Asked by chimingbell
Find the sum of the following (starts with 3, increment by 7, ends with 7N+3)
3+10+17+...+689+696+703
3+10+17+...+689+696+703
Answers
Answered by
Reiny
the term of 7N+3 tells me that you want the sum of n terms
sum(n) = (n/2)(2a + (n-1)d )
sum(n) = (n/2)(6 + (n-1)(7))
= (n/2)(6 + 7n - 7)
= n(7n - 1)/2
sum(n) = (n/2)(2a + (n-1)d )
sum(n) = (n/2)(6 + (n-1)(7))
= (n/2)(6 + 7n - 7)
= n(7n - 1)/2
Answered by
scott
notice that ... (703 + 3) = (696 + 10) = (689 + 17) ...
so ... sum = (703 + 3) * (101 / 2)
so ... sum = (703 + 3) * (101 / 2)
Answered by
chimingbell
How do you know n is 101?
Answered by
scott
703 - 3 = 700 ... 700 / 7 = 100
100 terms beyond the 1st term ... a total of 101
100 terms beyond the 1st term ... a total of 101
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