A farmer buys a used tractor for $Rs$ $12000 .$ He pays $Rs$ $6000$ cash and agrees to pay the balance in annual instalments of $Rs$ $500$ plus $12 \%$ interest on the unpaid amount. How much will the tractor cost him?
It is given 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 $5000 \ldots \ldots 12 \%$ of $500$
Thus, total interest to be paid
$=12 \%$ of $6000+12 \%$ of $5500+12 \%$ of $5000+\ldots \ldots+12 \%$ of $500$
$=12 \%$ of $(6000+5500+5000+\ldots .+500)$
$=12 \%$ of $(500+1000+1500+\ldots \ldots+6000)$
Now, the series $500,1000,1500 \ldots 6000$ 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 $=( Rs .12000+ Rs .4680)= Rs .16680$
If ${m^{th}}$ terms of the series $63 + 65 + 67 + 69 + .........$ and $3 + 10 + 17 + 24 + ......$ be equal, then $m = $
Let ${S_n}$ denotes the sum of $n$ terms of an $A.P.$ If ${S_{2n}} = 3{S_n}$, then ratio $\frac{{{S_{3n}}}}{{{S_n}}} = $
If ${n^{th}}$ terms of two $A.P.$'s are $3n + 8$ and $7n + 15$, then the ratio of their ${12^{th}}$ terms will be
The sum of all natural numbers between $1$ and $100$ which are multiples of $3$ is
Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=n(n+2)$