using-your-knowlege-on-Arithmetic-progressions-show-that-A-p-1-r-100-n-

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

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}

\dots . 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 \dots

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 \dots