How many terms of the A.P, 1+4+7+.......are required to make a sum of 590?
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
Many problems in math
To find out how many terms of an arithmetic progression (A.P.) are required to make a sum of 590, we need to determine the pattern of the A.P.
The given A.P. is 1, 4, 7,...
The first term (a) of the A.P. is 1, and the common difference (d) between consecutive terms is 4 - 1 = 3.
The formula to find the sum of an A.P. is given by:
Sn = (n/2)(2a + (n - 1)d)
Where:
Sn = the sum of the first n terms of the A.P.
a = the first term of the A.P.
d = the common difference between consecutive terms.
n = the number of terms.
We need to solve the equation Sn = 590 for n.
590 = (n/2)(2(1) + (n - 1)(3))
590 = (n/2)(2 + 3n - 3)
590 = (n/2)(3n - 1)
590 = (3n^2 - n)/2
Multiplying both sides by 2 (to eliminate the fraction):
1180 = 3n^2 - n
Rearranging the equation:
3n^2 - n - 1180 = 0
Now, we can solve this quadratic equation for n. We can factor or use the quadratic formula to find the values of n.