Question

find the sum of integers between 100 and 200 that are divisble by 9

Answer 1

Answer:

Step-by-step explanation:

a=100

d=9

l=200

$n=(\frac{l-a}{d} )+1$

$=(\frac{200-100}{9})+1$

$=\frac{100+9}{9}$

$s_{n}=\frac{n}{2}(a+l)$$s_n=\frac{n}{2} [2a+(n-1)d]$