Sum of First n Terms of an Arithmetic Progression
Imagine you are stacking bricks in a triangular pile. The bottom row has 10 bricks, the next row has 9, then 8, and so on, until the top row has 1 brick. How many bricks are there in total? You could count them one by one, but there is a much smarter way.
The Intuition: Pairing Terms
An Arithmetic Progression (A.P.) is a sequence where each term differs from the previous one by a fixed number, called the common difference (d). For example: 2,5,8,11,14,… has d=3.
Now, suppose we want the sum of the first n terms of an A.P. Let the first term be a and the common difference be d. The terms are:
a, a+d, a+2d, a+3d, …, a+(n−1)d
The trick is to write the sum forward and then backward, and add them.
Let Sn be the sum of the first n terms.
Sn=a+(a+d)+(a+2d)+⋯+[a+(n−1)d]
Write the same sum in reverse order:
Sn=[a+(n−1)d]+[a+(n−2)d]+⋯+a
Now add these two equations term by term. The first pair gives a+[a+(n−1)d]=2a+(n−1)d. The second pair gives (a+d)+[a+(n−2)d]=2a+(n−1)d. Every pair gives the same value 2a+(n−1)d. There are n such pairs.
2Sn=n×[2a+(n−1)d]
Therefore:
Sn=2n[2a+(n−1)d]
Sn=2n[2a+(n−1)d]
This is the sum of the first n terms of an A.P.
A More Natural Form
Notice that the last term of the A.P. is l=a+(n−1)d. So we can also write:
Sn=2n(a+l)
This is the average of the first and last terms, multiplied by the number of terms. That is exactly what the brick-pile intuition suggests: the average number of bricks per row is the average of the top and bottom rows, and the total is that average times the number of rows.
For an A.P., the sum is simply the number of terms times the average of the first and last terms. This works because the terms are equally spaced — the average of any symmetric pair is the same.
Example
Find the sum of the first 20 terms of the A.P.: 3,7,11,15,…
Here a=3, d=4, n=20.
Using the formula:
S20=220[2(3)+(20−1)(4)]=10[6+76]=10×82=820
Or using the last term: l=3+(20−1)(4)=3+76=79, so S20=220(3+79)=10×82=820.
A common mistake is to use n as the number of terms but forget that the nth term is a+(n−1)d, not a+nd. Always check: the first term corresponds to n=1, so the nth term has (n−1) steps of d.
Why This Matters
The sum formula is not just a memorised equation. It captures a deep idea: when numbers increase steadily, the total is the average of the extremes times the count. This same idea appears in physics (distance under constant acceleration), in finance (sum of instalments), and in many counting problems. Once you see the pairing trick, you never need to memorise the formula — you can derive it in seconds.