Question

Question 9
Jaspal Singh repays his total loan of Rs. 118000 by paying every month starting with the first instalment of Rs. 1000. If he increases the instalment by Rs. 100 every month, what amount will be paid by him in the ${30}^{th}$ instalment? What amount of loan does he still have to pay after the ${30}^{th}$ instalment?

Answer 1

Given that,
Jaspal Singh takes total loan = Rs. 118000
He repays his total loan by paying every month.
His first instalment = Rs. 1000
Second instalment = 1000 + 100 = Rs. 1100
Third instalment = 1100 + 100 = Rs. 1200 and so on.
Let its ${30}^{th}$ instalment be n.
Thus, we have 1000, 1100, 1200, …… which form an AP, with
First term (a) = 1000 and common difference (d) = 1100 – 1000 = 100
$\because$ nth term of an AP, $T_n$ = a+ (n-1)d
For ${30}^{th}$ instalment, $T_{30}$ = 1000 + (30 - 1)100
= 1000 + 29 $\times$ 100
= 1000 + 2900 = 3900
So, Rs. 3900 will be paid by him in the ${30}^{th}$ instalment.
He paid total amount upto 30 instalments in the following form.
1000 + 1100 + 1200 + . . . + 3900
First term (a) = 1000 and last term (l) = 3900
$\therefore$ Sum of 30 instalments, $S_{30}=\frac{30}{2}[a+l]$
$[\because sumoffirstntermsofanAPis,S_n=\frac{n}{2}[a+1],wherel=lastterm]$
$\Rightarrow S_{30}=15(1000+3900)$
= 15 $\times$ 4900 = Rs. 73500
$\therefore$ Total amount he still have to pay after the ${30}^{th}$ instalment.
= (Amount of loan) - (Sum of 30 instalments)
= 118000 - 73500 = Rs. 44500
Hence, Rs. 44,500 still have to pay after the ${30}^{th}$ instalment.