CAT 2020 Exam Shift 2 Solved Paper

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Question : 55 of 76

Marks: +1, -0

Let the

{m}

-th and

{n}

-th terms of a geometric progression be

\;{3}/{4}

and 12 , respectively, where $m

$-2$
2
6
$-4$

Solution:

T_{n}=12

T_{m}=3 \/ 4

\;{T_{n}}/{T_{m}}=\;{a r^{n-1}}/{a r^{m-1}}=\;{12}/{{3}/{4}}

r^{n-m}=16=(± 2)^{4}=(± 4)^{2}

To get the minimum value for

r+n-m, r

should be minimum.

∴ r=-4

n-m=2

∴

Required answer

=-2

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