How many terms of the A.P, 1+4+7+.......are required to make a sum of 590?

Answer 1

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

Answer 2

Many problems in math