Question

find the sum of all three digits no., which are divisible by 7.

Answer 1

The list of 3 digit numbers divisible by 7 are :

105, 112, ........994.

The above list forms an AP with first term, a = 105 and common difference, d = 7

Let an = 994

so,  a + (n - 1)d = 994

 ⇒ 105 + (n - 1)7 = 994

 ⇒ n = 128

now, sum of all the 3 digit numbers divisible by 7 is given by,

S128 = (128/2) [2×105 +(128 - 1)7]

 = 64 [210 + 889] = 64 × 1099 = 70336

Answer 2

you can solve this by using AP..
find out the first number and last number divisible by 7 in 3 digit number..
so, the first number is 105
and the last number is 994...
now, by using the formula of AP we will solve it...
here, a=105
d=7
then, for n th term
an = a+(n-1)d
=>994=105+(n-1)7
=>889=(n-1)7
=>127=n-1
=>n=128
so, by using summation formula..
sum = n/2 (a+l)
where l is the last term..
so, sum =128/2 (105+994)
= 70336
is the answer...