Question Number 45399 by Rio Michael last updated on 12/Oct/18
$${using}\:{your}\:{knowlege}\:{on}\:{Arithmetic}\:{progressions}, \\ $$$${show}\:{that}\:\:{A}=\:{p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{{n}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
$${principle}={p} \\ $$$${rate}\:{of}\:{interest}={r\%} \\ $$$${time}=\mathrm{1}{year} \\ $$$${interest}=\frac{{p}×\mathrm{1}×{r}}{\mathrm{100}}=\frac{{pr}}{\mathrm{100}} \\ $$$${amount}\:{A}={p}+\frac{{pr}}{\mathrm{100}}={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{1}} {at}\:{the}\:{end}\:{of}\:\mathrm{1}{st}\:{year} \\ $$$${inthe}\:{begining}\:{of}\:{second}\:{year}\: \\ $$$${principle}={p}+\frac{{pr}}{\mathrm{100}} \\ $$$${time}=\mathrm{1}{year} \\ $$$${rate}={r\%p} \\ $$$${interest}=\left(\frac{{pr}}{\mathrm{100}}+{p}\right)×\frac{\mathrm{1}×{r}}{\mathrm{100}} \\ $$$${amountA}=\left({p}+\frac{{pr}}{\mathrm{100}}\right)+\left(\frac{{pr}}{\mathrm{100}}+{p}\right)×\frac{\mathrm{1}×{r}}{\mathrm{100}} \\ $$$${A}=\left({p}+\frac{{pr}}{\mathrm{100}}\right)\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right) \\ $$$$\:\:={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{2}} \:{amount}\:{at}\:{the}\:{end}\:{of}\:\mathrm{2}{nd}\:{year} \\ $$$${principle}\:{inthe}\:{begining}\:{of}\:\mathrm{3}{rd}\:{year} \\ $$$${principle}={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{2}} \\ $$$${rate}={r\%} \\ $$$${time}=\mathrm{1}{year} \\ $$$${interst}={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{2}} ×\mathrm{1}×\frac{{r}}{\mathrm{100}} \\ $$$${amount}={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{2}} +{p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{2}} \:×\frac{{r}}{\mathrm{100}} \\ $$$$\:\:{p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{2}} \left\{\mathrm{1}+\frac{{r}}{\mathrm{100}}\right\}={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{3}} \\ $$$$….{thus}\:{A}={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{3}} \:\:\:{amount}\:{at}\:{the}\:{end}\:{of}\:\mathrm{3}{rd}\:{year} \\ $$$$ \\ $$$${thus}\:{A}={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{{n}} \:{at}\:{the}\:{end}\:{of}\:{nth}\:{year}… \\ $$$$ \\ $$$$ \\ $$
Commented by Rio Michael last updated on 12/Oct/18
$${thanks} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18
$${most}\:{welcome}… \\ $$