If $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in $A.P.,$ prove that $a, b, c$ are in $A.P.$
It is given that $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in $A.P.$
$\therefore b\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)$
$\Rightarrow \frac{b(a+c)}{a c}-\frac{a(b+c)}{b c}=\frac{c(a+b)}{a b}-\frac{b(a+c)}{a c}$
$\Rightarrow \frac{b^{2} a+b^{2} c-a^{2} b-a^{2} c}{a b c}=\frac{c^{2} a+c^{2} b-b^{2} a-b^{2} c}{a b c}$
$\Rightarrow b^{2} a-a^{2} b+b^{2} c-a^{2} c=c^{2} a-b^{2} a+c^{2} b-b^{2} c$
$\Rightarrow a b(b-a)+c\left(b^{2}-a^{2}\right)=a\left(c^{2}-b^{2}\right)+b c(c-b)$
$\Rightarrow a b(b-a)+c(b-a)(b+a)=a(c-b)(c+b)+b c(c-b)$
$\Rightarrow(b-a)(a b+c b+c a)=(c-b)(a c+a b+b c)$
$\Rightarrow b-a=c-b$
Thus, $a, b$ and $c$ are in $A.P.$
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?
Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1,6,11$,
. . . .and $Y$ be set consisting of the first $2018$ terms of the arithmetic progression $9, 16, 23$,. . . . . Then, the number of elements in the set $X \cup Y$ is. . . .
The sequence $\frac{5}{{\sqrt 7 }}$, $\frac{6}{{\sqrt 7 }}$, $\sqrt 7 $, ....... is
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The $8^{\text {th }}$ common term of the series $S _1=3+7+11+15+19+\ldots . .$ ; $S _2=1+6+11+16+21+\ldots .$ is $.......$.