maths c2

username2944572

14

....

(edited 6 years ago)

Reply 1

Science Finger

3

Sorry could you post the attachment again or something as it just says 'attachment not found'

Reply 2

Qer

16

Original post by years101

Attachment not found

how do i work out b?

i need to work the 1st term which is a right?

in part a you have find difference
for b multiply that difference with forth term that is given

Reply 3

yanzzz123

2

Once you've worked out the common ratio, you could use that times the fourth term to get the fifth term. However, you might need the first term to solve c)

Reply 4

an_atheist

15

Original post by years101

For prt a, the common ratio is found by dividing successive terms.
Part b, multiply 144 by your answer to a
For part c, The sum of a GP is a(1-r^n)/(1-r). When r is between -1 and 1, the sum to infinity can be found, as r^n becomes negligibly small, so the 1-r^n term can be removed. You will have to find the first term though

Reply 5

an_atheist

15

Original post by years101

to get a do i sub r into ar^4 or any of them and then find the answer and then divide by 3/4^4 ?

If you're using the fifth term then yes. I'd use the third or fourth (ar^2 and ar^3) because the numbers are simpler and you know the starting point is correct.
But yes, that is how you find a.

Reply 6

Science Finger

3

Ok as different exam boards use different letters and symbols for formulas this might be a different formula to what your used to
so n=ar^(k-1), where n=term, a=starting term and k=number of terms, r= common ratio

a) so 144=ar^(3-1) and 108=ar^(4-1)
therefore 144=ar^2 and 108=ar^3
(144/r^2)=a and (108/r^3)=a
combined (144/r^2)=(108/r^3)
rearranged, 144r^3=108r^2
simplified, 144r=108
r=(108/144)
r=0.75

b) as r=0.75, input back into starting equations
144=a(0.75^2)
144/0.75^2=a
a=256
check
108=a(0.75^3)
108/0.75^3=a
a=256

Reply 7

an_atheist

15

Original post by Science Finger

Ok as different exam boards use different letters and symbols for formulas this might be a different formula to what your used to
so n=ar^(k-1), where n=term, a=starting term and k=number of terms, r= common ratio

a) so 144=ar^(3-1) and 108=ar^(4-1)
therefore 144=ar^2 and 108=ar^3
(144/r^2)=a and (108/r^3)=a
combined (144/r^2)=(108/r^3)
rearranged, 144r^3=108r^2
simplified, 144r=108
r=(108/144)
r=0.75

b) as r=0.75, input back into starting equations
144=a(0.75^2)
144/0.75^2=a
a=256
check
108=a(0.75^3)
108/0.75^3=a
a=256

First thing, you didnt answer b correctly. You found a (the value is correct), bt yo're meant to be finding the fifth term of the progression
Also, afaik we are not supposed to post worked solutions

Reply 8

an_atheist

15

Original post by years101

so i did this

ar^3 = a(3/4)^3

(3/4)^3 is 27/64 so do i divide by 3/4)^3 now?

Yes, then you stick the numbers in the formula.

Reply 9

an_atheist

15

Original post by years101

i got a = 1
when i did this:

a(3/4)^3

27/64 divided by 3/4^3 = 1

i think i did it wrong

r does equal 3/4. the third term of the sequence is 144, so ar^2=144. Does that help?

Reply 10

Luke7456

20

sigh could do with something relatively easy after getting a headache from what I have been doing all day time to utterly mess this up to help my self esteem even more.

Okay here goes

if the third term is 144 and the fourth term is 108 then
remember that AR^n-1

so the third term is AR^2=144
and the fourth term is AR^3=108
part A
divide the 4th term by the third term AR^3/AR^2=108/144
the As cancel out and R^3/R^2 just equals R
so R=108/144=0.75

B the fifth term would be the fourth term multiplied by the common ration which is 108*0.75 which= 81

C sum to infinity is A/1-r

we know that r= 0.75
so A /0.25
so now we have to find A well if A(0.75)^2=144 then A(0.75*0.75)=144 so
A(0.5625)=144

A=144/0.5625= 256
sum to infinity is therefore 256/0.25
which = 1024.

maths c2

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