# Sum of Arithmetic Progression

Sum of arithmetic progression is very interesting because of its application in real life from basic arithmetic to Olympiad math levels. When dealing with the sum of arithmetic series the most important number is the middle term. I have read a story about a little boy how he easily sum up the sum of the first 100 numbers when he was 10 years old. That kid is no other than Carl Friedrich Gauss.

Formula:

Let,

$S_n$– Sum of series

$a_m$– middle term

$d$– common difference

$n$– number of terms

$a_1$– first term

$a_n$– last term

Formula Derivation:

Given the middle term:

$S_n=a_m\cdot n$  – 1st formula

Middle term can be calculated by the formula

$a_m=\displaystyle\frac{a_1+a_n}{2}$ $S_n=a_m\cdot n$

$S_n=\displaystyle\frac{n(a_1+a_n)}{2}$ –  2nd formula

We also learned from the previous post that $a_n=a_1+(n-1)d$

From 2nd formula,

$S_n=\displaystyle\frac{n(a_1+a_n)}{2}$ $S_n=\displaystyle\frac{n}{2}[a_1+a_1+(n-1)d]$

$S_n=\displaystyle\frac{n}{2}[2a_1+(n-1)d]$3rd formula

Sample Problem 1:

The first term of an arithmetic sequence is 4, the last term is 60. The second term is obtained by adding 4. Find the sum of the series.

Solution:

We have three formula available, we need to find the best formula according to the given. Better to list down the given.

$a_1=4$, $a_n=60$, $d=4$

The three formulas need the number of terms and that is what we need to find out.

Using the formula for nth term,

$a_n=a_1+(n-1)d$ $60=4+(n-1)4$ $n=15$

Solving for sum of the series, we can solve one of the three formulas.

$S_n=\displaystyle\frac{n(a_1+a_n)}{2}$ $S_n=\displaystyle\frac{15(4+60)}{2}$ $S_n=480$

Sample Problem 2:

The nth term of an arithmetic series is given by the formula $a_n=5+2n$. Find the sum of the first 100 terms.

Solution:

We’re doomed right? Where is our a1,an,d,n?

With the previous post there is the same problem and we know that the coefficient of n is the common difference. Thus, d=2.

Solving for $a_1$ :

Let n=1

$a_n=5+2n$ $a_1=5+2(1)$ $a_1=7$

Solving for $a_{100}$

Let n=100

$a_n=5+2n$ $a_{100}=5+2(100)$ $a_{100}=205$

Solving for the sum of the first 100th terms:

$S_n=\displaystyle\frac{n(a_1+a_n)}{2}$ $S_{100}=\displaystyle\frac{100(7+205)}{2}$ $S_{100}=10,600$

Worked Problem 3:

Find the sum of the first 50 positive multiples of 3.

Solution:

The first multiple of 3 is obviously 3, then 6, 9, 12,…

The last term then is 3(50)=150.

$S_n=\displaystyle\frac{n(a_1+a_n)}{2}$ $S_{50}=\displaystyle\frac{50(3+150)}{2}$ $S_{50}=3825$

Worked Problem 4:

The logs of lumber are piled in a shape of a triangle where the tip of the pile has one lumber. The next row has two, three and so on. If the bottom part of the pile have 15 lumbers. How many logs of lumber are there?

Solution:

The solution is like finding the sum of first 15 counting numbers. There is a special formula for that which is ${n+1\choose 2}$

$S_n={n+1\choose 2}=\displaystyle\frac{(n+1)n}{2}$ $S_n=\displaystyle\frac{(15+1)15}{2}$ $S_n=120$

Worked Problem 5: Adapted from Engr. Joselito Torculas of Adamson University Team

Find the value of $a_2+a_4+a_6+\ldots + a_{96}+a_{98}$

If $a_1,a_2,a_3,\ldots$is an arithmetic progression with common difference 1 and

$a_1+a_2+a_3+\dots+a_{97}+a_{98}=137$

Solution:

$a_1+a_2+a_3+\dots+a_{97}+a_{98}=137$

We know that $a_2=a_1+1 \to a_1=a_2-1$, generally $a_{n-1}=a_n-1$

$(a_2-1)+a_2+(a_4-1)+a_4+\dots+(a_{98}-1)+a_{98}=137$ $2(a_2+a_4+a_6+\dots+a_{96}+a_{98})-49=137$ $2(a_2+a_4+a_6+\dots+a_{96}+a_{98})=186$ $a_2+a_4+a_6+\dots+a_{96}+a_{98}=93$