Q.

The largest terms of the sequence 1503,4524,9581,16692,… is

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a

16692

b

4524

c

491529

d

641529

answer is C.

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Detailed Solution

Tn=n2500+3n3=f(n)f1(n)=n1000−3n3500+3n32 For the maxima or minf1(n)=0n=100031/3 Now 6<100031/3<7∴T7 is the largest term.  the largest term of the sequence =49500+1029=491529
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