# What is the sum of the arithmetic sequence 136, 124, 112 …, if there are 36 terms?

Jul 7, 2016

I got $- 2664$ for your sequence, which is also written as ${a}_{n} = 136 - 12 n$, and Wolfram Alpha shows this.

I did this without using any complicated formulas, nor did I use a calculator, so you should be able to do this on a test with or without a calculator.

Sequence:

$136 , 124 , 112 , . . . , \left(136 - 12 n\right)$, starting at $n = 0$.

For this arithmetic sequence, we have the recursive sequence

${a}_{k + 1} = {a}_{k} - 12$, where the first term is ${a}_{0} = 136$.

We can rewrite this as the sum from $n = 0$ to $n = 35$ for a total of $36$ terms:

${\sum}_{n = 0}^{35} {a}_{n} = \setminus m a t h b f \left({\sum}_{n = 0}^{35} \left(136 - 12 n\right)\right)$

= (136 - 12*0) + (136 - 12*1) + cdots + (136 - 12*35) = stackrel("Your goal")(overbrace(???)

But...

$\implies 136 - 12 \cdot 0 + 136 - 12 \cdot 1 + \cdots + 136 - 12 \cdot 35$

$= \stackrel{\text{36 numbers")overbrace((136*136cdots136)) stackrel("36 times}}{\overbrace{- 12 \cdot 0 - 12 \cdot 1 - \cdots - 12 \cdot 35}}$

$= \left(136 \cdot 36\right) - \left(12 \cdot 0 + 12 \cdot 1 + \cdots + 12 \cdot 35\right)$

$= \textcolor{b l u e}{\stackrel{\text{Part 1")overbrace((136*36)) - stackrel("Part 2}}{\overbrace{12 \cdot \left(0 + 1 + 2 + \cdots + 35\right)}}}$

Now all we need to do is get:

• $12 \cdot \left(0 + 1 + 2 + \cdots + 35\right)$
• $136 \cdot 36$
• The result of subtracting the former from the latter.

PART 1

To get the sum of $0 + 1 + \cdots + 35$, we can just use the shortcut.

For $a + b + \cdots + y$ where each term is a consecutive counting number and $a = 0$ or $1$, the sum is equal to $\frac{y + z}{2}$.

So, since the sequence ends at $35$, look at $\frac{35 \cdot 36}{2}$ to get:

$\implies \left(0 + 1 + \cdots + 35\right) = 35 \cdot \frac{36}{2}$

$= 35 \cdot 18$

$= 35 \cdot 20 - 35 \cdot 2 = 700 - 70 = 630$

Next, we can multiply this by $12$ to get:

$\textcolor{g r e e n}{12 \cdot \left(0 + 1 + \cdots + 35\right)}$

$630 \cdot 12$

$= 630 \cdot 10 + 630 \cdot 2$

$= 6300 + 1260 = \textcolor{g r e e n}{7560}$

PART 2

That's one part of this. Now for $136 \cdot 36$.

$\implies \textcolor{g r e e n}{136 \cdot 36}$

$= 136 \cdot 10 \cdot 4 - 136 \cdot 4$

$= 1360 \cdot 4 - 136 \cdot 4$

$= 2720 \cdot 2 - 272 \cdot 2$

$= 5440 - 544$

$= 5440 - 440 - 104 = \textcolor{g r e e n}{4896}$

Finally, the sum turns out to be:

$\textcolor{b l u e}{136 + 124 + 112 + \cdots}$

$= {\sum}_{n = 0}^{35} \left(136 - 12 n\right)$

$= 4896 - 7560$

$= - \left(7560 - 4896\right)$

= -(color(white)([("",color(black)(" "7" "stackrel(1)5" "stackrel(1)6" "stackrel(1)0)),(color(black)(-),color(black)(""^(5)cancel(4)""^(9)cancel(8)""^(10)cancel(9)" "6))]))

$= \textcolor{b l u e}{- 2664}$