Question 1119318: Find the sum of all multiples of 14 between 100 and 2000, inclusive Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52805) (Show Source):
The first term is 112 = 14*8.
The sum is 112 + 126 + 140 + . . . .
To find the last term and the number of terms, calculate the number of intervals of the length 14 starting from 112 and ending 2000.
The number of such intervals is the INTEGER PART of the number = 134.86, i.e. 134.
Thus the number of terms multiple to 14 from 100 (or 112) to 2000 is 135.
Check: 112 + 134*14 = 1988.
Now use the formula for the sum of the first n terms of an arithmetic progression
= .
In your case = 112; n = 135; = 1988.
Hence, the sum S = = 141750.
It is your answer.
Another option that some students like is to find the sum of all the positive multiples of 14 less than 2000 and subtract the sum of the multiples that are less than 100.
2000/14 = 142 6/7, so the largest multiple of 14 less than 2000 is 142*14 = 1988.
(Note: if you are familiar with 7*11*13 = 1001, you can quickly determine, without needing the above calculation, that 2002 is a multiple of 14, making 1988 the largest multiple of 14 less than 2000.)
So we can find the required sum as
(14+28+...+1988) - (14+28+...+98)
The first sum there is 142 terms with an average of (14+1988)/2 = 1001, making a total of 142142.
The second term is 7 terms with an average of (14+98)/2 = 56, making a total of 392.
