Question

Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying instalments every month. If the instalment for the first month is Rs. 1,000 and it increases by Rs. 100 every month, what amount will she pay as the ${30}^{th}$ instalment of loan ? What amount of loan she still has to pay after the ${30}^{th}$ instalment ?$\left[4Marks\right]$

Answer 1

Let us assume that the loan is cleared in 'n' months.
Then, The amounts are in AP with first term (a) = 1000, the common difference (d) = 100 and the Sum of amounts (Sn) = Rs. 118000
So, according to the question.
$S_n$ = $\frac{n}{2}${2a + (n - 1)d} $\left[0.5Mark\right]$
118000 = $\frac{n}{2}${2 $\times$ 1000 + (n - 1)100}
236000 = n {2000 - 100 + 100n}
236000 = 1900n + 100$n^2$
100$n^2$ + 1900n - 236000 = 0
Dividing the above by 100, we get.
$n^2$ + 19 - 2360 = 0
$n^2$ + 59n - 40n - 2360 = 0$\left[0.5Mark\right]$
n(n + 59) - 40(n +59) = 0
(n - 40) (n + 59) = 0
n = 40 or n = -59
Since the value cannot be negative so, n = 40 is correct.
Therefore, the loan will be cleared in 40 months. $\left[0.5Mark\right]$

Now, the amount to be paid by him in the 30th installment is calculated as under.
$a_{30}$ = a + 29d where a = 1000 and d = 100
$a_{30}$ = 1000 + (29 $\times$ 100) = 1000 + 2900
$a_{30}$ = 3900
So, the amount to be paid by him in the 30th installment is Rs. 3900$\left[0.5Mark\right]$

Now, the amount of remaining loan that he has to pay after the 30th installment is calculated as under.
$S_{30}$ = $\frac{n}{2}${2a + (n - 1)d}
= $\frac{30}{2}${2 $\times$ 1000 + (30 - 1)100}
= 15{2000 + 2900}
= 15 $\times$ 4900
= Rs. 73500
So, the total amount paid after the 30th installment is Rs. 73500.$\left[1Mark\right]$

The remaining amount of loan = total loan amount - amount paid after the 30th installment
= 118000 - 73500
= Rs. 44500
So, he still has to pay Rs. 44500 after the payment of the 30th installment.
$\left[1Mark\right]$