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Question

If k ϵ R+ and the middle term of (k2+2)8 is 1120, then value of k is

A
3
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B
2
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C
1
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D
4
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Solution

The correct option is B 2
KϵR+

The general term, (r+1)th term of the expansion
(K2+2)8 is Tr+1=8Cr(K2)8r2r

In the above expansion there are 9 terms, hence the middle term will be the 5th term (r=4)

T5=8C4(K2)8424

=8!4!4!K424.24

=8×7×6×54×3×2K4

=70K4

Given,

70K4=1120

K4=112/7

K4=16

K4=24

K=2

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