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Sequences & Series

I just wanted to check my answer for a question,

Kath places £100 pound into a bank account to save for a family holiday. Each subsequent month she increases her payments by 5%.
Assuming the bank does not pay interest, Find:

a.) The amount of money after 9 months: I got 1102.656432

b.)Show that n > log4/log1.05: I got to a stage where -1.05^n > -4 and assumed it is the same as 1.05^n > 4, then continued to use logs to show that n > log4/log1.05 or am i wrong to do that, otherwise there seems nothing else you can do, i know when you divide by a negative you flip the sign, but if divided by -1 and both sides of -1.05^n > -4 that means i would end up with 1.05^n < 4, which leads me with a wrong answer at the end.

c.)Maggie begins to save at the same time as Kath. She initially places £50 into the accountm and plant ot increaser her payments by a constant amount each month. Given that she would like to reach a total of £6000 in 29 months by how much should she increase her payment by each month? I don't know that to do for this, I had a go and got 1, don't think its right though. (Do you use part b to help here?)
Reply 1
(a) is fine

Do you have the question for (b) ?
Reply 2
(c) Try using the Sn formula for arithmetic sequences. Solve Sn =6000 to find the common difference d.
Reply 3
Part b states: month n is the first month in which there is more than £6000 in the account.Show that n> log4/log1.05I got to -1.05^n > -4, which I presumed is the same as 1.05^n > 4, and got the answer, but am I wrong, as apparently you are meant to flip the sign when dividing by a negative, could you talk me through the steps if you did it, thanks
Reply 4
I got d as 11.21, is that correct?
Original post by vc94
(c) Try using the Sn formula for arithmetic sequences. Solve Sn =6000 to find the common difference d.
Reply 5
Original post by Henwi
I got d as 11.21, is that correct?


Yes, 11.21
Reply 6
Original post by Henwi
Part b states: month n is the first month in which there is more than £6000 in the account.Show that n> log4/log1.05I got to -1.05^n > -4, which I presumed is the same as 1.05^n > 4, and got the answer, but am I wrong, as apparently you are meant to flip the sign when dividing by a negative, could you talk me through the steps if you did it, thanks


Sn = 100((1.05^n) - 1) /(1.05-1) > 6000

Rearrange this, should work!
Reply 7
I think you got your numbers mixed up, as the geometric sequence formula is: a(1-r^n)/1-r

In which i did 100(1-1.05^n)/1-1.05 > 6000, and ran into the same thing, Ill show you what happens:

1.) times the 6000 by (1-1.05) or -0.05 which = -300

100(1-1.05^n) > -300

2.) Divide by 100 to get:

1-1.05^n > -3

3.)minus 1 from both sides to get:

-1.05^n > -4

4.) This is where I got confused and guessed that -1.05^n > -4 is the same as 1.05^n > 4

and then continued to get n > log4/log1.05 using logs :smile:

Could you tell me if i was righ to presume the step i took in part 4?



Original post by vc94
Sn = 100((1.05^n) - 1) /(1.05-1) > 6000

Rearrange this, should work!
Reply 8
For the geometric sum you can use
a(1-r^n)/1-r or a(r^n -1)/r-1 they are equivalent.
The second version is probably easier when r>1 so you avoid messing with signs.
Reply 9
After multiplying by (1-1.05) which is negative, you should get
1-1.05^n < -3
then
-1.05^n < -4
1.05^n > 4
hi people just had to ask you something, i am still a bit confused how you got part a)... because I got £155.13.. which means its truly wrong.
no you are right

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