Post by: Whirlpool12 on March 23, 2015, 03:11:02 pm
I have an incredibly hard Spesh 1/2 teacher at MHS, and he wanted us to redo the questions that we did wrong in the test without asking him for help. Consequently, i'm not entirely sure on how to do some questions, (even after i've tried them myself) so it would be greatly appreciated if you could help! :)
Q1. For an arithmetic sequence a1, a2, a3, ..., it is given that ax = y and ay=x for some x, y (is a subset) of Z.
Assuming x < y, write the following terms of the sequence in terms of x and y.
a) ax+y
b) ay-x
Q2. Prove that, for all n (subset of N), n(n+1)(n+2)(n+3) is divisible by 24. (NOT BY INDUCTION)
Q3. A besieged fortress is held by 5700 soldiers, and there is enough food for each of them for 66 days. If 20 soldiers die each day, for how many days will the food last?
Thanks, i hope you guys can help! :)
Post by: izzywantsa97 on March 23, 2015, 05:36:43 pm
Post by: Special At Specialist on March 23, 2015, 09:23:09 pm
Q1. Find the sum of all integers between 200 and 400 that are divisible by 6.
I've never properly studied this topic before, but I'll attempt the first question for you:
Divisible by 6 = {204, 210, 216, 222, ..., 396}
We know that the sum of integers from 1 to n is n(n + 1) / 2
We also know that the sum of integers from 1 to m is m(m + 1) / 2
Thus we know that the sum of integers from m to n is n(n + 1) / 2 - m(m + 1) / 2 (+m)
(The (+m) on the end is if we want to include m)
If we want to go in steps of 2, we can just double the sum:
Sum from 2 to 10 in steps of 2 = 2 * sum from 1 to 5 in steps of 1
So 2 + 4 + 6 + 8 + 10 = 2*5*6/2 = 30
Sum from 6 to 30 in steps of 6 = 6 * sum from 1 to 5 in steps of 1
So 6 + 12 + 18 + 24 + 30 = 6*5*6/2 = 90
Sum from 36 to 60 in steps of 6 = 6 * sum from 6 to 10 in steps of 1
= 6 * (sum from 1 to 10 in steps of 1 - sum from 1 to 5 in steps of 1)
So 36 + 42 + 48 + 54 + 60 = 6 * (10*11/2 - 5*6/2) = 3*(110 - 30) = 3*80 = 240
Now that we have figured all of that out, we are ready for our question:
204 / 6 = 34
396 / 6 = 66
So the sum of all integers between 200 and 400 that are divisible by 6:
= sum from 204 to 396 in steps of 6
= 6 * sum from 34 to 66 in steps of 1
= 6 * (sum from 1 to 66 in steps of 1 - sum from 1 to 33 in steps of 1)
= 6 * (66*67/2 - 33*34/2)
= 3 * (66*67 - 33*34)
= 99 * (2*67 - 34)
= 99 * (134 - 34)
= 99 * 100
= 9900
Post by: Whirlpool12 on March 23, 2015, 11:23:32 pm
By any chance do you know how to solve the other three? :)
Post by: StressedAlready on March 24, 2015, 05:34:13 pm
Hi guys,
I have an incredibly hard Spesh 1/2 teacher at MHS, and he wanted us to redo the questions that we did wrong in the test without asking him for help. Consequently, i'm not entirely sure on how to do some questions, (even after i've tried them myself) so it would be greatly appreciated if you could help! :)
Q1. For an arithmetic sequence a1, a2, a3, ..., it is given that ax = y and ay=x for some x, y (is a subset) of Z.
Assuming x < y, write the following terms of the sequence in terms of x and y.
a) ax+y
b) ay-x
Q2. Prove that, for all n (subset of N), n(n+1)(n+2)(n+3) is divisible by 24. (NOT BY INDUCTION)
Q3. A besieged fortress is held by 5700 soldiers, and there is enough food for each of them for 66 days. If 20 soldiers die each day, for how many days will the food last?
Thanks, i hope you guys can help! :)
Isn't this typically a further 3/4 topic? *May be wrong*
I did 1/2 GMA last year and legit, this entire unit was just CAS calculator work with an inserted program. Pester your teacher... or fellow math faculty teachers.
I can however, attempt Q3:
Q3. A besieged fortress is held by 5700 soldiers, and there is enough food for each of them for 66 days. If 20 soldiers die each day, for how many days will the food last?
This is an arithmetic sequence, right?
Do you know of the values a, d, and n? a being the number your started with, d the number you decrease/increase by and n being the number of days in this case?
Isn't there an arithmetic sequence formula? Wouldn't you just input these values into the equation?
Post by: keltingmeith on March 25, 2015, 07:59:51 pm
Isn't this typically a further 3/4 topic? *May be wrong*
Sequences and series ARE studied in Further 3/4 (depending on module selection). "Specialist 1/2" isn't actually specialist 1/2, it's actually the sequence General Mathematics Advanced 1/2. GMA is intended to be just like General Mathematics (the precursor to further), but touches on some more difficult concepts (which pop in specialist), which is why a lot of schools give GMA to people who want to do specialist, and call it "Specialist 1/2".
Post by: kinslayer on March 26, 2015, 12:36:07 am
Let
We know that the binomial coefficient is an integer, so the numerator must be divisible by 4! = 24.
Post by: kinslayer on March 26, 2015, 12:59:02 am
Using the sum formula here: http://en.wikipedia.org/wiki/Arithmetic_progression
Solve for
This is a quadratic, so you will get two solutions. Both are positive; take the smaller one, as for larger n there would be a negative amount of food left.
You should get n = 76 days.
Post by: kinslayer on March 26, 2015, 01:04:51 am
The first of these is 204. Now we have another arithmetic progression with n = 33, a = 204, d = 6. Using the same formula as in Q3 (with positive difference this time) we have
Post by: kinslayer on March 26, 2015, 01:25:59 am
Using the formulae for the terms of the arithmetic sequence we have
part a):
part b)