Question 307207
So your goal is to get 11,360 rubas total right? I'm not sure why you've made it this complicated, but why not just use an arithmetic sequence as you did in plan 2?


One arithmetic sequence that generates 20, 25, 30, ..., 330, 335 is {{{a[n]=5n+15}}} where 'n' starts at {{{n=1}}}. Notice that if {{{n=3}}}, then {{{a[3]=5(3)+15=30}}} which is the 3rd term.



Recall that the sum of the first n terms of an arithmetic sequence is {{{S=(n(a+a[n]))/2}}}, where 'a' is the first term, {{{a[n]}}} is the nth term and 'n' is the number of terms you want to sum up. In this case, there are 64 terms meaning that {{{n=64}}}, the first term is {{{a=20}}} and the last term is {{{a[n]=335}}}.



Plug these values in to get {{{S=(64(20+335))/2=(64(355))/2=22720/2=11360}}}



So the sum of the terms 20, 25, 30, ..., 330, 335 is 11,360