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Maths IGCSE exam tomorrow- please help

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My IGCSE maths exam is tomorrow and I noticed that hard sequences questions now show up at the end of some of the recent past papers (the 2011 ones)
I don't really understand them, especially the parts where you're asked to work out the nth term.
I've attached the papers, the questions are at the very end of both exams.
Please explain how you got the answer
Any help would be appreciated!
 

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I also don't get 14 b in paper 2
the answer should be 4π²/T²
but how to get this answer?
 
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for 14.b the answer is suppose to be 4π^2l/T^2
Firstly you need to remove the root meaning all values beofre the root is doubled, T^2=(2)^2π^2 (l/g)
Next.. move g up to the other side as it is multiply so g*T^2=4π^2l
then move T^2 to divide down.. so the answer is g=4π^2l/T^2
Next sequence in p2 Q18, value of T23, here you just substitute the 'n' with '23', and calculate it you would get 4324.
and (b) (i) you just replace T(n) with T(2) and then subtract it with T(1) meaning you have to substitute all the ns with 2 subtract with all the ns with 1,
you will get 4, and 9.. next the formula. we see the number 4 and 9 are square numbers of 2 and 3.. so if n=1, how would u get square of 2? so (n+1)^2
usually in sequence u most probably have to do trial and error method till you can get it
last question C, it says we have to compare Vn and Tn so when n=1 T=1, and V=4
this shows that it is in a ratio of 1:4, so 1/6 * 4, you will get 2/3n(n+1)(2n+1)
 
Messages
119
Reaction score
23
Points
28
for 14.b the answer is suppose to be 4π^2l/T^2
Firstly you need to remove the root meaning all values beofre the root is doubled, T^2=(2)^2π^2 (l/g)
Next.. move g up to the other side as it is multiply so g*T^2=4π^2l
then move T^2 to divide down.. so the answer is g=4π^2l/T^2
Next sequence in p2 Q18, value of T23, here you just substitute the 'n' with '23', and calculate it you would get 4324.
and (b) (i) you just replace T(n) with T(2) and then subtract it with T(1) meaning you have to substitute all the ns with 2 subtract with all the ns with 1,
you will get 4, and 9.. next the formula. we see the number 4 and 9 are square numbers of 2 and 3.. so if n=1, how would u get square of 2? so (n+1)^2
usually in sequence u most probably have to do trial and error method till you can get it
last question C, it says we have to compare Vn and Tn so when n=1 T=1, and V=4
this shows that it is in a ratio of 1:4, so 1/6 * 4, you will get 2/3n(n+1)(2n+1)

for 14 b : But don't we just square to remove the square root?- why do we also have to multiply by 2?

I already understand parts a and b in question 18 (I should've mentioned that)
Now I understand the other parts as well, thanks
 
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yeah. so we square number 2 and pi and T.. so 2 square will be 4.. there is no need of multiplication by 2 anywhere
 
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for 14.b the answer is suppose to be 4π^2l/T^2
Firstly you need to remove the root meaning all values beofre the root is doubled, T^2=(2)^2π^2 (l/g)
Next.. move g up to the other side as it is multiply so g*T^2=4π^2l
then move T^2 to divide down.. so the answer is g=4π^2l/T^2
Next sequence in p2 Q18, value of T23, here you just substitute the 'n' with '23', and calculate it you would get 4324.
and (b) (i) you just replace T(n) with T(2) and then subtract it with T(1) meaning you have to substitute all the ns with 2 subtract with all the ns with 1,
you will get 4, and 9.. next the formula. we see the number 4 and 9 are square numbers of 2 and 3.. so if n=1, how would u get square of 2? so (n+1)^2
usually in sequence u most probably have to do trial and error method till you can get it
last question C, it says we have to compare Vn and Tn so when n=1 T=1, and V=4
this shows that it is in a ratio of 1:4, so 1/6 * 4, you will get 2/3n(n+1)(2n+1)
for the last part instead if we multiply 2/3 can we do 1/1.5?? we get the same ans.. will cie except?
 
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