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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Shamshad Ali buys a scooter for Rs. 22,000. He pays Rs. 4,000 cash and agrees to pay the balance in annual installment of Rs. 1,000 plus 10% interest on the unpaid amount. How much will the scooter costs him?

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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}$$(a+t_n)$
Given initial payment $= Rs. 4,000$
The balance amount $=22,000-4000= Rs.18000$
Also given that the payment in each installment $=1000+$ 10% interest on the unpaid amount.
$\Rightarrow\:$ payment in $1^{st} $ year $=1000+\large \frac{10}{100}$$.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 $=4000+35100=Rs. 39100$
answered Apr 9, 2014 by rvidyagovindarajan_1
edited Apr 9, 2014 by rvidyagovindarajan_1
 

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