# Sum of an arithmetric progression

• Aug 5th 2011, 02:10 AM
Blizzardy
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...

If we pretend that each pair of pulls is one pull then we have a common difference of $\displaystyle d=-0.02$ and a first term of $\displaystyle a=x$. 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 $\displaystyle 2n$ pulls as $\displaystyle n$ 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 $\displaystyle a$ and $\displaystyle d$ you should get the answer required
$\displaystyle S_n = \dfrac{n}{2}(2a+(n-1)d)$