Question

How do I find the sum of the arithmetic sequence 3, 5, 7, 9,... 21? $$$$

Answer 1

Hint: We recall the arithmetic sequence otherwise known as arithmetic progression (AP ) of whose term at the $n^{\text{th}}$ position is given by $x_n=a+(n-1)d$. We find the position of the term 21 in the given arithmetic sequence and then find the up to $n^{\text{th}}$ term using the formula $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$$$$$

Complete step-by-step solution:
We know that an arithmetic sequence otherwise known as arithmetic progression, abbreviated as AP is a type sequence where the difference between any two consecutive numbers is constant. If $\left(x_n\right)=x_1,x_2,x_3,...$ is an arithmetic sequence, then $x_2-x_1=x_3-x_2...$ . The difference between two terms is called common difference and denoted $d$ where$d=x_2-x_1=x_3-x_2...$. If the first term is $x_1=a$ then the $n^{\text{th}}$ term is given by
$$x_n=a+(n-1)d$$
The sum up to $n^{\text{th}}$ term of arithmetic sequence is given by
$$S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$$
We are given the following arithmetic sequence in the question
$$3,5,7,...21$$
We see that the first term is $x_1=a$ and the last term is 21. The common difference is $d=7-5=5-3=...=21-19=2$. Let 21 be the $n^{th}$ term of the given arithmetic sequence. So we use the formula for $n^{\text{th}}$ term and have
$$\begin{array}{rl}& \Rightarrow x_n=a+(n-1)d\\ & \Rightarrow 21=3+(n-1)2\\ & \Rightarrow 18=(n-1)2\\ & \Rightarrow n-1={\displaystyle \frac{18}{2}}=9\\ & \Rightarrow n=10\end{array}$$
So 21 is ${10}^{th}$ term of the arithmetic sequence. The required sum is the sum of terms of given arithmetic sequence up to ${10}^{th}$terms that is
$$\begin{array}{rl}& S_{10}={\displaystyle \frac{10}{2}}[2\cdot 3+(10-1)2]\\ & \Rightarrow S_{10}=5(6+18)\\ & \Rightarrow S_{10}=5\times 24=120\end{array}$$
So the required sum is 120.

Note: We can alternatively solve using the formula for sum in an arithmetic sequence with first term $a$ to last term $l$ as $S={\displaystyle \frac{n}{2}}(a+l)$. Here we have $a=3,l=21$ and we have to obtain position of the last term $n=10$ using the $n^{th}$ term formula. So we have the required sum as $S={\displaystyle \frac{10}{2}}(3+21)=5\times 24=120$. We can also use sum of first $n$ odd numbers formula that is $1+3+5...+(2n-1)=n^2$ to find the sum up to $n={11}^{th}$ term as $1+3+5...21={11}^2=121$, then the required sum is $3+5...21=121-1=120$.