Question

A man repays a loan of Rs.65000 by paying rs. 400 in first month and then increasing the payment by rs.300 every month. How long will it take for him to clear the loan​

Answer 1

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20 months

Step-by-step explanation:

Given  

 A man repays a loan of rs 65000 by paying rs 400 in the first month and then increasing the payment by rs 300 every month. How long will it take for him to clear the loan?  

Let the loan be cleared in n months.

Now the amount is in A.P .

Let the first payment be 400

So second will be 400 + 300  

Third will be 400 + 300 + 300 and so on.

So S = 400 + 700 + 1000 +….and so on.

a = 400. d = 300

Now sum to n terms is given by

Sn = [n/2 (2 a + (n – 1) d]

65,000 = [n/2 (800 + (n – 1)300]

1,30,000 = 500 n + 300 n^2

300 n^2 + 500 n – 1,30,000 = 0

We can use n = - b ± √b^2 – 4 a c

                          --------------------------

                                    2 a

                 n = - 500 ±√250000 + 15,60,00000 / 600  

                  n = - 500 ± 12,500 / 600

                  n = 20

 Thus the loan will be cleared in 20 months.

Answer 2

$\huge\star\underline\mathfrak\red{Answer:-}$

20 months

Given  

A man repays a loan of rs 65000 by paying rs 400 in the first month and then increasing the payment by rs 300 every month. How long will it take for him to clear the loan?  

Let the loan be cleared in n months.

Now the amount is in A.P .

Let the first payment be 400

So second will be 400 + 300  

Third will be 400 + 300 + 300 and so on.

So S = 400 + 700 + 1000 +….and so on.

a = 400. d = 300

Now sum to n terms is given by

Sn = [n/2 (2 a + (n – 1) d]

65,000 = [n/2 (800 + (n – 1)300]

1,30,000 = 500 n + 300 n^2

300 n^2 + 500 n – 1,30,000 = 0

We can use n = - b ± √b^2 – 4 a c

                          --------------------------

                                    2 a

                 n = - 500 ±√250000 + 15,60,00000 / 600  

                  n = - 500 ± 12,500 / 600

                  n = 20

 Thus the loan will be cleared in 20 months.

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