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
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
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.