TS ICET 23 May 2018 Shift 1 Paper

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Question : 130 of 200

Marks: +1, -0

If the

21^{\st

and

22^{\nd

terms in the expansion of

(1 + x) ^44

are equal then x =

$1/7$
$1/8$
$7/8$
$1/56$

Solution:

Given 21st term = 22nd term

⇒(^{44} C _{20}) ×( x ^{20})=(^{44} C _{21}) ×( x ^{21})

x=21/24=7/8

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