Question

What is the sum of all integers which are multiples of 7 and are between 100 and 1000?

Answer 1

the sum is
105+112+...994
now this an A.P with a=105 d=7 last term l=994
 no. of terms n=127
so it sum is $S= \frac{n(a+l)}{2}= \frac{128(105+994)}{2}=564886$

Answer 2

First, find the sum of all numbers from 100 to 1000: ==> (7+1)*1000= 8000  Sum of all the numbers divisible by 7: ==> 564886