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^{\text {a }} $ instalment? What amount of loan does he still have to pay after the $ 30^{\text {th }} $ instalment?

Given:

Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. 

He increases the instalment by Rs 100 every month.

To do:

We have to find the amount paid by him in the \( 30^{\text {a }} \) instalment and the amount of loan he still has to pay after the \( 30^{\text {th }} \) instalment.

Solution:

Total loan amount $=Rs.\ 118000$

First installment $= Rs.\ 1000$

Second installment $=Rs.\ 1000 + 100$

$= Rs.\ 1100$

Third installment $= Rs.\ 1100 + 100$

$= Rs.\ 1200$ and so on

This is in AP.

Here

First term $a =1000$

Common difference $d=100$

We know that,

$a_{n}=a+(n-1) d$

This implies,

$a_{30}=1000+(30-1)(100)$

$=1000+29(100)$

$=1000+2900$

$a_{30}=3900$
Therefore, the amount paid by Jaspal Singh in 30th instalment is $Rs.\ 3900$

Loan amount paid in 30 instalments $=S_{30}$
The sum of the first \( n \) terms of an AP is given by $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{30}=\frac{30}{2}[2(1000)+(30-1)(100)]$

$=15[2000+2900]$

$=15(4900)$

$=73500$

Total amount paid till 30 installments $S_{30}=Rs.\ 73500$

Amount of loan to be paid after 30 installments $=$ Total loan $-$ Loan amount paid till 30 installments

$=Rs.\ 118000-73500$

$=Rs.\ 44500$

Therefore, the amount of loan to be paid by Jaspal Singh after 30 installments is $Rs.\ 44500$.

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Updated on: 10-Oct-2022

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