Geometric progression question help..

Hi guys, pretty new to advanced maths after a long career break and find i am struggling quite regularly on an HNC i am working on for work.

One of the questions i have is regarding geometric progression, it is as follows.

A firm starts work with 110 employees for the 1st week. The number

of the employees rises by 6% per week. How many persons will be

employed in the 20th week if the present rate of expansion continues.

Now i have worked it out as S(20) = 110 x (1-1.06^20) divide by 1-1.06

As the answer comes to 4046 with my workings. Obviously very wrong.

Thanks for any help.

Re: Geometric progression question help..

Hi, Mac.

Yes, that number does seem too large, doesn't it?

It appears you used a formula to calculate the sum of the first 20 terms of the progression, S(20). This isn't relevant to your situation, where you just seek the 20th term.

The first term is 110. To get the next you would multiply this by 1.06. So the *second* term would be 110*(1.06)^(*2*-1).

I wrote it that way to demonstrate an example of the general formula for the *nth* term, which is 110*(1.06)^(*n*-1).

So for your situation, the 20th term would be 110*(1.06)^(20-1) = 332.8, a much more reasonable number for the situation.

Hope this helps, and if you like here is the relevant formula [see below] used when working with geometric progression problems like this one.

a(n) = a*r^(n-1), where 'a' is the first term [110 itt], and 'r' is the common ratio [1.06 itt]

Re: Geometric progression question help..

Quote:

Originally Posted by

**macmclaren** One of the questions i have is regarding geometric progression, it is as follows.

A firm starts work with 110 employees for the 1st week. The number

of the employees rises by 6% per week. How many persons will be

employed in the 20th week if the present rate of expansion continues.

Try $\displaystyle 110(1.06)^{19}$. WHY?

Re: Geometric progression question help..

Ah i see exactly where i was going wrong, fantastic. Thanks guys :-)