Question

A man borrows Rs. 4000 and agrees to repay with a total interest of Rs. 680 in 12 monthlyinstalments. Each instalment being less than the preceding one by Rs. 20. Find the first installment and the last instalment.​

Answer 1

Step-by-step explanation:

money he borrow =4000total interest=680

total money he has to pay after 12 months =4000+680=4680

it is given that :

each instalment is less than preceding one by rs 20

d=-20 n=12. s12=4680

now sn =n/2(2a+(n-1)d)

s12 = 12/2(2a(12-1)-20

4680=6(2a-220)

780=2a-220

560=2a

a=280

a12=a+(n-1)d

a12=280-220

a12=60

Answer 2

First installment = 500 Rs.

Last installment = 280 Rs.

Given:

A man borrows Rs. 4000 and agrees to repay with a total interest of Rs. 680 in 12 monthly installments.

Each installment being less than the preceding one by Rs. 20.

To find:

The first installment and the last installment.

Explanation:

Total payable amount = borrowed money + Interest

Total payable amount = 4,000 + 680 = 4,680

Each installment being less than the preceding one by Rs. 20.

That is it can be obeys arithmetic progression with common difference= -20

Total sum of money after 12 month = 4,680 Rs.

In arithmetic progression,

Total sum = $\frac{n}{2}$ (2a +(n-1)d)

a = First term

d = common difference

n= number of terms

Total sum = $\frac{n}{2}$ (2a +(n-1)d)

4.680 = $\frac{12}{2}$(2a +(12-1)(-20))

  $\frac{4,680}{6}$ = 2a +(11)(-20)

    780 = 2a - 220

    1,000 = 2a

     a = 500

That is first installment = 500 Rs.

Last installment = a+(n-1)d = 500+(12-1)(-20) = 280 Rs.

To learn more...

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