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Question

If sum of n terms of a sequence is given by Sn=3n25n+7 and tr represents its rth term, then

A
t7=34
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B
t2=7
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C
t10=34
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D
t8=40
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Solution

The correct option is A t7=34
Sn=3n2+75n
tn=SnSn1
=(3n25n+7)[3(n1)25(n1)+7]
=3n25n+7[3(n22n+1)5n+5+7]
=3n25n+73n2+6n3+5n12
=6n+7312
tn=6n+715
tn=6n8
so,
t2=6×28=4
t7=6×78=428=34
t8=6×88=488=40
t10=6×108=608=52
thus,
t7=34 and t8=40 are correct options.

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