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a level maths sequences question help

stuck on part b
Original post by o_reo
stuck on part b


Post what you've done so far.
Reply 2
Original post by Muttley79
Post what you've done so far.

well I know this isn't right but what I originally did was say 41/50 was the common ratio and just put it into the sum to infinity = a/(1-r) where a=d and then I got 5 and 5/9 d
so I did it wrong but not sure what Im meant to do as they dont give any other information
Remember that when it bounces it travels twice the height of the bounce because it has to go up then back down again
Original post by o_reo
well I know this isn't right but what I originally did was say 41/50 was the common ratio and just put it into the sum to infinity = a/(1-r) where a=d and then I got 5 and 5/9 d
so I did it wrong but not sure what Im meant to do as they dont give any other information

Think about it a bit more - you drop it from a certain height it bounces up and then goes back down - so you want total distance ...

d ------------ + 41d/50 + 41d/50 + ...
initial drop + first up + first down + second up etc
(edited 3 years ago)
Reply 5
Original post by Muttley79
Think about it a bit more - you drop it from a certain height it bounces up and then goes back down - so you want total distance ...

d ------------ + 41d/50 + 441d/50 + ...
initial drop + first up + first down + second up etc

so what would the common ratio be
Original post by o_reo
so what would the common ratio be


Without working it out, I assume your initial common ratio for the bounce heights, of 41/50, is correct. You need to massage the sum slightly to get it into the form of a geometric progress plus/times a couple of bits.

Muttley79 had a typo, and your sum would be:

d+d4150+d4150+d(4150)2+d(4150)2+d +d\frac{41}{50} +d\frac{41}{50}+d\left(\frac{41}{50}\right)^2+d\left(\frac{41}{50}\right)^2+\cdots
(edited 3 years ago)
Reply 7
Original post by ghostwalker
Without working it out, I assume your initial common ratio for the bounce heights, of 41/50, is correct. You need to massage the sum slightly to get it into the form of a geometric progress plus/times a couple of bits.

Muttley79 had a typo, and your sum would be:

d+d4150+d4150+d(4150)2+d(4150)2+d +d\frac{41}{50} +d\frac{41}{50}+d\left(\frac{41}{50}\right)^2+d\left(\frac{41}{50}\right)^2+\cdots

ok thank you!! I get it now

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