Question 1 25 marks a stocks of company xyz are initially...

Question 1 (25 marks) a) Stocks of company XYZ are initially issued at price of RM18. The value of the stock grows by 20% every year. i. Show that the value of a stock follows a geometric sequence. [2 Marks] ii. Calculate the value of the stock ten years after the initial public offering. [5 Marks] b) In an electrical circuit, resistors are connected in series. The resistance of each resistor increases by a constant value of 5$$\Omega$$. The first resistor has a resistance of 10$$\Omega$$. i. Show that the total resistance of the first n resistors forms an arithmetic series. [1 Mark] ii. Calculate the total resistance if 8 resistors are connected in series. [4 Marks] c) Identify the initial term and the common difference of the following arithmetic sequence and thus write down the term 20th of this sequence. 5, 9, 13, 17, 21,... [5 Marks]

Transcript

00:01 We have a few questions on ap gp.

00:05 In part a, we're given that there are four positive numbers.

00:09 So let's put up the four numbers like this.

00:14 And the first three forms a gp with the first term a and a common ratio r.

00:22 So if the first term is a, the second term will be ar and the third will be ar square.

00:29 Now the last three numbers form a ap.

00:33 So that means over here we have an ap with a common difference of 9 so over here if i were to list out the last three numbers i'll have ar ar plus 9 and ar plus 9 plus 9 so ar plus 18 we're also given that the last number is 3 times the first so we have ar plus 18 the last number over here is 3 times the first number so it's equals to 3a so let's put that as equation one and so we need one more equation since there are two variable here so we can take this one is equal to this so we have a r square plus is equals to ar plus nine second equation so let's for the first equation let's express r in terms of the rest so it will be like this okay let's sub r into two we will have this okay so now let's expand the left side and on the right side you can see that this cancels off you are left with 3a minus 9 so and here you can see that you can cancel this with this and so the left side would be 9a square minus 108 a plus 3 to 4.

02:53 I'm going to bring my list a up to the right side.

02:57 You will get this.

03:00 Okay.

03:01 So you can solve the a.

03:05 So after this, you can solve for the a.

03:08 And you will get your a is 12 or a is 4 .5.

03:16 Just bring all the a square and the a all to the left side equals to zero and solve as a quadratic equation.

03:23 You will get this.

03:24 Now, which one to choose? now, it will come down to what is our r.

03:30 Now, sub -a equals 12 into here to get your r, your r will be 3 over 2.

03:38 Or sub -a is 4 .5 into here, you will get r -equest to minus 1.

03:43 Now, you will reject this pair because all the numbers are positive, so you cannot have r -equest to minus 1, otherwise this part will become negative.

03:55 So reject since all terms are positive so therefore our a is 12 so that's for part a alright for part b let me erase alright let's look at part b now we are given the first term of an ap is 21 so let's put back that and the last term so there are many terms is 22 and the sum of all the terms is so we want to find the common difference.

04:40 So we are going to let d be the common difference.

04:43 Now, for the sum of n terms, it's actually n over 2, the first term, plus the last term, where a is the first term.

04:55 And our n -term is a plus n -minus 1, d, where this is the common difference.

05:04 So our s -n is 2 -0 -2 -1.

05:07 So we have this.

05:11 Okay, so our first term is 21.

05:13 Our last term is 22 and this is equals to 2021 so we have 43 n is 4042 i bring the 2 up to the right side yeah so for 3 n is 444 our n is 94 so the 94 term will be a which is 21 plus n minus 1 will be 94 minus 1 my 94 term is 22 so this whole thing is equal to 22 so solving this whole part here, my d is 1 over 93.

06:00 So that's what part.

06:05 All right.

06:06 So let's look at part c.

06:14 For part c, we have a gp's infinite sum.

06:18 So as infinity, which is equals to a over 1 minus r, where a is the first term and r is the common ratio.

06:27 And it's 60.

06:30 So we will have a equals to 6.

06:34 60, 1 minus r.

06:37 Now, we are also given that the sound of the first two terms is 45, so we have a plus ar is 45.

06:46 Okay, i'm going to sub this a into here...