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

##### 1 Answer

I got

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, . . . , (136 - 12n)# , 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

#sum_(n=0)^(35) a_n = \mathbf(sum_(n=0)^(35) (136 - 12n))#

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

But...

#=> 136 - 12*0 + 136 - 12*1 + cdots + 136 - 12*35#

#= stackrel("36 numbers")overbrace((136*136cdots136)) stackrel("36 times")overbrace(- 12*0 - 12*1 - cdots - 12*35)#

#= (136*36) - (12*0 + 12*1 + cdots + 12*35)#

#= color(blue)(stackrel("Part 1")overbrace((136*36)) - stackrel("Part 2")overbrace(12*(0+1+2+cdots+35)))#

Now all we need to do is get:

#12*(0+1+2+cdots+35)# #136*36# - The result of subtracting the former from the latter.

**PART 1**

To get the sum of

For

So, since the sequence ends at

#=> (0 + 1 + cdots + 35) = 35*36/2#

#= 35*18#

#= 35*20 - 35*2 = 700 - 70 = 630#

Next, we can multiply this by

#color(green)(12*(0 + 1 + cdots + 35))#

#630 * 12#

#= 630*10 + 630*2#

#= 6300 + 1260 = color(green)(7560)#

**PART 2**

That's one part of this. Now for

#=>color(green)(136*36)#

#= 136*10*4 - 136*4#

#= 1360*4 - 136*4#

#= 2720*2 - 272*2#

#= 5440 - 544#

#= 5440 - 440 - 104 = color(green)(4896)#

Finally, the sum turns out to be:

#color(blue)(136 + 124 + 112 + cdots)#

#= sum_(n=0)^(35) (136 - 12n)#

#= 4896 - 7560#

#= -(7560 - 4896)#

#= -(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))]))#

#= color(blue)(-2664)#