Question

How to find sum of numbers divisible by 8 from 150-300?

Answer 1

the sequence is 152,160.....296
$tn = a + (n - 1)d \\ 296 = 152 + (n - 1) \times 8 \\ 296 - 152 = (n - 1)8 \\ 144 \div 8 = n - 1 \\ 18 + 1 = n \\ n = 19$
There are 19 terms.
$sn = \frac{n}{2} (t1 + tn)d \\ s19 = \frac{19}{2} (152 + 296) \ \\ \frac{19}{2} \times 448 \\ 4256$
Hope this helps you:-)))
don't forget to do it brainleist.