1. ## [SOLVED] Sequence Question

Question:
Each year a company gives a grant to a charity. The amount given each year increases by $5\%$ of its value in the preceding year. The grant in 2001 was $\5000$. Find:

(i) the grant given in 2011,
(ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive.

Attempt:

$a = \5000$
$r = 5\% = 0.05$

$2001 \rightarrow 2011 = 10 years$

$U_{11} = a.r^{10}$
$U_{11} = 5000 \times 0.05^{10} = 4.88^{-10}$

I don't think that is possible ! How can I solve this question?

2. Maybe $r=1.05$

3. Hello,

r is not 5%, but 1.05, because you keep the previous amount.
Let's take an example :

The first year, they give 5000$. The following year, they will give 5000 increased of 5%. This means that they will give 5000+5%*5000=5000+.05*5000=5000*1.05 The following year, they will give 5000*1.05 added up by 5%. This means that they will give 5000*1.05+5%*(5000*1.05)=(5000*1.05)*1.05=5000*1.0 5² Is it clear for you now ? 4. Originally Posted by looi76 (ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive. $S_n = \frac{a(1 - r^n)}{1 - r}$ $S_n = \frac{5000(1 - 1.05^{10})}{1 - 1.05}$ $S_n = \62889.46$ Where did I go wrong? Answer in text book is$71034

5. Originally Posted by looi76
$S_n = \frac{a(1 - r^n)}{1 - r}$

$S_n = \frac{5000(1 - 1.05^{10})}{1 - 1.05}$

$S_n = \62889.46$

Where did I go wrong? Answer in text book is \$71034
Hi looi76,

When the question says inclusive they mean both the year 2001 and 2011 should be included. So actually there are 11 years in all. So your n=11...

$S_{11} = \frac{5000(1 - 1.05^{11})}{1 - 1.05} = \71033.93$