Question

Find the sum of the integers between $100$ and $200$ that are divisible by $6.$

Answer 1

Integers between $100$ and $200$ divisible by $6$ are-

$102,108,114,...198$

Above series is in $A.P.$ with first term $102$ and common difference $6$.

$a=102$

$d=6$

Last term $l=198$

$t_n=a+(n-1)d198=102+(n-1)6(n-1)6=198-102=96(n-1)6=96$

$n-1=16⟹n=17\therefore \text{Sum}=\frac{n}{2}[a+l]=\frac{17}{2}[102+198]=\frac{17}{2}\times 300=150\times 17=2550$