#### KAIO7

Which one ?

##### Member
Which one ?
Well I did the entire chapter from the textbook and those were the 10 questions I didn’t get. So all of them but you can’t be bothered doing them all choose any one. Thank you in advance.

#### Drongoski

##### Well-Known Member
OK - I'll do Q2

The general term of this A.P. is 5k + 3

The 5 in the 5k indicates that each time k increases by 1, the term increases by 5; this means the common difference is 5. But you don't need this info to work out the sum.

The 1st term (with k=10) a = 5x10 + 3 = 53
the last term (k=30) l = 5x30 + 3 = 153
Total number of terms = 30 - 10 + 1 = 21

.: the sum = (21/2) (53 + 153) = 2163

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#### Drongoski

##### Well-Known Member
Well I did the entire chapter from the textbook and those were the 10 questions I didn’t get. So all of them but you can’t be bothered doing them all choose any one. Thank you in advance.
Maybe you should rephrase that as "are unable to do". You are presenting a rather long list of questions.

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#### KAIO7

##### Member
I apologise if this comes across as being rude or blunt, but based on the long list of questions you have presented, I’d say that you have a gap in your understanding of series and sequences. I suggest you revise the main theory and formulae and redo some of the examples you did in class, then try attempting these questions.

Anyway, I’ll provide you with some guidance on how to do problems that involve finding the common difference and I’ll reference the solution of 1a in my
explanation
Let’s first look at this simple example, 1+3+5+7+9+11+13

It’s quite obvious that we are adding 2 each time, so 2 becomes our common difference. In maths we like to generalise things, so how can I mathematically find the common difference and then come up with a general formula ? Well It turns out it’s actually simple, all I need is any 3 consecutive terms of the series. For example3,5 and 7. The reason I did that is because I need to ensure that the common difference is actually ‘common’ for the whole series. So 7-5=5-3=2
So to generalise this, so it can work for any AP, we’ll re-write it like this; T(n+1)-T(n)=T(n)-T(n-1) where T(n) denotes the nth term of the series.

Now if the series is specifed to be an arithemtic series in the question, then you only need to take two consecutive terms, so in the example I’ve provided, we can deduce that from 7-5 or 9-7 or 5-3, etc..

I should also mention that there is a difference between series and sequences, a series is basically something like the example I’ve provided where as a sequence is just an ordered set of numbers, so 1,3,5,7,9,... is an example f a sequence

Now, your question states that what you’re dealing with is an AP, so what you need to do is to take any two consecutive terms,
For example, let’s take log_a (54) and log_a (18)
To find the common difference, we’ll use T(2)-T(1)=log_a(18) - log_a (54)
This can be simplified using the log rules, so it becomes log_a(18/54) which is log_a(1/3) and that’s the common difference.

Now to find a formula for the nth term, we should recall this: T(n)= a+(n-1)*d where a is the first term in the series and d is the common difference.
We just found the common difference and we know what the first term is, so all you have to is substitute them into our formula and simplify futher using the log rules.

If this type of guided explanation seems to help you, then maybe you should watch a couple of YouTube videos that explain series and sequences using a step by step approach.

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#### KAIO7

##### Member
Here are some generous hints for questions 9 & 10.

For a series to have a limiting sum, r has to be between 1 and -1. In an intuitive sense, this means that the terms in the series are getting smaller and smaller until their value is ‘too small’ to affect the overall sum. To achieve this, I need to be multiplying the terms in the series by a number that will make them smaller ( as in they approach zero), and well that number has to obviously be a fraction, that’s why r is in between 1 and -1.

Now the formula for the limiting sum of a geometric series is a/(1-r) where a is the first terms and r is the ratio, |r|<1.

For question 9, they want you to verify whether the series has a limiting sum or not ( hint: use the fact the r has to be between 1 and -1)
And then if it does have a limiting sum, you need to find it.
Second hint: 7/sqrt(7) can be rewritten as sqrt(7) (rationalise the denominator).

Q10a
S=a/(1-r) is the sum of all terms of the series including the first term.
But the question says that the sum of all terms except the first is equal to 5a
So S-a=5a and we know what S is. I’ll leave the algebra to you.

##### Member
I figured out Q4, Q6 and Q9. Still working on the rest

#### KAIO7

##### Member
I figured out Q4, Q6 and Q9. Still working on the rest
Great! Keep it up

##### Member
Im really stuck with 1a, I get that d = -loga 3 but how do you work out Tn??? I get some funky answer

##### Member
I still don't get Q3, Q7, Q8 and Q10. I figured out the other 6 the remaining four I'm stuck. Anyone willing to do those 4 showing all working out.

#### KAIO7

##### Member
$\bg_white T_n=a+(n-1)\times d$

Now before I find $\bg_white T_n$, I want you to note that $\bg_white -log_a(3)=-1\times (log_a(3))$ and thats the same as $\bg_white log_a(\frac{1}{3})$ using the log rules.

Now to find $\bg_white T_n$ we just use the formula, $\bg_white T_n= log_a(54)+(n-1)\times -log_a(3)$

Expanding this out, $\bg_white T_n= log_a(54)-n\times log_a(3)+ log_a(3)$

Now we can simplify further using the log rules, $\bg_white T_n= log_a(54\times 3)-nlog_a(3)$

Therefore, $\bg_white T_n= log_a(162)-nlog_a(3)$

Let's test if what we got is correct, for example if I'm looking for $\bg_white T_2$ which is basically the second term in our sequence. when we substitute n=2 into the formula we obtained, we should get the second term which is $\bg_white log_a(18)$

$\bg_white T_2=log_a(162)-2log_a(3)$ which is the same as $\bg_white log_a(\frac{162}{3^2})$

simplify that and you'll get $\bg_white T_2=log_a(18)$ and that's correct meaning our formula works.

So $\bg_white T_n= log_a(162)-nlog_a(3)$ and that's the same as $\bg_white log_a(2\times 3^4)-nlog_a(3)$

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#### KAIO7

##### Member
For Q7, the first question that we should ask is: What sort of series am I dealing with ? well it's an AP because the common difference is obviously 7 ( Refer to my first answer if you need more explanation in regards to determining the nature of the progression in a series)

Now the second thing that should come to mind are the two formulae that allow us to find the sum of an AP

$\bg_white S_n=\frac{n}{2}(2a+(n-1)\times d)$

$\bg_white S_n=\frac{n}{2}(a+l)$ where l is the last term of the series. (This is actually the same as the first but just expressed differently)

I can't explicitly determine what my last term is going to be, and my first formula doesn't explicitly deal with the common difference which is what the question is technically asking us about, so It'll be wiser to use the first.

$\bg_white S_n= \frac{n}{2}(2\times 7+(n-1)\times7)$

$\bg_white S_{n+1}=\frac{n+1}{2}(2\times7+ ((n+1)-1)\times7)$

We want to prove that $\bg_white S_{n+1}-S_n$ is 7

So let's use the expressions we obtained:
$\bg_white S_{n+1}-S_n$ = $\bg_white (\frac{n+1}{2}(14+ 7n)) - (\frac{n}{2}(14+7(n-1))$

simplifying this: $\bg_white S_{n+1}-S_n$= $\bg_white ( 7(n+1)+\frac{7n(n+1)}{2})-(7n+\frac{7n(n-1)}{2})$

Terms are going to cancel each other out leaving $\bg_white S_{n+1}-S_n =7$ which means the sum of the first n+1 terms differs from the sum of the first n term by 7.

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#### KAIO7

##### Member
Q8 is another application of the limiting sum of a series. So here is what we have, the ball is being dropped from a height of 8 meters, when it hits the ground, and bounces back up, what do you think would happen? well it wouldn't obviously bounce back to my hand, its height from where it bounced would decrease. I'm being told in the question that each time it bounces back up, the height decreases and becomes half of that in the previous bounce, so let's convert that to numbers:
Distance travelled: Down and up
$\bg_white a_1=8+4$
$\bg_white a_2=(8+4)\times\frac{1}{2}$

$\bg_white a_3=a_2\times\frac{1}{2}=((12)\times\frac{1}{2}) \times\frac{1}{2}$

$\bg_white a_4=a_3\times\frac{1}{2}=(12\times \frac{1}{2} \times \frac{1}{2}) \times \frac{1}{2}$

and so on.

Now to answering the first part of the question: the ball travelled 8 meters from my hand to the ground and then bounced back up by half of that, i.e(4 meters)
so the total distance travelled is $\bg_white a_1= 8+4=12$

If I want to look at the distance travelled overall, I'd end up with a series with geometric progression, as I'm multiplying each term by 1/2

$\bg_white D=12+6+3+....$

$\bg_white \frac{T_n}{T_{n-1}} =0.5$ which means it's indeed a geometric series with r=1/2

Obviously you can do the algebra yourself using the definition of T_n in a geometric series; $\bg_white T_n=ar^{n-1}$

As for the second part, this is where limiting sum shows up. If you have a close look at the series, you'll notice that the terms are decreasing and approaching zero, which means 'eventually' the value of the terms would be too small consider in the overall sum. Also this is geometric series with where r is in between 1 and -1, which means it satisfies the conditions to have a limiting sum

So the distance traveled eventually is equivalent to the limiting sum; $\bg_white S=\frac{a}{1-r}$

$\bg_white S=\frac{12}{1-\frac{1}{2}}= 24$
Therefore the overall distance travelled by the ball is 24

The answer was edited because of a carless error.

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##### Member
I got the same solution as you but the solution is said to be Tn= log a (2) + (4-n) log a (3). How did they get this

##### Member
$\bg_white T_n=a+(n-1)\times d$

Now before I find $\bg_white T_n$, I want you to note that $\bg_white -log_a(3)=-1\times (log_a(3))$ and thats the same as $\bg_white log_a(\frac{1}{3})$ using the log rules.

I got the same solution as you but the solution is said to be Tn= log a (2) + (4-n) log a (3). How did they get this???
Now to find $\bg_white T_n$ we just use the formula, $\bg_white T_n= log_a(54)+(n-1)\times -log_a(3)$

Expanding this out, $\bg_white T_n= log_a(54)-n\times log_a(3)+ log_a(3)$

Now we can simplify further using the log rules, $\bg_white T_n= log_a(54\times 3)-nlog_a(3)$

Therefore, $\bg_white T_n= log_a(162)-nlog_a(3)$

Let's test if what we got is correct, for example if I'm looking for $\bg_white T_2$ which is basically the second term in our sequence. when we substitute n=2 into the formula we obtained, we should get the second term which is $\bg_white log_a(18)$

$\bg_white T_2=log_a(162)-2log_a(3)$ which is the same as $\bg_white log_a(\frac{162}{3^2})$

simplify that and you'll get $\bg_white T_2=log_a(18)$ and that's correct meaning our formula works.

So $\bg_white T_n= log_a(162)-nlog_a(3)$

I got the same solution as you but the solution is said to be Tn= log a (2) + (4-n) log a (3). How did they get this

#### Drongoski

##### Well-Known Member
Q8

Shouldn't it be?:

d1 = 8+4 = 12
d2 = 0.5 x 12 = 6
d3 = 0.5 x 6 = 3

etc

.: a = 12 r = 0.5

.: Sinf = 12/(1 - 0.5) = 24

or

after 1st drop of 8: total = 8 + (2x4 + 2x2 + 2x1 + . . . ) = 8 + 8/(1 - 0.5) = 8 + 16 = 24

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#### KAIO7

##### Member
Q8

Shouldn't it be?:

d1 = 8+4 = 12
d2 = 0.5 x 12 = 6
d3 = 0.5 x 6 = 3

etc

.: a = 12 r = 0.5

.: Sinf = 12/(1 - 0.5) = 24

or

after 1st drop of 8: total = 8 + (2x4 + 2x2 + 2x1 + . . . ) = 8 + 8/(1 - 0.5) = 8 + 16 = 24
Yes, apologies for the carless error. Thought the question was asking about the height of the ball from the ground expressed as the distance travelled, will edit that now. Thanks for pointing that out.

#### KAIO7

##### Member
I got the same solution as you but the solution is said to be Tn= log a (2) + (4-n) log a (3). How did they get this
$\bg_white T_n=log_a(2)+4log_a(3)-nlog_a(3)$
which is the same as $\bg_white log_a(2)+log_a(3^4)-nlog_a(3)$
and if you use the log rule on more time you get $\bg_white log_a(2\times 3^4)-nlog_a(3)$
And that's the same as the result we both got.