Question

What is the sum of all natural numbers between 100 and 500 that are exactly divisible by 3?

Answer 1

The natural numbers between 100 and 500 that are exactly divisible by 3 are 102,105,......495,498.
a=102
d=3
tn=a+(n-1)d
498=102+(n-1)3
$498 - 102 = 3n - 3 \\ 396 = 3n - 3 \\ 399 = 3n \\ n = \frac{399}{3} \\ n = 133$
therefore are 133 terms.
$sn = \frac{n}{2} (t1 + tn) \\ \frac{133}{2} (102 + 498) \\ \frac{133}{2} \times 600 \\ 133 \times 300 \\ 39900$
the sum of natural numbers between 100 and 500 divisible by 3 is 39900.
hope this helps you:-))))
don't forget to do it brainleist.