# SOLUTION: Let n=1+2+3+5+.....+999 and m=2+4+6+....+1000. Then m-n equals ?

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 Question 629901: Let n=1+2+3+5+.....+999 and m=2+4+6+....+1000. Then m-n equals ? Answer by htmentor(789)   (Show Source): You can put this solution on YOUR website![I assume you mean n=1+3+5+..., in other words only odd numbers] We can do this problem formally by determining the sums of arithmetic sequences, etc. But if we recognize that each time another term is added to the sequences, the sum of m grows larger by 1 compared to n. For example, the sum of the first 3 terms is 9 and 12. Since we've added 3 terms sum m is larger by 3*1 = 3 Since the sequences have 500 terms, subtracting the two sums gives 500*1 = 500 If you'd like the formal solution, here it is: The n-the term of an arithmetic sequence can be written a(n) = a + (n-1)d, where a=the 1st term, d=the common difference In both cases, the common difference is 2, so we can write a(n) = 2n - 1 [for sequence n] a(n) = 2n [for sequence m] The sum of the 1st n terms of an arithmetic sequence = (n/2)(a+a(n)) The sum of n = 250(1+999) = 250000 The sum of m = 250(2+1000) = 250500 We see the difference is 500, as we determined above.