multiples

ruplun wrote:Find the sum of all the integers which are multiples of 7 and lie between 200 and 400

200/7 gives the remainder = 4, which means the smallest integer between 200 and 400, divisible by 7 = 203.
Similarly, 400/7 gives a remainder of 1, which means that the largest integer between 200 and 400, divisible by 7 = 399.

So, a series can be formed from this: 203, 210, ..., 399, which is obviously an arithmetic sequence.
Here, the first term, a = 203, common difference, d = 7
nth term of an arithmetic sequence, a(n) = a + (n - 1)d
Since 399, is the last term, so by the above formula, 399 = 203 + (n - 1)7
Solving the above equation we get, n = 29

Now, sum of the integers in an arithmetic sequence = (n/2)[a + a(n)] = (29/2)[203 + 399] = 8729

ArpanaAmishi wrote:Hi Anurag,

How did you drive this

'200/7 gives the remainder = 4, which means the smallest integer between 200 and 400, divisible by 7 = 203.
Similarly, 400/7 gives a remainder of 1, which means that the largest integer between 200 and 400, divisible by 7 = 399. '

Sorry to keep you asking stupid questions.

200/7 gives the remainder = 4, which means if we add 3 to 200 (we added 3 as we want to divide by 7 and remainder = 4), then it will leave a remainder of 0 when divided by 7.

Similarly, 400/7 gives a remainder of 1, which means if we subtract 1 from 400 = 399, then this integer will be completely divisible by 7.

I hope that helps.