A farmer buys a used tractor for Rs 12000.

Solution:

It is given that the farmer pays Rs 6000 in cash.

Therefore, unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

According to the given condition, the interest paid annually is

$12 \%$ of $6000,12 \%$ of $5500,12 \%$ of $5000, \ldots, 12 \%$ of 500

Thus, total interest to be paid $=12 \%$ of $6000+12 \%$ of $5500+12 \%$ of $5000+\ldots+12 \%$ of 500

$=12 \%$ of $(6000+5500+5000+\ldots+500)$

$=12 \%$ of $(500+1000+1500+\ldots+6000)$

Now, the series 500, 1000, 1500 … 6000 is an A.P. with both the first term and common difference equal to 500.

Let the number of terms of the A.P. be n.

$\therefore 6000=500+(n-1) 500$

$\Rightarrow 1+(n-1)=12$

$\Rightarrow n=12$

$\therefore$ Sum of the A.P $=\frac{12}{2}[2(500)+(12-1)(500)]=6[1000+5500]=6(6500)=39000$

Thus, total interest to be paid $=12 \%$ of $(500+1000+1500+\ldots+6000)$

$=12 \%$ of $39000=$ Rs 4680

Thus, cost of tractor = (Rs 12000 + Rs 4680) = Rs 16680