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Question

Let the rth term, tr, of a series is given by tr=r1+r2+r4. Then limnnr=1tr is

A
14
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B
1
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C
12
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D
none of these
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Solution

The correct option is C 12
As Tr=r1+r2+r4=122r1+r2+r4=12(1r2r+11r2+r+1)=12(1r(r1)+11r(r+1)+1)
Then
nr=1Tr=nr=112(1r(r1)+11r(r+1)+1)=12(11n(n+1)+1)
Therefore, Ltnnr=1Tr=Ltn12(11n(n+1)+1)=12
Hence, option 'C' is correct.

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