Asked by Joey
How many terms of the A.P, 1+4+7+.......are required to make a sum of 590?
Answers
Answered by
Reiny
You have an AS with
a = 1, d = 3, Sum(n) = 590
(n/2)(2a + (n-1)d) = sum(n)
(n/2)(2 + 3n - 3) = 590
n(3n - 1) = 1180
3n^2 - n - 1180 = 0
solve for n using your favourite methods, rejecting the negative answer
Hint: since n has to be a whole number, it will have to factor
a = 1, d = 3, Sum(n) = 590
(n/2)(2a + (n-1)d) = sum(n)
(n/2)(2 + 3n - 3) = 590
n(3n - 1) = 1180
3n^2 - n - 1180 = 0
solve for n using your favourite methods, rejecting the negative answer
Hint: since n has to be a whole number, it will have to factor
Answered by
Kumar Siddharth Yadav
Many problems in math
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