Thank you for reporting, we will resolve it shortly
Q.
Which term of the series $ 3+8+13+18+.... $ is 498:
Bihar CECEBihar CECE 2002
Solution:
If the difference of the consecutive terms is constant, then the series is in AP.
Given series is $ 3\text{ }+\text{ }8\text{ }+\text{ }13\text{ }+\text{ }18\text{ }+....+\text{ }498 $
Here, $ a=3,\,d=5,\,\,l=498 $
As we know
$ {{t}_{n}}=l=a+(n-1)d $
$ \Rightarrow $ $ 498=3+(n-1)\,5 $
$ \Rightarrow $ $ (n-1)=\frac{495}{5} $
$ \Rightarrow $ $ n-1=99$
$\Rightarrow n=100 $