Question

The 99th term in the series 3,5,7,9,11….. is

Answer 1

$a=3\\d=2\\so, tn=a+(n-1)d\\t99= a+(98)d\\\\ t99=3+(98)2\\t99=3+196\\=199$

Answer 2

Answer:

201

Step-by-step explanation:

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