by using the knowledge of sequence and series show that the compound interest is given by An=P(1+((RT)/(100)))^n


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 41288 by mondodotto@gmail.com last updated on 04/Aug/18

$by using the knowledge of$ $sequence and series show that$ $the compound interest is given by$ $An = P {(1 + \frac{RT}{100})}^{n}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18
	$begining of the end of year end of year year(principle) − interest − amount 1) P _ ((P×R×1)/(100)) − P+((P×R×1)/(100)) 2) P(1+(R/(100))) − P(1+(R/(100)))×((R×1)/(100)) − P(1+(R/(100))).(R/(100))+P(1+(R/(100))) 3)P(1+(R/(100)))^2 − P(1+(R/(100)))^2 ×(R/(100)) −P{1+(R/(100))}^2 ×(R/(100))+P{1+(R/(100))}^2 thus thus for first year principle(p)=p amount(A)=P(1+(R/(100))) for 2nd year (p)=p(1+(R/(100))) amount(A)=p(1+(R/(100)))^2 for 3rd year(p)=p(1+(R/(100)))^2 amount(A)=p(1+(R/(100)))^3 thus for n year amount(A)=p(1+(R/(100)))^n$
	$b e g i n i n g o f t h e e n d o f y e a r e n d o f y e a r$ $y e a r (p r i n c i p l e) - i n t e r e s t - a m o u n t$ $1) P_\frac{P \times R \times 1}{100} - P + \frac{P \times R \times 1}{100}$ $2) P (1 + \frac{R}{100}) - P (1 + \frac{R}{100}) \times \frac{R \times 1}{100} - P (1 + \frac{R}{100}) . \frac{R}{100} + P (1 + \frac{R}{100})$ $3) P {(1 + \frac{R}{100})}^{2} - P {(1 + \frac{R}{100})}^{2} \times \frac{R}{100} - P {1 + \frac{R}{100}}^{2} \times \frac{R}{100} + P {1 + \frac{R}{100}}^{2}$ $t h u s$ $t h u s f o r f i r s t y e a r p r i n c i p l e (p) = p a m o u n t (A) = P (1 + \frac{R}{100})$ $f o r 2 n d y e a r (p) = p (1 + \frac{R}{100}) a m o u n t (A) = p {(1 + \frac{R}{100})}^{2}$ $f o r 3 r d y e a r (p) = p {(1 + \frac{R}{100})}^{2} a m o u n t (A) = p {(1 + \frac{R}{100})}^{3}$ $t h u s f o r n y e a r a m o u n t (A) = p {(1 + \frac{R}{100})}^{n}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum