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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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A farmer buys a used tractor for Rs. 12,000. He pays Rs. 6,000 cash and agrees to pay the balance in annual installments of Rs. 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

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

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1 Answer

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Toolbox:
  • $n^{th}$ 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 $=6000+10680=Rs.16680$

 

answered Apr 8, 2014 by rvidyagovindarajan_1
edited Apr 9, 2014 by rvidyagovindarajan_1
 

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