# A man starts repaying a loan as the first instalment of $Rs.100$. If he increases the

instalment by $Rs.5$ every month. What is the amount he will pay in the ${30^{th}}$ instalment?

Last updated date: 18th Mar 2023

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Answer

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Hint: Express the given problem statement in series form.

As, we are given that,

The first instalment of the loan is $Rs.100$.

And each month the instalment increases by $Rs.5$.

So, the second instalment will be \[Rs.\left( {100 + 5} \right) = Rs.105\].

According to the given conditions instalments that man has to pay is in the series form as follows,

$Rs100,{\text{ }}Rs105,{\text{ }}Rs{\text{ }}110,{\text{ }}Rs115...................$so on.

So, from the above we can easily see that the instalments form an A.P. series.

And the first term of the A.P. will be, $a = Rs.100$.

With a common difference of, second term – first term$ = 105 - 100 = d = Rs.5$.

As, the instalments form an A.P.

We know that the formula to find ${n^{th}}$ term of an A.P is ${a_n}$ which is given by.

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

Now we must find the ${30^{th}}$ instalment.

So substituting, $n = 30$.

Then, the ${30^{th}}$ instalment will be ${a_{30}}$.

Then, ${a_{30}} = a + \left( {n - 1} \right)d$ .

Now, substituting the values of $a,d,n{\text{ }}$in the above equation. We get,

${a_{30}} = 100 + \left( {30 - 1} \right)5 = 100 + 145 = 245$.

Hence, the amount that the man has to pay at the end of ${30^{th}}$ instalment will be $Rs.{\text{ }}245.$

Note: Whenever we come up with these types of problems then first we have to check whether the terms form any A.P, G.P or H.P. If they form A.P, G.P or H.P then accordingly use the \[{n^{th}}\] term formula to find the solution.

As, we are given that,

The first instalment of the loan is $Rs.100$.

And each month the instalment increases by $Rs.5$.

So, the second instalment will be \[Rs.\left( {100 + 5} \right) = Rs.105\].

According to the given conditions instalments that man has to pay is in the series form as follows,

$Rs100,{\text{ }}Rs105,{\text{ }}Rs{\text{ }}110,{\text{ }}Rs115...................$so on.

So, from the above we can easily see that the instalments form an A.P. series.

And the first term of the A.P. will be, $a = Rs.100$.

With a common difference of, second term – first term$ = 105 - 100 = d = Rs.5$.

As, the instalments form an A.P.

We know that the formula to find ${n^{th}}$ term of an A.P is ${a_n}$ which is given by.

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

Now we must find the ${30^{th}}$ instalment.

So substituting, $n = 30$.

Then, the ${30^{th}}$ instalment will be ${a_{30}}$.

Then, ${a_{30}} = a + \left( {n - 1} \right)d$ .

Now, substituting the values of $a,d,n{\text{ }}$in the above equation. We get,

${a_{30}} = 100 + \left( {30 - 1} \right)5 = 100 + 145 = 245$.

Hence, the amount that the man has to pay at the end of ${30^{th}}$ instalment will be $Rs.{\text{ }}245.$

Note: Whenever we come up with these types of problems then first we have to check whether the terms form any A.P, G.P or H.P. If they form A.P, G.P or H.P then accordingly use the \[{n^{th}}\] term formula to find the solution.

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