a level maths sequences question help

Watch
o_reo
Badges: 16
Rep:
?
#1
Report Thread starter 1 month ago
#1
stuck on part b
Attached files
0
reply
Muttley79
Badges: 20
Rep:
?
#2
Report 1 month ago
#2
(Original post by o_reo)
stuck on part b
Post what you've done so far.
0
reply
o_reo
Badges: 16
Rep:
?
#3
Report Thread starter 1 month ago
#3
(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
0
reply
16CharsIsTooShor
Badges: 13
Rep:
?
#4
Report 1 month ago
#4
Remember that when it bounces it travels twice the height of the bounce because it has to go up then back down again
0
reply
Muttley79
Badges: 20
Rep:
?
#5
Report 1 month ago
#5
(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
Last edited by Muttley79; 1 month ago
0
reply
o_reo
Badges: 16
Rep:
?
#6
Report Thread starter 1 month ago
#6
(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
0
reply
ghostwalker
  • Study Helper
Badges: 17
#7
Report 1 month ago
#7
(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 +d\frac{41}{50} +d\frac{41}{50}+d\left(\frac{41}  {50}\right)^2+d\left(\frac{41}{5  0}\right)^2+\cdots
Last edited by ghostwalker; 1 month ago
0
reply
o_reo
Badges: 16
Rep:
?
#8
Report Thread starter 1 month ago
#8
(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 +d\frac{41}{50} +d\frac{41}{50}+d\left(\frac{41}  {50}\right)^2+d\left(\frac{41}{5  0}\right)^2+\cdots
ok thank you!! I get it now
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (398)
56.29%
I don't have everything I need (309)
43.71%

Watched Threads

View All