1. ## Application of Ap/GP

i already try for question no Q1 as below, but the answer is not same from text book. so i hope some body can help me here.

Q1. A man deposits RM 500 in a bank. The interest receive of 5% per year is kept in bank. Find
(a) the number of years it will take for the total savings to exceed RM750.
(the ans: 9 but my ans 11)

(b) the total saving after 8 years.

Q2. On 1 st January each year, a man puts RM 50 in a bank whick give 4% interest each year. What is the value of his saving on 31 st December of the tenth year?

2. Q1. 1st year: $500\cdot 1.05$, 2nd year: $(500\cdot 1.05)\cdot 1.05$, etc. So $n$th year: $500\cdot (1.05)^n$, therefore this should hold: $500\cdot (1.05)^n\ge 750\leftrightarrow (1.05)^n\ge 1.5$. Take the logarithm of both sides (base 1.05, you can without reversing the sign, because it is greater than 1): $\log_{1.05}(1.05)^n\ge \log_{1.05}1.5$ by definition it is: $n\ge\log_{1.05}1.5\approx 8.31039$. So the first integer after...
Q2. Using the same idea: end of 10th year: $50\cdot (1.04)^{10}=1446.27$

3. Originally Posted by james_bond
Q1. 1st year: $500\cdot 1.05$, 2nd year: $(500\cdot 1.05)\cdot 1.05$, etc. So $n$th year: $500\cdot (1.05)^n$, therefore this should hold: $500\cdot (1.05)^n\ge 750\leftrightarrow (1.05)^n\ge 1.5$. Take the logarithm of both sides (base 1.05, you can without reversing the sign, because it is greater than 1): $\log_{1.05}(1.05)^n\ge \log_{1.05}1.5$ by definition it is: $n\ge\log_{1.05}1.5\approx 8.31039$. So the first integer after...
Q2. Using the same idea: end of 10th year: $50\cdot (1.04)^{10}=1446.27$
i will try to calculate it back..it consider under G.P?

4. It is a geometric progression. (Not sure what you're asking.)

5. Originally Posted by james_bond
It is a geometric progression. (Not sure what you're asking.)
yes. it what i mean.

ans
Q1
(a) 9
(b) rm 738.70

Q2
rm 624.30

6. Hi, sorry for not replying sooner.
Q1(b): $n=8$, so: $500\cdot (1.05)^8\approx 738.728$
Q2: sorry I misread and misunderstood the question and wrote some stupid ****. Sorry for that.
1st year (31st december) he has: $50\cdot 1.04$, next year he has: $50\cdot 1.04^2+50\cdot 1.04$, etc. 10th year: $50\cdot\left(1.04^{10}+1.04^9+\ldots +1.04\right)=50\cdot\frac{1\cdot(1.04^{10+1}-1.04^1)}{1.04-1}\approx 624.318$.