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Question

Let the rth term Tr of a series is given by Tr=r1+r2+r4, then limnnr=1Tr equals?

A
1
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B
14
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C
12
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D
none of these
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Solution

The correct option is A 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, limnnr=1Tr=limn12(11n(n+1)+1)=12
Hence, option 'C' is correct.

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