Question

What is the formula for the sum of an arithmetic sequence?

Answer 1

`S_n = n/2(2a+(n-1)d)`

Explanation:

Suppose we have an AP with first term `a` and common difference `d`, then we can write the sum of the first `n` terms as:

`S_n = a + (a+d) + (a+2d) + ... + (a+(n-1)d)`

Writing the same sum, but in reverse, we get:

`S_n = (a+(n-1)d) + ... (a+2d) + (a+d) + a`

If we add both of these we get:

`2S_n = (2a+(n-1)d) + (2a+(n-1)d) + ... + (2a+(n-1)d)`
`\ \ \ \ \ = n(2a+(n-1)d)`

Leading to the standard AP summation formula

`S_n = n/2(2a+(n-1)d)`

Answer 2

The sum of an arithmetic sequence is given by `S_n=sum_(i=1)^n a_i=n/2(a_1+a_n)`.

Explanation:

Let `S_n` be the sum of the arithmetic sequence and the `n^"th"` term of the sequence be `a_n=a_1+d(n-1)` where `d` is the common difference.
`S_n=a_1+(a_1+d)+(a_1+2d)+...+(a_n-d)+a_n`
`S_n=a_n+(a_n-d)+(a_n-2d)+...+(a_1+d)+a_1`
Adding up the two equations term by term we get
`2S_n=(a_1+a_n)+(a_1+a_n)+...+(a_1+a_n)+(a_1+a_n)`
*Notice how the `d-d` and `2d-2d` all cancel out
Since we have `n` terms of `(a_1+a_n)` on the right side, we can re-write the equation as
`2S_n=n(a_1+a_n)`
`:.S_n=n/2(a_1+a_n)`