Question

Partial Sums of an Arithmetic Sequence A partial sum of an arithmetic sequence is given. Find the sum.

$1+5+9+\dots +401$

Short answer

The required partial sum is $20301$.

Step-by-step solution

Step 1. Given information.

The given arithmetic sequence is:

$1+5+9+\dots +401$

Step 2. Write the concept.

The $n\text{th}$ term of an arithmetic sequence is:

$a_n=a+\left(n-1\right)d$

The partial sum of $n$ term of an arithmetic sequence is:

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Where $a$ is the first term and $d$ is a common difference.

Step 3. Determine the first term and common difference.

In the given sequence the first term is $a_1=1$and the second term is $a_2=5$. So, the common difference is:

$\begin{array}{c}d=a_2-a_1\\ =5-1\\ =4\end{array}$

Step 4. Determine the number of terms.

It is given that the $n\text{th}$term of the sequence is $401$.

$\begin{array}{c}{a}_n=401\\ a+\left(n-1\right)d=401\\ 1+\left(n-1\right)\left(4\right)=401\\ 1+4n-4=401\\ 4n-3=401\\ 4n=401+3\\ n=\frac{404}{4}\\ n=101\end{array}$

Step 5. Determine the partial sum.

Substitute $a=1,d=4,n=101$in the above formula to find the partial sum.

$\begin{array}{c}{S}_{101}=\frac{101}{2}\left[2\left(1\right)+\left(101-1\right)\left(4\right)\right]\\ =\frac{101}{2}\left[2+\left(100\right)\left(4\right)\right]\\ =\frac{101}{2}\left[2+400\right]\\ =\frac{101}{2}\left[402\right]\\ =101\left(201\right)\\ =20301\end{array}$

Thus, the required partial sum is $20301$.