interest being compounded half yearly.

Question

What is the compound interest on $15000/-$ at $12\%$ P.a for one and half years, interest being compounded half yearly? (Ans: $C I = 2865.24$)

I Selected this Procedure to solve: If a sum of $x$ is lent for $h$ years at the rate of $y\%$ per annum and the compounding is done for every $n$ months ($m$ a year), then the amount will be equal to

$$CI = x\left(1 + \frac{\frac{y}{\frac{12}{m}}}{100}\right)^{h \cdot \frac{12}{m}}$$.

My try: $R = 12\%$, $n =$ one and half years which means $18$ months $= 3/2$ years \begin{align*} C I & = 15000\left[\left(1 + \frac{\frac{12}{\frac{18}{6}}}{100}\right)^{\frac{3}{2} \cdot3}\right] – 15000\\ & = 15000\left(1 + \frac{1}{25}\right)^{\frac{9}{2}} – 15000 \end{align*}

So am I doing correct way?

Answer 1

12% per annum compounded every 6 months is 3 payments of 6%.

$15000\cdot (1+0.06)^3=17865.24$

So $17865.24-15000.00 = 2865.24$

Answer 2

If an initial deposit $P_0$ earns $100r\%$ interest compounded $N$ times a year, the principal $P(t)$ at time $t$ is given by the formula $$P(t) = P_0\left(1 + \frac{r}{N}\right)^{Nt}$$ We are given $P_0 = 15000$, $100r\% = 12\% \implies r = 0.12$, $t = 1.5$, and interest is compounded half yearly $\implies N = 2$. Hence, the amount of interest earned is \begin{align*} P\left(\frac{3}{2}\right) - P_0 & = 15000\left(1 + \frac{0.12}{2}\right)^{2 \cdot 1.5} - 15000\\ & = 15000[(1 + 0.06)^3 - 1]\\ & = 15000(1.06^3 - 1)\\ & = 2865.24 \end{align*} Your first formula is incorrectly labeled CI (for compound interest) when you were actually attempting to compute the principal at time $t$. However, it is incorrect since you need to divide $12$ by $n$ in order to find $m$, the number of times interest is compounded annually. $$m = \frac{12}{n}$$ Moreover, the number of times interest in $h$ years is compounded is $$h \cdot \frac{12}{n} = hm$$ Observe that $$\frac{y}{\frac{12}{n}} = y \cdot \frac{n}{12}$$ so that $$\frac{\frac{y}{\frac{12}{n}}}{100} = \frac{y \cdot \frac{n}{12}}{100} = y \cdot \frac{n}{12} \cdot \frac{1}{100} = \frac{y}{100} \cdot \frac{n}{12}$$ Hence, you should have obtained the formula $$P(t) = x\left(1 + \frac{y}{100} \cdot \frac{n}{12}\right)^{hm}$$ for the principal at time $t$. Consequently, you should have obtained the compound interest formula $$CI(t) = x\left[\left(1 + \frac{y}{100} \cdot \frac{n}{12}\right)^{hm} - 1\right]$$ Substituting $x = 15000$, $y = 12$, $n = 6$, $t = 3/2$, $h = 3/2$, and $m = \frac{12}{6} = 2$ yields \begin{align*} CI\left(\frac{3}{2}\right) & = 15000\left[\left(1 + \frac{12}{100} \cdot \frac{6}{12}\right)^{\frac{3}{2} \cdot 2} - 1\right]\\ & = 15000\left[\left(1 + 0.12 \cdot \frac{1}{2}\right)^3 - 1\right]\\ & = 15000\left[\left(1 + \frac{0.12}{2}\right)^3 - 1\right]\\ & = 15000[(1 + 0.06)^3 - 1]\\ & = 15000[1.06^3 - 1]\\ & = 2865.24 \end{align*} which agrees with the result we obtained above.

Answer 3

$A(1.5)=A(O)(1+\frac{0.12}{2})^{(1.5*2)}$

$A(1.5)=15000(1+\frac{0.12}{2})^{(1.5*2)}=17865.24$

$I_{[1.5,0]}=17865.24-15000=2865.24$