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Question

3,1(3),13(3).....
The 19th term of the above sequence is .

A
3318
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B
3152
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C
37326
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Solution

The correct option is A 3318
In the given sequence,
t2t1=133=13.
Common ratio, r=13
and
first term, a=3
Now, a19=ar191.
=3×(13)18=3318

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