series and seqeunces (1 Viewer)

stressedadfff · Sep 23, 2021

hey guys, for this question, i get that you use the sum formula for aps but wouldnt n + 21? can someone send a solution pls im confused...

Screen Shot 2021-09-23 at 9.12.50 pm.png

jimmysmith560 · Sep 23, 2021

Would this help?

Part (a)

Part (b)

stressedadfff · Sep 23, 2021

jimmysmith560 said:
Would this help?

Part (a)

View attachment 32271

Part (b)

View attachment 32274

omg yes thank you could you please explain why you didnt just sub n=21 weeks tho

laterz laterz · Sep 23, 2021

jimmysmith560 said:
Would this help?

Part (a)

View attachment 32271

Part (b)

View attachment 32274

hey jimmy ur part b is wrong

n is 21 and not 11 (last line of the question)

jimmysmith560 · Sep 23, 2021

laterz laterz said:
hey jimmy ur part b is wrong

n is 21 and not 11 (last line of the question)

Interesting, thank you for pointing it out! (not my solutions by the way, but those of the paper from which this question is sourced

).

jimmysmith560 · Sep 23, 2021

It seems as though the solution for part (b) above is correct? @laterz laterz

CM_Tutor · Sep 23, 2021

@jimmysmith560, it is correct.

jimmysmith560 · Sep 23, 2021

Hivaclibtibcharkwa said:
How did you find the source of this question and the paper??

This question is from the Shore 2015 Mathematics Trial HSC Examination, meaning it is from the old syllabus. It's available online.

CM_Tutor said:
@jimmysmith560, it is correct.

Thank you @CM_Tutor for confirming!

jimmysmith560 · Sep 23, 2021

Hivaclibtibcharkwa said:
Yea but I don’t see where the person has given this information, the source. Did you reverse image search it or something?

I just searched the question on Google. I think it would be quite difficult to find it through a reverse image search, especially if the image is of somewhat low quality.

stressedadfff · Sep 23, 2021

jimmysmith560 said:
Interesting, thank you for pointing it out! (not my solutions by the way, but those of the paper from which this question is sourced ).

yes thats what i realised too, i dont really get it but i feel like its right?

jimmysmith560 · Sep 23, 2021

stressedadfff said:
yes thats what i realised too, i dont really get it but i feel like its right?

It's correct, CM_Tutor confirmed it

As Hivaclibtibcharkwa mentioned, you cannot sub in n = 21 because of the changing difference from 2 to -2 after the 11th week, meaning they have to be added separately. Logically, there is a reason behind the fact that you're asked to find how much snow falls in the 11th week in part (a).

CM_Tutor · Sep 24, 2021

stressedadfff said:
yes thats what i realised too, i dont really get it but i feel like its right?

If you think through what is described, we have

Snow fall in week 1 = 5 cm
Snow fall in week 2 = 5 + 2 = 7 cm
Snow fall in week 3 = 7 + 2 = 9 cm
Snow fall in week 4 = 9 + 2 = 11 cm
... pattern continues ...
Snow fall in week 9 = 19 + 2 = 21 cm
Snow fall in week 10 = 21 + 2 = 23 cm
Snow fall in week 11 = 23 + 2 = 25 cm

We are told that the pattern changes at the middle week of the season from increasing snow fall to decreasing snow fall.
In a 21 week season, week 11 is the middle week as there are 10 weeks before it and 10 weeks after it. So, we then get...

Snow fall in week 12 = 25 - 2 = 23 cm
Snow fall in week 13 = 23 - 2 = 21 cm
Snow fall in week 14 = 21 - 2 = 19 cm
... pattern continues ...
Snow fall in week 19 = 11 - 2 = 9 cm
Snow fall in week 20 = 9 - 2 = 7 cm
Snow fall in week 21 = 7 - 2 = 5 cm

Our total snow fall over the season is then

$5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5$

This entire series is NOT an AP as there it is not a constant common difference. So, we need to manipulate it to make one or more series that can be found using AP formulae. This can be done in a number of ways, and whichever you adopt will lead to the correct answer... though the last one is the most efficient:

Approach 1(a): Divide the series where the common difference changes

$5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5$
$= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23 + 25\bigg) + \bigg(23 + 21 + 19 + ... + 9 + 7 + 5\bigg)$

Each set of brackets contains an AP:

First bracket $=5 + 7 + 9 + ... + 23 + 25$ is an AP of 11 terms (weeks 1 to 11, inclusive) which has $a = 5$ , $d = 2$ , and $n = 11$ .

$S_{11} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{11}{2}\big[2 \times 5 + (11-1) \times 2\big] = \cfrac{11}{2} \times 30 = 165$

Second bracket $=23 + 21 + 19 + ... + 7 + 5$ is an AP of 10 terms (weeks 12 to 21, inclusive) which has $a = 23$ , $d = -2$ , and $n = 10$ .

$S_{10} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{10}{2}\big[2 \times 23 + (10-1) \times -2\big] = 5 \times 28 = 140$

$\begin{align*} \text{Thus, total snowfall}\ &= 5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5 \\ &= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23 + 25\bigg) + \bigg(23 + 21 + 19 + ... + 9 + 7 + 5\bigg) \\ &= S_{11} + S_{10} \\ &= 165 + 140 \\ &= 305\ \text{cm} \end{align*}$

Approach 1(b): Slightly different division of the series

$5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5$
$= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + \bigg(25 + 23 + 21 + ... + 9 + 7 + 5\bigg)$

Each set of brackets contains an AP:

First bracket $=5 + 7 + 9 + ... + 23$ is an AP of 10 terms (weeks 1 to 10, inclusive) which has $a = 5$ , $d = 2$ , and $n = 10$ .

$S_{10} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{10}{2}\big[2 \times 5 + (10-1) \times 2\big] = 5 \times 28 = 140$

Second bracket $=25 + 23 + 21 + ... + 7 + 5$ is an AP of 11 terms (weeks 11 to 21, inclusive) which has $a = 25$ , $d = -2$ , and $n = 11$ .

$S_{11} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{11}{2}\big[2 \times 25 + (11-1) \times -2\big] = \cfrac{11}{2} \times 30 = 165$

$\begin{align*} \text{Thus, total snowfall}\ &= 5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5 \\ &= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + \bigg(25 + 23 + 21 + ... + 9 + 7 + 5\bigg) \\ &= S_{10} + S_{11} \\ &= 140 + 165 \\ &= 305\ \text{cm} \end{align*}$

Approach 2: Less obvious but quicker...

$\begin{align*} \text{Total Snowfall }\ &= 5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5 \\ &= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + 25 + \bigg(23 + 21 + ... + 9 + 7 + 5\bigg) \\ &= 2\bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + 25 \end{align*}$

This rearrangement works by recognising that the increasing pattern of weeks 1 to 10 is also the pattern for weeks 12 to 21, but in reverse. Thus, we have that $5 + 7 + 9 + ... + 19 + 21 + 23 = 23 + 21 + 19 + ... + 9 + 7 + 5$ . We now have only one set of brackets and a single AP, plus an extra term that needs to be added:

$5 + 7 + 9 + ... + 23$ is an AP of 10 terms (weeks 1 to 10, forwards, which is the same as weeks 12 to 21, backwards - inclusive in each case) which has $a = 5$ , $d = 2$ , and $n = 10$ .

$S_{10} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{10}{2}\big[2 \times 5 + (10-1) \times 2\big] = 5 \times 28 = 140$

$\begin{align*} \text{Thus, total snowfall}\ &= 5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5 \\ &= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + 25 + \bigg(23 + 21 + ... + 9 + 7 + 5\bigg) \\ &= 2\bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + 25 \\ &= 2 \times S_{10} + 25 \\ &= 2 \times 140 + 25 \\ &= 305\ \text{cm} \end{align*}$

stressedadfff · Sep 24, 2021

CM_Tutor said:
If you think through what is described, we have

Snow fall in week 1 = 5 cm

Snow fall in week 2 = 5 + 2 = 7 cm

Snow fall in week 3 = 7 + 2 = 9 cm

Snow fall in week 4 = 9 + 2 = 11 cm

... pattern continues ...

Snow fall in week 9 = 19 + 2 = 21 cm

Snow fall in week 10 = 21 + 2 = 23 cm

Snow fall in week 11 = 23 + 2 = 25 cm

We are told that the pattern changes at the middle week of the season from increasing snow fall to decreasing snow fall.
In a 21 week season, week 11 is the middle week as there are 10 weeks before it and 10 weeks after it. So, we then get...

Snow fall in week 12 = 25 - 2 = 23 cm

Snow fall in week 13 = 23 - 2 = 21 cm

Snow fall in week 14 = 21 - 2 = 19 cm

... pattern continues ...

Snow fall in week 19 = 11 - 2 = 9 cm

Snow fall in week 20 = 9 - 2 = 7 cm

Snow fall in week 21 = 7 - 2 = 5 cm

Our total snow fall over the season is then

$5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5$

This entire series is NOT an AP as there it is not a constant common difference. So, we need to manipulate it to make one or more series that can be found using AP formulae. This can be done in a number of ways, and whichever you adopt will lead to the correct answer... though the last one is the most efficient:

Approach 1(a): Divide the series where the common difference changes

$5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5$
$= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23 + 25\bigg) + \bigg(23 + 21 + 19 + ... + 9 + 7 + 5\bigg)$

Each set of brackets contains an AP:

First bracket $=5 + 7 + 9 + ... + 23 + 25$ is an AP of 11 terms (weeks 1 to 11, inclusive) which has $a = 5$ , $d = 2$ , and $n = 11$ .

$S_{11} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{11}{2}\big[2 \times 5 + (11-1) \times 2\big] = \cfrac{11}{2} \times 30 = 165$

Second bracket $=23 + 21 + 19 + ... + 7 + 5$ is an AP of 10 terms (weeks 12 to 21, inclusive) which has $a = 23$ , $d = -2$ , and $n = 10$ .

$S_{10} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{10}{2}\big[2 \times 23 + (10-1) \times -2\big] = 5 \times 28 = 140$

$\begin{align*} \text{Thus, total snowfall}\ &= 5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5 \\ &= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23 + 25\bigg) + \bigg(23 + 21 + 19 + ... + 9 + 7 + 5\bigg) \\ &= S_{11} + S_{10} \\ &= 165 + 140 \\ &= 305\ \text{cm} \end{align*}$

Approach 1(b): Slightly different division of the series

$5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5$
$= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + \bigg(25 + 23 + 21 + ... + 9 + 7 + 5\bigg)$

Each set of brackets contains an AP:

First bracket $=5 + 7 + 9 + ... + 23$ is an AP of 10 terms (weeks 1 to 10, inclusive) which has $a = 5$ , $d = 2$ , and $n = 10$ .

$S_{10} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{10}{2}\big[2 \times 5 + (10-1) \times 2\big] = 5 \times 28 = 140$

Second bracket $=25 + 23 + 21 + ... + 7 + 5$ is an AP of 11 terms (weeks 11 to 21, inclusive) which has $a = 25$ , $d = -2$ , and $n = 11$ .

$S_{11} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{11}{2}\big[2 \times 25 + (11-1) \times -2\big] = \cfrac{11}{2} \times 30 = 165$

$\begin{align*} \text{Thus, total snowfall}\ &= 5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5 \\ &= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + \bigg(25 + 23 + 21 + ... + 9 + 7 + 5\bigg) \\ &= S_{10} + S_{11} \\ &= 140 + 165 \\ &= 305\ \text{cm} \end{align*}$

Approach 2: Less obvious but quicker...

$\begin{align*} \text{Total Snowfall }\ &= 5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5 \\ &= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + 25 + \bigg(23 + 21 + ... + 9 + 7 + 5\bigg) \\ &= 2\bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + 25 \end{align*}$

This rearrangement works by recognising that the increasing pattern of weeks 1 to 10 is also the pattern for weeks 12 to 21, but in reverse. Thus, we have that $5 + 7 + 9 + ... + 19 + 21 + 23 = 23 + 21 + 19 + ... + 9 + 7 + 5$ . We now have only one set of brackets and a single AP, plus an extra term that needs to be added:

$5 + 7 + 9 + ... + 23$ is an AP of 10 terms (weeks 1 to 10, forwards, which is the same as weeks 12 to 21, backwards - inclusive in each case) which has $a = 5$ , $d = 2$ , and $n = 10$ .

$S_{10} = \cfrac{n}{2}\big[2a + (n-1)d\big] = \cfrac{10}{2}\big[2 \times 5 + (10-1) \times 2\big] = 5 \times 28 = 140$

$\begin{align*} \text{Thus, total snowfall}\ &= 5 + 7 + 9 + 11 + ... + 21 + 23 + 25 + 23 + 21 + 19 + ... + 9 + 7 + 5 \\ &= \bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + 25 + \bigg(23 + 21 + ... + 9 + 7 + 5\bigg) \\ &= 2\bigg(5 + 7 + 9 + 11 + ... + 21 + 23\bigg) + 25 \\ &= 2 \times S_{10} + 25 \\ &= 2 \times 140 + 25 \\ &= 305\ \text{cm} \end{align*}$

thank you i understand!

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series and seqeunces (1 Viewer)

stressedadfff

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jimmysmith560

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