Progression

The first terms are

$a_1=5$ , $a_2=10$ and $a_3=15$

common difference, $d=15-10=10-5=5$

Computing for the last three terms, we have

$a_{100}=a_1+99d=5+99(5)=500$

$a_{99}=a_1+98d=5+98(5)=495$

$a_{98}=a_1+97d=5+97(5)=490$

Finally, the sum of the last three terms is

$500+495+490=$ $\boxed{1485}$

The $n^{th}$ term for this sequence is $a_n = 5n$ .

$a_{98} + a_{99} + a_{100}$

$= 5(98 + 99 + 100)$

$= 5(300 - 3)$

$= 1500 - 15$

$= 1485$