Question

A man borrows $Rs8000$ and agrees to repay with a total interest of $Rs1360$ in $12$ monthly installment. Each installment being less then preceding one by $Rs40$. Find the amount of the first and last installment.

Answer 1

The amount of money that man borrows is $Rs8000$ and the interest he has to pay is $Rs1360$.

The total money he has to pay is,

$8000+1360=Rs9360$

Since, he has to pay in 12 installments with a difference of 40 in each installment, then, it will form an A.P. where $d=-40$, $n=12$ and $S_n=9360$.

Therefore,

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

$9360=\frac{12}{2}\left[2a+\left(12-1\right)\left(-40\right)\right]$

$9360=6\left(2a-440\right)$

$1560=2a-440$

$2a=2000$

$a=1000$

The last term in A.P. is,

$a_n=a+\left(n-1\right)d$

$=1000+\left(12-1\right)\left(-40\right)$

$=1000-440$

$=560$

Therefore, the first installment is $Rs1000$ and the last installment is $Rs560$.