# Application of Ap/GP

• Jan 29th 2009, 11:40 PM
nikk
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?

• Jan 30th 2009, 12:33 AM
james_bond
Q1. 1st year: $\displaystyle 500\cdot 1.05$, 2nd year: $\displaystyle (500\cdot 1.05)\cdot 1.05$, etc. So $\displaystyle n$th year: $\displaystyle 500\cdot (1.05)^n$, therefore this should hold: $\displaystyle 500\cdot (1.05)^n\ge 750\leftrightarrow (1.05)^n\ge 1.5$. Take the logarithm of both sides (base 1.05, you can without reversing the sign, because it is greater than 1): $\displaystyle \log_{1.05}(1.05)^n\ge \log_{1.05}1.5$ by definition it is: $\displaystyle n\ge\log_{1.05}1.5\approx 8.31039$. So the first integer after...
Q2. Using the same idea: end of 10th year: $\displaystyle 50\cdot (1.04)^{10}=1446.27$
• Jan 30th 2009, 02:17 AM
nikk
Quote:

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

i will try to calculate it back..it consider under G.P?
• Jan 30th 2009, 02:28 AM
james_bond
It is a geometric progression. (Not sure what you're asking.)
• Jan 30th 2009, 06:44 PM
nikk
Quote:

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
• Feb 2nd 2009, 12:50 AM
james_bond
Hi, sorry for not replying sooner.
Q1(b): $\displaystyle n=8$, so: $\displaystyle 500\cdot (1.05)^8\approx 738.728$
Q2: sorry I misread and misunderstood the question and wrote some stupid ****. Sorry for that.
1st year (31st december) he has: $\displaystyle 50\cdot 1.04$, next year he has: $\displaystyle 50\cdot 1.04^2+50\cdot 1.04$, etc. 10th year: $\displaystyle 50\cdot\left(1.04^{10}+1.04^9+\ldots +1.04\right)=50\cdot\frac{1\cdot(1.04^{10+1}-1.04^1)}{1.04-1}\approx 624.318$.