# What is the sum of the arithmetic sequence 9,14,19... if there are 34 terms?

May 25, 2016

Built the equation for the sum of an arithmetic sequence and used it to derive the solution.

$\textcolor{g r e e n}{= 128 + 2975 = 3111}$

#### Explanation:

Let the sequence position be $i$
Let the sequence position of the last term be $n$

Let the difference between each term be $d$

Let the ith term in the sequence be ${a}_{i}$
Let the last term in the sequence be ${a}_{n}$

So
${a}_{i} \to {a}_{1} = 9$
${a}_{i} \to {a}_{2} = 14$
${a}_{i} \to {a}_{3} = 19$
a_i->a_n=a_34=?" "larr "the last term in the sequence"

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$\textcolor{b l u e}{\text{Investigating the 'common difference' between each term}}$

$14 - 9 = 5$
$19 - 14 = 5$

$\textcolor{g r e e n}{\text{the difference between each successive term is } d = 5}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the value of the 34th term}}$

So if ${a}_{i} \to {a}_{1} = 9 \text{ }$ then $\text{ } {a}_{i} \to {a}_{0} = 9 - 5 = 4$

So
${a}_{1} = {a}_{0} + d$
${a}_{2} = {a}_{0} + d + d \text{ "->" } {a}_{0} + 2 d$
${a}_{3} = {a}_{0} + d + d + d \text{ "->" } {a}_{0} + 3 d$

so ${a}_{i} = {a}_{0} + \left(i \times d\right)$

and ${a}_{n} = {a}_{0} + \left(n \times d\right)$

So ${a}_{n} = {a}_{34} = {a}_{0} + \left(34 \times d\right) \text{ "->" } 4 + \left(34 \times 5\right) = 174$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the sum of the 34 terms}}$

Consider again the:
${a}_{1} = {a}_{0} + d$
${a}_{2} = {a}_{0} + d + d \text{ "->" } {a}_{0} + 2 d$
${a}_{3} = {a}_{0} + d + d + d \text{ "->" } {a}_{0} + 3 d$

The sum of ${a}_{1} \to {a}_{3}$ is:

${a}_{1} \to {a}_{0} + d$
${a}_{2} \to {a}_{0} + 2 d$
${a}_{3} \to \underline{{a}_{0} + 3 d} \text{ "larr" add}$
$\text{ } 3 {a}_{0} + 6 d$

Which is the same as:

$3 {a}_{0} + \left(d + 2 d + 3 d\right)$

$3 {a}_{0} + d \left(1 + 2 + 3\right)$

$3 {a}_{0} + d \left(3 \times \text{Mean value}\right)$

$\left(i \times {a}_{0}\right) + d \left(i \times \frac{\text{first count+last count}}{2}\right)$

$\left(i \times {a}_{0}\right) + d \left(i \times \frac{1 + i}{2}\right)$
'.....................................................
So for a count of $n$ terms we have

$\textcolor{g r e e n}{\left(n {a}_{0}\right) + d \left(n \times \frac{1 + n}{2}\right)}$

$\textcolor{red}{\text{Notice that in the end I did not need the value of the last term}}$
'.................................................
$n = 34$
$d = 5$
${a}_{0} = 4$

sum of the sequence is $\left(34 \times 4\right) + 5 \left(34 \times \frac{1 + 34}{2}\right)$

$\textcolor{g r e e n}{= 136 + 2975 = 3111}$