TS ICET 23 May 2019 Shift 1 Paper

Question : 129 of 200

Marks: +1, -0

If 6th term and 13th term of a geometric progression are 24 and

3/{16}

respectively, then the 25th term is

Solution:

let a and r be the first term and the common ratio of the given progression,

t_{6}=a r^{5}=24

→(1)

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

→ 2

\;{(2)}/{(1)}=\;{a r^{12}}/{a r^{5}}=\;{3 \/ 16}/{24}

r^{7}=\;{1}/{128} ⇒ r=\;{1}/{2}

∴ a r^{5}=24 ⇒ a=32 × 24

∴ t_{25}=a r^{24}

=32 × 24 ×(\;{1}/{2})^{24}

=\;{3}/{2^{16}}

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