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 $=( Rs .12000+ Rs .4680)= Rs .16680$
Let $S_{1}$ be the sum of first $2 n$ terms of an arithmetic progression. Let, $S_{2}$ be the sum of first $4n$ terms of the same arithmetic progression. If $\left( S _{2}- S _{1}\right)$ is $1000,$ then the sum of the first $6 n$ terms of the arithmetic progression is equal to:
If ${\log _5}2,\,{\log _5}({2^x} - 3)$ and ${\log _5}(\frac{{17}}{2} + {2^{x - 1}})$ are in $A.P.$ then the value of $x$ is :-
If the $A.M.$ between $p^{th}$ and $q^{th}$ terms of an $A.P.$ is equal to the $A.M.$ between $r^{th}$ and $s^{th}$ terms of the same $A.P.$, then $p + q$ is equal to
The sum of $n$ arithmetic means between $a$ and $b$, is
The sum of the first $20$ terms common between the series $3 +7 + 1 1 + 15+ ... ......$ and $1 +6+ 11 + 16+ ......$, is