Question

A sum of Rs. 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interest from an AP? if so, find the interest at the end of 30 year making use of this fact.

Answer 1

hay!!


we know that the formula to calculate simple interest is given by


$simple \: interest \: = \frac{p \times r\times t}{100}$



$so \: the \: interest \: at \: the \: end \: of \: the \: 1st \: year \: = \frac{1000 \times 8 \times 1}{100 } = 80$


$the \: interest \: at \: the \: end \: of \: 2nd \: year \: = \frac{1000 \times 8 \times 2}{100} = 160$



$the \: interest \: of \: at \: the \: end \: of \: 3rd \: year \: = \frac{1000 \times 8 \times 3}{100} = 240$



Similarly, we can obtain the interest at the end of the 4th year 5th year and so on .


so, the interest (in RS) at the end of the 1st, 2nd, 3rd,...... years, respectively are 80,160,240,.......


it is an AP as the difference between the consecutive terms in the list is 80,i.e


d = 80. Also, a = 80


So, to find the interest at the end of 30 years will shall find
$a_{30}$


$now \: a_{30} = a + (30 - 1)d = 80 + 29 \times 80 = 2400$


so, the interest at the end of 30 years will be Rs. 2400


I hope it's help you


@Swarnimkumar21##

Answer 2

hiii. .....friend

The answer is here,

Principle Amount = 1000.

Rate of interest = 8%


Simple Interest => P×T×R/100.

Interest for 1st year => 1000×1×8/100

=> 80.

Interest for 2years => 1000×8×2/100.

=> 160.

Interest for 3 years => 1000×8×3/100.

=> 240.

So, The interest for 1st ,2nd , 3rd years & so on are in AP with common difference 80.

=> 80,160,240,..........

interest at the end of 30years,

=> an = a+(n-1)d

=> a30 = 80+(30-1)80.

=> 80+29 (80)

=> 30×80.

=> 2400.

So, The interest at the end of 30years is 2400.


:-)Hope it helps u.