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

Hint: We recall the arithmetic sequence otherwise known as arithmetic progression (AP ) of whose term at the ${{n}^{\text{th}}}$ position is given by ${{x}_{n}}=a+\left( n-1 \right)d$. 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 ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$\[\]

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
\[{{x}_{n}}=a+\left( n-1 \right)d\]
The sum up to ${{n}^{\text{th}}}$ term of arithmetic sequence is given by
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
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
\[\begin{align}
  & \Rightarrow {{x}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 21=3+\left( n-1 \right)2 \\
 & \Rightarrow 18=\left( n-1 \right)2 \\
 & \Rightarrow n-1=\dfrac{18}{2}=9 \\
 & \Rightarrow n=10 \\
\end{align}\]
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
\[\begin{align}
  & {{S}_{10}}=\dfrac{10}{2}\left[ 2\cdot 3+\left( 10-1 \right)2 \right] \\
 & \Rightarrow {{S}_{10}}=5\left( 6+18 \right) \\
 & \Rightarrow {{S}_{10}}=5\times 24=120 \\
\end{align}\]
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 $S=\dfrac{n}{2}\left( a+l \right)$. 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 $S=\dfrac{10}{2}\left( 3+21 \right)=5\times 24=120$. We can also use sum of first $n$ odd numbers formula that is $1+3+5...+\left( 2n-1 \right)={{n}^{2}}$ to find the sum up to $n={{11}^{th}}$ term as $1+3+5...21={{11}^{2}}=121$, then the required sum is $3+5...21=121-1=120$.