TS ICET 23 May 2018 Shift 2 Paper

Question : 129 of 200

Marks: +1, -0

If the

6^{\th}

and

13^{\th}

terms of a geometric progression are 24 and

3/16

respectively. then its

25^{\th

term is

Solution:

Let a and r be the first term and common ratio of the geometric progression.
Given

t _{6}= ar ^{5}=24

→(1)

t_{13}=a r^{12}={3}/{16}

→ (2)

\;{(2)}/{(1)}=r^{7}={\;{3}/{16}}/{24}={1}/{16 × 8}

r^{7}={1}/{2^{7}} ⇒ r={1}/{2}

a r^{5}=24 ⇒ a={24}/{r^{5}}={24}/{({1}/{2})^{5}}=(24)(2)^{5}

t_{25}=a r^{24}=(24)(2)^{5} × \;{1}/{2^{24}}

=3 × 2^{3} × 2^{5} × \;{1}/{2^{24}}

=3 × \;{1}/{2^{16}}=3 × 2^{-16}

Go to Question:

Prev Question

Next Question