# Answer to Question #20562 in Algebra for fardowsa

Question #20562

what is the nth term for this sequence?

7, 18, 35, 58, 87, 122

7, 18, 35, 58, 87, 122

Expert's answer

we can consider differences

18-7=11

35-18=17

58-35=23

87-58=29

122-87=35 if a1=7 then a2=7+11

a3=a2+11+6=7+11+11+6=a1+2*11+6

a4=a3+11+2*6 so a_n=a_(n-1)+11+(n-2)*6

or a_n=a1+(n-1)*11+(1+2+...+(n-2))*6=a1+11*(n-1)+(n-1)*(n-2)*6/2=

=a1+11(n-1)+3(n-1)*(n-2)=7+11n-11+3(n^2-3n+2)=11n-4+3n^2-9n+6=3n^2+2n+2, for

n>=2 so a1=7, an=3n^2+2n+2, but for n=1 a1=3*1^2+2*1+2=7 sofor every n a_n=3n^2+2n+2

18-7=11

35-18=17

58-35=23

87-58=29

122-87=35 if a1=7 then a2=7+11

a3=a2+11+6=7+11+11+6=a1+2*11+6

a4=a3+11+2*6 so a_n=a_(n-1)+11+(n-2)*6

or a_n=a1+(n-1)*11+(1+2+...+(n-2))*6=a1+11*(n-1)+(n-1)*(n-2)*6/2=

=a1+11(n-1)+3(n-1)*(n-2)=7+11n-11+3(n^2-3n+2)=11n-4+3n^2-9n+6=3n^2+2n+2, for

n>=2 so a1=7, an=3n^2+2n+2, but for n=1 a1=3*1^2+2*1+2=7 sofor every n a_n=3n^2+2n+2

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