The Student Room Group

Sequences and series

At the beginning of the year 2000 a company bought a new machine for £15 000. Each
year the value of the machine decreases by 20% of its value at the start of the year.
(a) Show that at the start of the year 2002, the value of the machine was £9600.

When the value of the machine falls below £500, the company will replace it.
(b) Find the year in which the machine will be replaced.
To plan for a replacement machine, the company pays £1000 at the start of each year
into a savings account. The account pays interest at a fixed rate of 5% per annum. The
first payment was made when the machine was first bought and the last payment will be
made at the start of the year in which the machine is replaced.
(c) Using your answer to part (b), find how much the savings account will be worth
immediately after the payment at the start of the year in which the machine is
replaced
Reply 1
Note sure where to start!?!
At the beginning of the year 2000 a company bought a new machine for £15 000. Each year the value of the machine decreases by 20% of its value at the start of the year.
(a) Show that at the start of the year 2002, the value of the machine was £9600.


beginning of 2000: value = £15,000
beginning of 2001: value = 15,000 x 80% = £12,000 (because the value has decreased by 20%, the new value will be 80% of the old value)
beginning of 2002: value = 12,000 X 80% = £9,600
instead of saying 80%, you can say 'x 4/5', it's the same thing.

That's something of a starting point. For part b, you keep going like that until the value is < 500. If you're good with logs, you can use them as well, but i don't think that's the 'proper' way of doing it.

For part c, you need to make a recurrence relation using the fact that at the beginning of each year, £1000 is added to the account, and at the end of that year, 5% is added on as interest. Then you do that for each year, up the the year you got in part b.

I'll try doing the question properly all the way through if you want, but i'm not very good at thse ones and will probably get the wrong answer and make things worse. I don't like they way stuff is attached to different points in the year, in confuses me.
[I have a chemistry project to write up. Doing maths online is a great way of avoiding it. That's sad, isn't it?]
Reply 3
Hello!

a) It's a geometric series, a = 15000 and r=0.8
so after 2 years, the machine is worth 15000 x 0.8^2 = 9600

b) In the n-th year, the machine is worth ar^(n-1) < 500
15000 x 0.8^(n-1) < 500
0.8^(n-1) < 1/30
(n-1) log 0.8 < log (1/30)
n-1 < (log 1/3)/(log 0.8)
n-1 < 15.24 so after 16 years.

(Check: 15000 x 0.8^16 = 422.21 )

c) Year 1 start: 1000
Year 1 end: 1.05 x 1000
Year 2 start: 1000 + 1.05x1000
Year 2 end: 1.05x(1000+ 1.05 x 1000) = 1.05x1000 + 1.05^2 x 1000
Year 3 start: 1000+ 1.05x 1000 + 1.05^2 x 1000

so year 5 start (say) =
1000 + 1.05x1000 + 1.05^2 x 1000 + 1.05^3 x 1000 + 1.05^4 x 1000
= 1000 (1 + 1.05 + 1.05^2 + 1.05^3 + 1.05^4)
this is a geometric series, a=1, r=1.05

so after 16 years,
amount is worth
1000 [sum to 16 terms of that geometric series]
= 1000 [1(1 - 1.05^16)/(1-1/05)]
= 1000 (21.57856)
= 21578.56

love danniella
12 years ago and still useful 😃
♥️