Given initial payment $= Rs. 4,000$

The balance amount $=22,000-4000= Rs.18000$

Also given that the payment in each installment $=1000+$ 10% interest on the unpaid amount.

$\Rightarrow\:$ payment in $1^{st} $ year $=1000+\large \frac{10}{100}$$.18000$

Payment in $2^{nd} $ year $=1000+\large \frac{10}{100}$$.(18000-1000)=1000+\large\frac{10}{100}$$.(17000)$

Payment in $3^{rd} $ year $=1000+\large \frac{10}{100}$$.(18000-2000)=1000+\large\frac{10}{100}$$.(16000)$

an so on

Total amount paid in all the installments is

$S=\bigg[1000+\large\frac{10}{100}$$.18000\bigg]+\bigg[1000+\large\frac{10}{100}$$.17000\bigg]+\bigg[1000+\large\frac{10}{100}$$.16000\bigg]+..........$

$=\big[1000+1000+.......n\:times.\big]+\large\frac{10}{100}$$\big[18000+17000+16000+.......t_n\big]$

$18000+17000+16000+........$ is an A.P. with $1^{st}$ term $a=18000$ and common difference $d=-1000$

Step 2

We have to find the no.of terms $n$ .

The last term in this series $t_n$ is 1000

We know that $t_n=a+(n-1)d$

$\Rightarrow\: 1000=18000+(n-1)(-1000)$

$\Rightarrow\:-17000=(n-1)(-1000)$

$\Rightarrow\:n=18$

We know that sum of $n$ terms of an A.P. $=\large\frac{n}{2}$$(a+t_n)$

$\therefore\:18000+17000+16000+.........1000=\large\frac{18}{2}$$\big[18000+1000\big]$

$\qquad\:=9(19000)=171000$

$\therefore$ $\large\frac{10}{100}$$.171000=17100$

Step 3

$1000+1000+....... n\:times$ $=1000\times18=18000$

$\therefore$ the total amount paid through installments $=18000+17100=35100$

$\therefore$ The tractor cost for the farmer $=4000+35100=Rs. 39100$