If 17th and 18th terms in the expansion of (2+a)50 are equal, then the value of a is:
A
0
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B
1
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C
2
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D
3
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Solution
The correct option is B1 General term Tr+1=50Cr×2(50−r)×ar
Now, T17=T18⇒T16+1=T17+1⇒50C16×234×a16=50C17×233×a17⇒a=2×50C1650C17⇒a=2×17(50−17+1)=1[∵nCr−1nCr=rn−r+1]