using your knowlege on Arithmetic progressions, show that A= p(1+(r/(100)))^n


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Question Number 45399 by Rio Michael last updated on 12/Oct/18

$u s i n g y o u r k n o w l e g e o n A r i t h m e t i c p r o g r e s s i o n s,$ $s h o w t h a t A = p {(1 + \frac{r}{100})}^{n}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
	$principle=p rate of interest=r% time=1year interest=((p×1×r)/(100))=((pr)/(100)) amount A=p+((pr)/(100))=p(1+(r/(100)))^1 at the end of 1st year inthe begining of second year principle=p+((pr)/(100)) time=1year rate=r%p interest=(((pr)/(100))+p)×((1×r)/(100)) amountA=(p+((pr)/(100)))+(((pr)/(100))+p)×((1×r)/(100)) A=(p+((pr)/(100)))(1+(r/(100))) =p(1+(r/(100)))(1+(r/(100)))=p(1+(r/(100)))^2 amount at the end of 2nd year principle inthe begining of 3rd year principle=p(1+(r/(100)))^2 rate=r% time=1year interst=p(1+(r/(100)))^2 ×1×(r/(100)) amount=p(1+(r/(100)))^2 +p(1+(r/(100)))^2 ×(r/(100)) p(1+(r/(100)))^2 {1+(r/(100))}=p(1+(r/(100)))^3 ....thus A=p(1+(r/(100)))^3 amount at the end of 3rd year thus A=p(1+(r/(100)))^n at the end of nth year...$
	$p r i n c i p l e = p$ $r a t e o f i n t e r e s t = r %$ $t i m e = 1 y e a r$ $i n t e r e s t = \frac{p \times 1 \times r}{100} = \frac{p r}{100}$ $a m o u n t A = p + \frac{p r}{100} = p {(1 + \frac{r}{100})}^{1} a t t h e e n d o f 1 s t y e a r$ $i n t h e b e g i n i n g o f s e c o n d y e a r$ $p r i n c i p l e = p + \frac{p r}{100}$ $t i m e = 1 y e a r$ $r a t e = r % p$ $i n t e r e s t = (\frac{p r}{100} + p) \times \frac{1 \times r}{100}$ $a m o u n t A = (p + \frac{p r}{100}) + (\frac{p r}{100} + p) \times \frac{1 \times r}{100}$ $A = (p + \frac{p r}{100}) (1 + \frac{r}{100})$ $= p (1 + \frac{r}{100}) (1 + \frac{r}{100}) = p {(1 + \frac{r}{100})}^{2} a m o u n t a t t h e e n d o f 2 n d y e a r$ $p r i n c i p l e i n t h e b e g i n i n g o f 3 r d y e a r$ $p r i n c i p l e = p {(1 + \frac{r}{100})}^{2}$ $r a t e = r %$ $t i m e = 1 y e a r$ $i n t e r s t = p {(1 + \frac{r}{100})}^{2} \times 1 \times \frac{r}{100}$ $a m o u n t = p {(1 + \frac{r}{100})}^{2} + p {(1 + \frac{r}{100})}^{2} \times \frac{r}{100}$ $p {(1 + \frac{r}{100})}^{2} {1 + \frac{r}{100}} = p {(1 + \frac{r}{100})}^{3}$ $. . . . t h u s A = p {(1 + \frac{r}{100})}^{3} a m o u n t a t t h e e n d o f 3 r d y e a r$ $t h u s A = p {(1 + \frac{r}{100})}^{n} a t t h e e n d o f n t h y e a r . . .$
	Commented by Rio Michael last updated on 12/Oct/18

	$t h a n k s$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18

	$m o s t w e l c o m e . . .$
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