Geometric Sequence to solve an Interest Problem

[AN630078]

Homework Statement

Hello, I have been revising over models and real-life situations using geometric sequences when I found he problem below. The first part was causing me some initial difficulty and I am uncertain whether the method I have implemented would be correct. Would anyone be able to advise me how to improve my workings in case I am faced with similar problems in future exercises.

£2000 is invested into an account and each year the sum is increased by 6% of the amount in the account.
Find how much will be in the account after ten years.
Find how long it will take the account to reach £5000

Relevant Equations

bn=2000*1.06^n-1

To find how much would be in the account after ten years, let the balance in the account at the start of year n be bn.
Then b1=2000
I believe that this a compound interest problem.
Common ratio r = 1.06
bn =2000*1.06^n−1
Thus, b10 =2000×1.06^9 = £3378.95791
The balance of the account at the start of year 10 is £3379 (to the nearest £)The balance exceeds £5000 at the start of year n if bn>5000
bn=2000*1.06^n-1
(Replace the inequality with an equation)
Solve bn=5000
2000*1.06^n-1=5000
1.06^n-1=5000/2000
1.06^n-1=5/2
If p^x=q then x=logp(q)
n-1=log1.06(5/2)
n=log1.06(5/2)+1
n=16.7252
The sequence is increasing so b16=2000*1.06^15=4793.1163<5000
and b17=2000*1.06^16=5080.70>5000
Therefore, the balance first exceeded £5000 at the beginning of year 17

 

[pasmith]

You have the right idea, but have not counted your years correctly.

The question asks firstly for the balance after 10 years. You have given the balance at start of the 10th year, which is the balance after only 9 years. (In general, you want to avoid "at the start of year X" and instead use "after (X-1) years".)

If you let the initial amount be $b_0$, then $b_n$ would be the balance after $n$ years rather than the balance after $n-1$ years, which is maybe less confusing.

If the question doesn't say "to the nearest pound" then you should round to the nearest penny.

In part two you again have the right idea, but it's best to restrict yourself either to natural logarithms or logarithms to base 10; you don't want an examiner to have to rmember for themselves that $\log_b a = \ln a / \ln b$. And again, express time periods as "after X years" rather than "at the start of year (X + 1)".

 

[AN630078]

You have the right idea, but have not counted your years correctly.

The question asks firstly for the balance after 10 years. You have given the balance at start of the 10th year, which is the balance after only 9 years. (In general, you want to avoid "at the start of year X" and instead use "after (X-1) years".)

If you let the initial amount be $b_0$, then $b_n$ would be the balance after $n$ years rather than the balance after $n-1$ years, which is maybe less confusing.

If the question doesn't say "to the nearest pound" then you should round to the nearest penny.

In part two you again have the right idea, but it's best to restrict yourself either to natural logarithms or logarithms to base 10; you don't want an examiner to have to rmember for themselves that $\log_b a = \ln a / \ln b$. And again, express time periods as "after X years" rather than "at the start of year (X + 1)".

Thank you very much for your reply.

So for the first part of the question would I find how much is in the account after 10 years by;
bn =2000*1.06^10
Thus, b10 =2000×1.06^10 = £3581.69539 ~ £3581.70

Then for part 2;
bn=2000*1.06^n
(Replace the inequality with an equation)
Solve bn=5000
2000*1.06^n=5000
1.06^n=5000/2000
1.06^n=5/2
Take natural logs of both sides;
In (1.06^n)=In(5/2)
n=In(5/2)/In(1.06)
n=15.725

n must be an integer so round to the neartest whole number = 16 and check;
When n=15, b15=2000*1.06^15=£4793.11638<£5000
When n=16, b16=2000*1.06^16=£5080.7033>£5000

Would this be correct?