Question

help me guys to solve this Q

Answer 1

Given initial payment =Rs.6,000=Rs.6,000

The balance amount =12,000−6000=Rs.6000=12,000−6000=Rs.6000

Also given that the payment in each installment =500+=500+ 12% interest on the unpaid amount.

⇒⇒ payment in 1st1st year =500+12100=500+12100.6000.6000

Payment in 2nd2nd year =500+12100=500+12100.(6000−500)=500+12100.(6000−500)=500+12100.(5500).(5500)

Payment in 3rd3rd year =500+12100=500+12100.(6000−1000)=500+12100.(6000−1000)=500+12100.(5000).(5000)

an so on

Total amount paid in all the installments is

S=[500+12100S=[500+12100.6000]+[500+12100.6000]+[500+12100.5500]+[500+12100.5500]+[500+12100.5000]+...........5000]+..........

=[500+500+........ntimes]+12100=[500+500+........ntimes]+12100[6000+5500+5000+.......tn][6000+5500+5000+.......tn]

6000+5500+5000+......tn.6000+5500+5000+......tn. is an A.P. with 1st1st term a=6000a=6000 and common difference d=−500d=−500

Step 2

We have to find the no.of terms nn .

The last term in this series tntn is 500

We know that tn=a+(n−1)dtn=a+(n−1)d

⇒500=6000+(n−1)(−500)⇒500=6000+(n−1)(−500)

⇒−5500=(n−1)(−500)⇒−5500=(n−1)(−500)

⇒n=12⇒n=12

We know that sum of nn terms of an A.P. =n2=n2(a+tn)(a+tn)

∴6000+5500+5000+.........500=122∴6000+5500+5000+.........500=122[6000+500][6000+500]

=6(6500)=39000=6(6500)=39000

∴∴ 1210012100.39000=4680.39000=4680

Step 3

500+500+.......ntimes500+500+.......ntimes =500×12=6000=500×12=6000

∴∴ the total amount paid through installments =6000+4680=10680=6000+4680=10680

∴∴ The tractor cost for the farmer =6000+10680=Rs.16680