Math, asked by muhammadanaf2623, 6 months ago

5)A man borrow 8000 and agrees to repay with a total interest of 1360 in 12
monthly instalments. Each instalment being less than the preceding one by ₹40. Find the amount of the first and last instalment.

Question of Class 2​

Answers

Answered by vikas301v
1

Answer:

The amount repaid=8000+1360=9360

The number of instalment is 12

n=12,Sn=S12=9360

Each instalment is 40 less

Sn=n[2a+(n-1)d]

...

2

S12=12[2a+(12-1)*-40)]

......

2

9360=6[2a+11*(-40)]

9360=6(2a-440)

9360=2a-440

...........

6

2a-440=1560

2a=1560+440

2a=2000

a=2000/2=1000

The last instalment =Tn

Tn=a+(n-1)d

T12=1000+(12-1)*(-40)

=1000+11*(-40)

=1000-440

T12=560

The first instalment is 1000 & last instalment is 560

Answered by hemajatt1206
8

Answer:

The amount of money that man borrows is Rs8000 and the interest he has to pay is Rs1360.

The total money he has to pay is,

8000+1360=Rs9360

Since, he has to pay in 12 installments with a difference of 40 in each installment, then, it will form an A.P. where d=−40, n=12 and S

n

=9360.

Therefore,

S

n

=

2

n

[2a+(n−1)d]

9360=

2

12

[2a+(12−1)(−40)]

9360=6(2a−440)

1560=2a−440

2a=2000

a=1000

The last term in A.P. is,

a

n

=a+(n−1)d

=1000+(12−1)(−40)

=1000−440

=560

Therefore, the first installment is Rs1000 and the last installment is Rs560.

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