How do you find the sum of the arithmetic sequence #4x+6x+...+22x#?

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Answer 1 · 2016-05-29T07:10:53

It is

#4x+6x+...22x=2x*[2+3+...+11]#

To find the sum of #2+3+...+11# or #(1+2+...+11)-1#

we can use #S=n*(n+1)/2# for #(1+2+...+11)# hence #S=11*(12)/2=66# and substract one hence

#(2+3+...+11)=65#

Finally

#4x+6x+...22x=2x*[2+3+...+11]=2x*65=130*x#