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

$120$

#### Explanation:

Given arithmetic sequence

$3 , 5 , 7 , 9 , \ldots \ldots \ldots , 21$

The first term $a = 3$ & a common difference $d = 5 - 3 = 7 - 5 = \setminus \cdots = 2$

If there are $n$ number of terms in the above A.P. then last term $l = 21$ will be the nth term given as

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

$21 = 3 + \left(n - 1\right) 2$

$n = 10$

hence the sum of given arithmetic progression ($A . P .$) up to $10$ terms is given general formula

${S}_{n} = \setminus \frac{n}{2} \left(a + l\right)$

${S}_{10} = \setminus \frac{10}{2} \left(3 + 21\right)$

$= 120$

Jul 11, 2018

${S}_{10} = 120$

#### Explanation:

$\text{the sum to n terms of an arithmetic sequence is}$

•color(white)(x)S_n=n/2[2a+(n-1)d]

$\text{where a is the first term and d the common difference}$

$\text{here } a = 3 , d = 7 - 5 = 5 - 3 = 2 , n = 10$

${S}_{10} = 5 \left[\left(2 \times 3\right) + \left(2 \times 9\right)\right]$

$\textcolor{w h i t e}{\times} = 5 \left(6 + 18\right) = 120$