# Thread: Sum of an arithmetric progression

1. ## Sum of an arithmetric progression

In a contest, an athlete competes against a robot to see who can climb up a 16m rope faster. The athlete climbs up x m in each of his first 2 pulls; (x - 0.02)m in each of his next 2 pulls; and every subsequent pair of pulls shows a similiar decrease of 2cm from the previous pair.

Show that the distance he travels after 2n pulls is 2nx - 0.02n(n - 1).

I think I am supposed to use the sum of arithmetric progression formula but I am kind of confused with the question...

2. ## Re: Sum of an arithmetric progression

If we pretend that each pair of pulls is one pull then we have a common difference of $d=-0.02$ and a first term of $a=x$. If we were to pick each term it'd be harder to establish a common difference making it harder to solve.

Since we're treating each 2 pulls as 1 pull then we can treat $2n$ pulls as $n$ pull we can treat this as the sum of an arithmetic sequence (total distance=sum in this case) using the standard sum formula, subbing in your values of $a$ and $d$ you should get the answer required

$S_n = \dfrac{n}{2}(2a+(n-1)d)$

3. ## Re: Sum of an arithmetric progression

Arithmetic sequences have the nice property that the nice property that the average of all the numbers in the sequence is equal to the average of the first and last numbers. His first pull was "x" and his nth was x- (n-1)(.02). The average of those numbers is [tex]\frac{x+ x- (n-1)(.02)}{2}= x- (n-1)(0.01). Now, to find the total distance, multiply that by n.