$\begin{array}{1 1}17680 \\ 16180 \\ 17180 \\ 16680 \end{array} $

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- $n^{th}$ term of an A.P. = $t_n=a+(n-1)d$
- Sum of $n$ terms of an A.P. $=\large\frac{n}{2}$$(a+t_n)$

Given initial payment $= Rs. 6,000$

The balance amount $=12,000-6000= Rs.6000$

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

$\Rightarrow\:$ payment in $1^{st} $ year $=500+\large \frac{12}{100}$$.6000$

Payment in $2^{nd} $ year $=500+\large \frac{12}{100}$$.(6000-500)=500+\large\frac{12}{100}$$.(5500)$

Payment in $3^{rd} $ year $=500+\large \frac{12}{100}$$.(6000-1000)=500+\large\frac{12}{100}$$.(5000)$

an so on

Total amount paid in all the installments is

$S=\bigg[500+\large\frac{12}{100}$$.6000\bigg]+\bigg[500+\large\frac{12}{100}$$.5500\bigg]+\bigg[500+\large\frac{12}{100}$$.5000\bigg]+..........$

$=\big[500+500+........n\:times\big]+\large\frac{12}{100}$$\big[6000+5500+5000+.......t_n\big]$

$6000+5500+5000+......t_n.$ is an A.P. with $1^{st}$ term $a=6000$ and common difference $d=-500$

Step 2

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

The last term in this series $t_n$ is 500

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

$\Rightarrow\: 500=6000+(n-1)(-500)$

$\Rightarrow\:-5500=(n-1)(-500)$

$\Rightarrow\:n=12$

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

$\therefore\:6000+5500+5000+.........500=\large\frac{12}{2}$$\big[6000+500\big]$

$\qquad\:=6(6500)=39000$

$\therefore$ $\large\frac{12}{100}$$.39000=4680$

Step 3

$500+500+....... n\:times$ $=500\times12=6000$

$\therefore$ the total amount paid through installments $=6000+4680=10680$

$\therefore$ The tractor cost for the farmer $=6000+10680=Rs.16680$

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