How do you find the sum of the arithmetic sequence 4 + 8 + 12 + ... + 312?

1 Answer
May 17, 2016

12324

Explanation:

Let the position count be #i#
Let any term be #a_i#

Let total count be #n#

So we have #a_1+a_2+a_3+...+a_n" "->" "4+8+12+...+312#

Notice that:
#8-4=4#
#12-8=4#

So really this is the sum of the 4 times table

#a_1=1xx4=4#
#a_2=2xx4=8#
#a_3=3xx4=13#
#a_n=nxx4=312 => n= 312/4 = 78#

#color(green)((1color(magenta)(xx4))+(2color(magenta)(xx4))+(3color(magenta)(xx4))+...+(78color(magenta)(xx4)))#

#color(magenta)(4)color(green)((1+2+3+..+78))#
'~~~~~~~~~~~~~~~~~~~
If you really wish to use mathematical notation
Given that the term #Sigma# means sum of and #Sigma_(i=1)^n a_i color(white)(.)# means sum of #a_1+a_2+...+a_n #

Then what we really have is #color(magenta)(4xx)color(green)(Sigma_(i=1)^78(i))#

The sum of #color(green)(1+2+3+...+78)" is count"xx"mean value"#

So we have

#color(green)(color(magenta)(4xx)Sigma_(i=1)^78(i))color(blue)(" "->" "color(magenta)(4)[78xx(1+78)/2] = color(magenta)(4xx)3081=12324)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Demonstrating the principle")#

#Sigma_(i=1)^3(i) =1+2+3 =6" " ->" " 3xx(1+3)/2 =6#

#Sigma_(i=1)^4(i) = 1+2+3+4 = 10" "->" "4xx(1+4)/2 = 10#

#Sigma_(i=1)^5(i)=1+2+3+4+5=15" "->" "5xx(1+5)/2=15#