Sequences and Series

1. Dec 26, 2008

locked.on

Any guidance or worked solutions would be appreciated

1. The problem statement, all variables and given/known data

For a potato race, a straight line is marked on the ground from a point A, and points B,C,D,... are marked on the line so that AB = BC = CD = ... = 2 metres. A potato is placed at each of the points B,C,D,...

A runner has to start from A and bring each potato by a separate journey back to a basket at A. Find the number of potatoes so that the total distance run during the race will be 480 metres.

2. Relevant equations

A) S = n/2 [2a + (n - 1) d]
B) S = n/2 [ first term + last term]

3. The attempt at a solution

I can calculate the answer by listing out the distances covered in successive trips to obtain the required distance. However, I am unable to show relevant working and this method will not work for larger sequences.

*Tn* = 2n(n+1), where *Tn* is the 'n' th term in the sequence.

The problem is I can't find a relevant equation for the common difference, let's say, *d*.

*d* = 4(n+1) ?

I can't use equation B) as the common difference is not a constant.

2. Dec 26, 2008

symbolipoint

The original description already gave the common difference of 2 for each term in the sequence.

3. Dec 28, 2008

yeongil

Actually, it's 4, because the runner starts at A, goes to one of the points to get a potato, and run back to A. To get the potato at point B the runner runs 4 meters, to get the potato at point C the runner runs 8 meters, etc.

OP: you don't need equation B at all. Equation A is actually a combination of equation B and the rule to find the nth term of a sequence: $$a_{n}=a_{1} + (n{-}1)d$$. Just plug in numbers into equation A and solve. (I got 15 potatoes; hope that's what you got.)

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