MHB Athlete's Laps: Solve 1st & 6th Day + Total

[gotah]

I have a Q that says:
An athlete must each day run three laps more than the day before. On the sixth day she runs 17laps
Calculate how many laps she runs 1.) on the first day 2.) in total by the end of the sixth day.

in 1.
I know that Un=U1+(n-1)d
where d=3
Un=17
n=6

i then get 17= U1 + (6-1)3, and I get U = 2

In 2 i use S=(n/2)(2*U1 + (n-1)d)
so S = (6/2)(2*2+(6-1)3) =57 (in the answer sheet it says that it's 42) what am I doing wrong?

 

[CaptainBlack]

I have a Q that says:
An athlete must each day run three laps more than the day before. On the sixth day she runs 17laps
Calculate how many laps she runs 1.) on the first day 2.) in total by the end of the sixth day.

in 1.
I know that Un=U1+(n-1)d
where d=3
Un=17
n=6

i then get 17= U1 + (6-1)3, and I get U = 2

In 2 i use S=(n/2)(2*U1 + (n-1)d)
so S = (6/2)(2*2+(6-1)3) =57 (in the answer sheet it says that it's 42) what am I doing wrong?

Nothing, your argument has the athlete running: 2, 5, 8, 11, 14, 17. Which satisfy the conditions of the problem and sums to 57.

Also your method is correct.

CB