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5uckerberg

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Here are some hints a which is the start is 3 and the common difference is 4.
 

5uckerberg

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The third part can be done innovatively well you can start it like this 4+8+12+...+76+80 and then combine the first and last term which will give 84 and do it 10 times as there are 20 terms and you are working with pairs. This gives 840 bit of course you need to take away 20 because the actual amount is one tonne less so therefore 840 take away 20 which gives us 820. See no reference sheet needed for part iii.
 
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5uckerberg

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Here are some hints a which is the start is 3 and the common difference is 4.
Use that to our advantage we can say that we can go with first term as 4 and then second term as 8 and etc. If you have mastery of 4 times tables you will notice that the 25th term is 100 but for this question it is instead 99 so therefore this will be the 26th term because we are using 4n-1 to our advantage here by first using the 4 times tables where we know that 4 times 26 is 104 but take away 1 gives 103 so thereby we will have the 26th year.

For part iii think like Gauss when he was told by his teacher to add the terms from 1 to 100.
 
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Eagle Mum

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3, 7, 11, 15... etc form an arithmetic sequence.

Are you familiar with the formula for calculating the sum of the first n terms of an arithmetic sequence? Do you understand how the formula was derived?

S = n [2a + (n-1)d] /2

S is the sum of the first n terms
n is the number of terms
a is the initial term
d is the difference between terms

For part b) of the question, we are asked to find n and are given:
a = 3
d = 4
S > 100

Substituting these values:
n [6 + (n-1)4] /2 > 100

Simplify the left side:
n (2n + 1) > 100

Substitute the inequality and solve the equivalent quadratic equation: 2n^2 + n - 100 = 0
n = (-1 + sq rt( 1 + 800)) / 4 ~ 6.8

Therefore, the next higher rounded integer n=7 is the solution to the inequality, so the answer is in the seventh year.

For part c) of the question, we are asked to find S and are given:

a = 3
d = 4
n = 20

Substituting these values:
S = 20 [6 + (19)4] /2
= 820
 
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Eagle Mum

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This link is to a 6 minute YouTube video explaining (proving) the above formula for calculating the sum of the first n terms of an arithmetic sequence.
If you don’t already know it, it’s well worth the watch. I skim watched a few and this one was the shortest video that explains it well.
 

5uckerberg

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3, 7, 11, 15... etc form an arithmetic sequence.

Are you familiar with the formula for calculating the sum of the first n terms of an arithmetic sequence? Do you understand how the formula was derived?

S = n [2a + (n-1)d] /2

S is the sum of the first n terms
n is the number of terms
a is the initial term
d is the difference between terms

For part b) of the question, we are asked to find n and are given:
a = 3
d = 4
S > 100

Substituting these values:
n [6 + (n-1)4] /2 > 100

Simplify the left side:
n (2n + 1) > 100

Substitute the inequality and solve the equivalent quadratic equation: 2n^2 + n - 100 = 0
n = (-1 + sq rt( 1 + 800)) / 4 ~ 6.8

Therefore, the next higher rounded integer n=7 is the solution to the inequality, so the answer is in the seventh year.

For part c) of the question, we are asked to find S and are given:

a = 3
d = 4
n = 20

Substituting these values:
S = 20 [6 + (19)4] /2
= 820
Nice way of working however I would go with 4+8+12+...+76+80 using the fact that the series uses a familiar pattern that the user knows of and then use the way how Gauss worked on this question and then take away 20 for the 20 terms.
 

Eagle Mum

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Nice way of working however I would go with 4+8+12+...+76+80 using the fact that the series uses a familiar pattern that the user knows of and then use the way how Gauss worked on this question and then take away 20 for the 20 terms.
I did appreciate your approach of trying to teach the concepts which I think will benefit other viewers, but I am mindful that this OP is sitting HSC this year, so I thought combining the YouTube video and showing how to directly apply the equation would be the most efficient reply for him. I did wonder whether this was the correct maths section on BoS to pose a question on arithmetic series as I thought this was covered in Yr 11 rather than in extn 1, though I’m not all that familiar with the curriculum details.
 

5uckerberg

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I did appreciate your approach of trying to teach the concepts which I think will benefit other viewers, but I am mindful that this OP is sitting HSC this year, so I thought combining the YouTube video and showing how to directly apply the equation would be the most efficient reply for him. I did wonder whether this was the correct maths section on BoS to pose a question on arithmetic series as I thought this was covered in Yr 11 rather than in extn 1, though I’m not all that familiar with the curriculum details.
The arithmetic series @Eagle Mum is taught very early in Year 12 Maths Advanced. I mean if the class is very quick at learning it can be taught around Year 11 term 3. This is a very easy concept so that is why it is put in early.
 

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