric progression is

{13}/{12}

and their product is - 1. Then the fourth term is

-16/9
-8/9
-8/3
-4/3

Solution:

Let

a, ar, ar^2

be the three terms.
Given

a(1+r+r^2)

and

=13/12

→(1)

(ar)^3=-1

ar = –1

→ (2)

\;{(1)}/{(2)} ⇒ 1+r+r^{2}=\;{-13 r}/{12}

⇒ 12 r^{2}+25 r+12=0

⇒(3 r+4)(4 r+3)=0

r=\;{-4}/{3}

or

{-3}/{4}

t_{4}=a r^{3}=ar(r^{2})=-1(\;{-4}/{3})^{2}

=\;{-16}/{9}

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