 QUESTION

# Jaspal Singh repays his total loan of Rs.118000 by 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 30th instalment? What amount of loan does he still have to pay after the 30th instalment?A. 35000B. 44500C. 43420D. 32450

Hint: Here, since the instalments starting from 1000 are paid 100 per month, we can consider the series 1000, 1100, 1200,…. where they are in AP with first term $a=1000$ and common difference, d= 100. Now, for calculating the 30th instalment paid, we have to calculate the nth term using the formula: ${{a}_{n}}=a+(n-1)d$, Now, find the sum of 30 instalments using the formula: ${{S}_{n}}=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$. Hence, the amount still to pay will be total instalments paid subtracted from the total amount.

Complete Step-by-Step solution:
Here, given that the total loan of Jaspal Singh = Rs.118000
First instalment = 1000
The instalments paid in each month = 100
Hence, the monthly instalments are:
First month = 1000
Second month = 1000+100=1100
Third month = 1100+100=1200
And so on till the 30th instalment.
Hence, we can say that the above terms form an AP with common difference, d = 100, n = 30.
So, the first term , $a=1000$
Now, we have to find the 30th instalment, that is, the 30th term of the series. We have the formula to find the nth term AP. i.e.
${{a}_{n}}=a+(n-1)d$
By applying this formula we obtain:
\begin{align} & {{a}_{30}}=1000+(30-1)100 \\ & {{a}_{30}}=1000+29\times 100 \\ & {{a}_{30}}=1000+2900 \\ & {{a}_{30}}=3900 \\ \end{align}
Hence, the 30th term of AP is 3900.
Therefore, we can say that the amount paid by Jaspal in the 30th instalment is Rs.3900
Now, we have to find the total installment.
That is, the total amount paid in 30 instalments is:
1000+1100+1200+……+3900
Now, we have to find the sum of the 30 terms.
Given a = 1000, n= 30, ${{a}_{30}}=3900$
Hence, the sum of n terms when the first term and the last term is known is given by the formula:
${{S}_{n}}=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$
Now, by applying this formula for n = 30 we get:
\begin{align} & {{S}_{30}}=\dfrac{30}{2}\left[ 1000+3900 \right] \\ & {{S}_{30}}=\dfrac{30}{2}\times 4900 \\ \end{align}
Next, by cancellation, we get:
\begin{align} & {{S}_{30}}=15\times 4900 \\ & {{S}_{30}}=73500 \\ \end{align}
Hence, the total amount paid in 30 instalments is Rs.73,500.
Now, we have to calculate the amount of loan Jaspal still has to pay after the 30th instalment.
Amount of loan to be paid = Total amount of loan – Amount paid in 30 instalments
Amount of loan to be paid = 118000 – 73500
Amount of loan to be paid = 44500
Hence, we can say that the amount of loan Jaspal still has to pay after the 3oth instalment is Rs.44,500.
Therefore, the correct answer for this question is option (b).

Note: Here, if you don’t know the nth term also you can find the sum using the formula:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$. But here you have to find the 30th term. So you can apply the other formula directly.