Question

What is the eighth term in an arithmetic series that has an 8th partial sum of 108 and a first term of 2

Answer 1

1st partial sum: a
2nd partial sum: a + (a+d) = 2a + d
3rd partial sum: a + (a+d) + (a+2d) = 3a + (d+2d) = 3a + d(1+2)
4th partial sum: a + (a+d) + (a+2d) + (a+3d) = 4a + d(1+2+3)
5th partial sum: a + (a+d) + (a+2d) + (a+3d) + (a+4d) = 5a + d(1+2+3+4)
...
nth partial sum: a + (a+d) + (a+2d) + ... + (a+(n-1)d)= na + d(1+2+3+...+(n-1))

The sum of the first n-1 whole numbers is 1 + 2 + 3 + ... + (n-1) = \(\dfrac{(n-1)n}{2}\)

Thus, nth partial sum: \(na + d\dfrac{(n-1)n}{2} = \dfrac{n}{2}(2a + (n-1)d)\)

Thus, nth partial sum: \(\dfrac{n}{2}(2a + (n-1)d) = \dfrac{n}{2}(a + [a+(n-1)d])\)

Thus, nth partial sum: \(\dfrac{n}{2}(First + Last)\)

For this problem, solve: \(\dfrac{8}{2}(2 + Last) = 108\)