# The 15th term of an arithmetic series is 52, and the sum of the first 15 terms is 405 how do you find the sum of the first 22 terms?

Mar 27, 2016

${S}_{22} = 869$

#### Explanation:

Recall the following:

$1. \textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{Arithmetic sequence} : {t}_{n} = a + \left(n - 1\right) \mathrm{dc} o l \mathmr{and} \left(w h i t e\right) \left(\frac{a}{a}\right) |}}}$

where:
${t}_{n} =$term number
$a =$first term
$n =$number of terms
$d =$common difference

$2. \textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{Arithmetic series} : {S}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right) \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
${S}_{n} =$sum of $n$ numbers, starting at $a$
$n =$number of terms
$a =$first term
$d =$ common difference

Determining the First Term and Common Difference
$1$. Start by using the arithmetic sequence formula to express the ${15}^{\text{th}}$ term mathematically. Label the simplified equation as equation $1$.

${t}_{n} = a + \left(n - 1\right) d$

$52 = a + \left(15 - 1\right) d$

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{Equation} \textcolor{w h i t e}{i} 1} : 52 = a + 14 d$

$2$. Use the arithmetic series formula to express the sum of the first $15$ terms. Label the simplified equation as equation $2$.

${S}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$

$405 = \frac{15}{2} \left(2 a + \left(15 - 1\right) d\right)$

$405 = \frac{15}{2} \left(2 a + 14 d\right)$

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{Equation} \textcolor{w h i t e}{i} 2} : 405 = 15 a + 105 d$

$3$. Use elimination to subtract equation $2$ from equation $1$. Start by eliminating the terms with the variable, $a$. However, since equation $1$ does not have the same coefficient for $a$ as equation $2$, multiply equation $1$ by $\textcolor{t e a l}{15}$.

$\textcolor{t e a l}{15} \left(52\right) = \textcolor{t e a l}{15} \left(a + 14 d\right)$

$780 = 15 a + 210 d$

$4$. Now that equation $1$ has the same coefficient for $a$ as equation $2$, subtract equation $2$ from equation $1$.

$\textcolor{w h i t e}{\times x} 780 = 15 a + 210 d$
$\frac{- \left(405\right) = - \left(15 a + 105 d\right)}{\textcolor{b r o w n}{375 = 0 a + 105 d}}$

$5$. Solve for $d$ in $\textcolor{b r o w n}{375 = 0 a + 105 d}$.

$375 = 0 a + 105 d$

$375 = 105 d$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} d = \frac{25}{7} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$6$. Substitute $d = \frac{25}{7}$ into either equation $1$ or $2$ to determine the value of $a$. In this case, we'll use equation $1$.

$780 = 15 a + 210 d$

$780 = 15 a + 210 \left(\frac{25}{7}\right)$

$780 = 15 a + 750$

$30 = 15 a$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} a = 2 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Determining the Sum of the First 22 Terms
$1$. Since you now have the values of $a$ and $d$, you can use the arithmetic series formula to find the sum of the first $22$ terms. Thus, substitute your known values into the formula.

${S}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$

${S}_{22} = \frac{22}{2} \left(2 \left(2\right) + \left(22 - 1\right) \left(\frac{25}{7}\right)\right)$

$2$. Solve for ${S}_{22}$.

${S}_{22} = 11 \left(4 + 21 \left(\frac{25}{7}\right)\right)$

${S}_{22} = 11 \left(4 + 75\right)$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {S}_{22} = 869 \textcolor{w h i t e}{\frac{a}{a}} |}}}$