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Question

If in the expansion of (21/3+131/3)n, the ratio of 6th term from beginning and from the end is 1/6, then the value of n is

A
5
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B
7
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C
13
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D
None of these
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Solution

The correct option is C 13
As we know that in an expansion (a+b)n, the general term is given as-
Tr+1=nCranbnr
In the given expansion-
(21/3+131/3)n
a=21/3,b=131/3,n=?
Therefore,
nC5(21/3)5(131/3)n5nCn5(21/3)n5(131/3)5=16(Given)
1(21/3)n10(31/3)n10=16
1(6)n103=1(6)1
n103=1
n10=3
n=10+3=13

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