Question 1: If $\displaystyle \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in A.P. }$, prove that:

$\displaystyle \text{i) } \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \text{ are in A.P. }$

$\displaystyle \text{ii) } a(b+c), b(c+a), c(a+b) \text{ are in A.P. }$

$\displaystyle \text{i) } \text{Given } \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$

$\displaystyle \Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$

$\displaystyle \Rightarrow 2ac = ab + bc$

$\displaystyle \text{To prove: } 2 \big( \frac{c+a}{b} \big) = \big( \frac{b+c}{a} \big) + \big( \frac{a+b}{c} \big)$

$\displaystyle \Rightarrow 2ac(a+c) = bc(b+c) +ab(a+b)$

$\displaystyle \Rightarrow ( ab+bc)(a+c) = b^2c + bc^2 +a^2b + ab^2$

$\displaystyle \Rightarrow a^2b + abc + abc + bc^2 = b^2c + bc^2 +a^2b + ab^2$

$\displaystyle \Rightarrow a^2b + 2abc + bc^2 = b ( bc+ab) + b^2c^2 + a^2b$

$\displaystyle \Rightarrow a^2b + 2abc + bc^2 = a^2b + 2abc + bc^2$

Hence proved.

$\displaystyle \text{ii) } \text{Given } \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$

$\displaystyle \Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$

$\displaystyle \Rightarrow 2ac = ab + bc$

$\displaystyle \text{To prove: } 2b(c+a) = a( b+c) + c( a+b)$

$\displaystyle \Rightarrow 2bc + 2ba = ab + ac + ca + cb$

$\displaystyle \Rightarrow 2bc + 2ba = ab + 2ac + bc$

$\displaystyle \Rightarrow 2bc + 2ab = ab + ab + bc + bc$

$\displaystyle \Rightarrow 2ab + 2bc = 2ab + 2bc$

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$\displaystyle \text{Question 2: If } a^2, b^2, c^2 \text{ are in A.P. , prove that } \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \text{ are in A.P. }$

$\displaystyle \text{Given } a^2, b^2, c^2 \text{ are in A.P. }$

$\displaystyle 2b^2 = a^2 + c^2$

$\displaystyle \Rightarrow b^2 - a^2 = c^2 - b^2$

$\displaystyle \Rightarrow (b-a)(b+a) = ( c-b)(c+b)$

$\displaystyle \Rightarrow \frac{(b-a)}{(c+b)} = \frac{(c-b)}{(b+a)}$

$\displaystyle \Rightarrow \frac{(b-a)}{(c+b)(c+a)} = \frac{(c-b)}{(b+a)(c+a)}$

$\displaystyle \Rightarrow \frac{1}{c+b} - \frac{1}{b+c} = \frac{1}{a+b} - \frac{1}{c+a}$

$\displaystyle \Rightarrow \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \text{ are in A.P. }$

$\displaystyle \Rightarrow \frac{a+b+c}{b+c}, \frac{a+b+c}{c+a}, \frac{a+b+c}{a+b} \text{ are in A.P. }$

$\displaystyle \Rightarrow \frac{a}{b+c} +1, \frac{b}{c+a} +1, \frac{c}{a+b} +1 \text{ are in A.P. }$

$\displaystyle \Rightarrow \frac{a}{b+c}, \frac{b}{c+a},\frac{c}{a+b} \text{ are in A.P. }$

$\displaystyle \\$

Question 3: If $\displaystyle a, b, c \text{ are in A.P. }$, then show that:

$\displaystyle \text{i) } a^2(b+c), b^2(c+a), c^2(a+b)$ are also in A.P.

$\displaystyle \text{ii) } b+c-a, c+a-b, a+b-c \text{ are in A.P. }$

$\displaystyle \text{iii) } bc-a^2, ca-b^2, ab-c^2 \text{ are in A.P. }$

$\displaystyle \text{i) } \text{Given } a, b, c \text{ are in A.P. }$ $\displaystyle \text{Therefore } 2b = a+ c$

$\displaystyle \text{To prove: } 2b^2(a+c) = a^2(b+c) + c^2(a+b)$

$\displaystyle \Rightarrow 2b^2 (2b) = a^2b + a^2c +c^2 a +c^2b$

$\displaystyle \Rightarrow 4b^3 = ac( a+c) + b(a^2 +c^2)$

$\displaystyle \Rightarrow 4b^3 = ac(2b) + b[ (a+c)^2 - 2ac]$

$\displaystyle \Rightarrow 4b^3 = 2abc + b(a+c)^2- 2abc$

$\displaystyle \Rightarrow 4b^3 = b(2b)^2$

$\displaystyle \Rightarrow 4b^3 = 4b^3$ Hence proved.

$\displaystyle \text{ii) } \text{Given } a, b, c \text{ are in A.P. }$ $\displaystyle \text{Therefore } 2b = a+ c$

$\displaystyle \text{To prove: } 2(c+a-b) = ( b+c-a) + ( a + b - c)$

$\displaystyle \Rightarrow 2 ( 2b - b) = 2b$

$\displaystyle \Rightarrow 2b = 2b$ Hence proved.

$\displaystyle \text{iii) } \text{Given } a, b, c \text{ are in A.P. }$ $\displaystyle \text{Therefore } 2b = a+ c$

$\displaystyle \text{To prove: } 2 ( ca-b^2) = ( bc-a^2) + ( ab - c^2)$

$\displaystyle \text{RHS } = ( bc-a^2) + ( ab - c^2)$

$\displaystyle = c ( b-c) + a(b-a)$

$\displaystyle = c \Big[ \frac{a+c}{2} -c \Big] + a \Big[ \frac{a+c}{2} -a \Big]$

$\displaystyle = c \Big( \frac{a-c}{2} \Big) + a \Big( \frac{c-a}{2} \Big)$

$\displaystyle = \frac{ac-c^2 + ac - a^2}{2}$

$\displaystyle = ac - \frac{1}{2} (a^2 +c^2)$

$\displaystyle = ac - \frac{1}{2} \Big[ (a+c)^2 - 2ac \Big]$

$\displaystyle = ac - \frac{1}{2} \Big[ 4b^2 - 2ac \Big]$

$\displaystyle = ac - 2b^2 +ac$

$\displaystyle = 2ac - 2b^2 = \text{ LHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 4: If } \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \text{ are in A.P. } ,\text{ prove that }$

$\displaystyle \text{i) } \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in A.P. }$

$\displaystyle \text{ii) } bc, ca, ab \text{ are in A.P. }$

$\displaystyle \text{i) } \text{Given } \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c} \text{ are in A.P. }$

$\displaystyle \Rightarrow \frac{c+a}{b} - \frac{b+c}{a} = \frac{a+b}{c} - \frac{c+a}{b}$

$\displaystyle \Rightarrow \frac{ac+a^2-b^2-bc}{ab} = \frac{ab+b^2-c^2-ac}{bc}$

$\displaystyle \Rightarrow \frac{(a+b)(a-b) + c(a-b)}{ab} = \frac{(b+c)(b-c) +a(b-c)}{bc}$

$\displaystyle \Rightarrow \frac{(a-b)(a+b+c)}{ab} = \frac{(b-c)(a+b+c)}{bc}$

$\displaystyle \Rightarrow \frac{a-b}{ab} = \frac{b-c}{bc}$

$\displaystyle \Rightarrow \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$

$\displaystyle \Rightarrow \frac{2}{b} = \frac{1}{a} +\frac{1}{c}$

$\displaystyle \therefore \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in A.P. }$

ii) From i) since $\displaystyle \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in A.P. }$

$\displaystyle \Rightarrow \frac{1}{b}- \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$

$\displaystyle \Rightarrow \frac{a-b}{ab} = \frac{b-c}{bc}$

$\displaystyle \Rightarrow \frac{a-b}{a } = \frac{b-c}{c}$

$\displaystyle \Rightarrow (a-b)c = a(b-c)$

$\displaystyle \Rightarrow ac-bc = ab-ac . \text{ Hence proved.}$

$\displaystyle \\$

Question 5: If $\displaystyle a, b, c \text{ are in A.P. }$, then prove that:

$\displaystyle \text{i) } (a-c)^2 = 4 ( a-b)(b-c)$

$\displaystyle \text{ii) } a^2 + c^2 +4ac = 2 ( ab +bc + ca)$

$\displaystyle \text{iii) } a^3 + c^3 + 6abc = 8b^3$

$\displaystyle \text{i) } \text{Given } a, b, c \text{ are in A.P. }$

$\displaystyle \therefore 2b = a+c \Rightarrow b = \frac{a+c}{2}$

$\displaystyle \text{To prove: } (a-c)^2 = 4(a-b)(b-c)$

$\displaystyle \text{RHS } = 4(a-b)(b-c)$

$\displaystyle = 4 \Big( a - \frac{a+c}{2} \Big) \Big( \frac{a+c}{2} -c \Big)$

$\displaystyle = 4 \Big( \frac{a-c}{2} \Big) \Big( \frac{a-c}{2} \Big)$

$\displaystyle = (a-c)^2 = \text{ LHS Hence proved. }$

$\displaystyle \text{ii) } \text{Given } a, b, c \text{ are in A.P. }$

$\displaystyle \therefore 2b = a+c \Rightarrow b = \frac{a+c}{2}$

$\displaystyle \text{To prove: } a^2 +c^2+4ac = 2(ab+bc+ca)$

$\displaystyle \text{RHS } = 2(ab+bc+ca)$

$\displaystyle = 2 \Big[ a \Big( \frac{a+c}{2} \Big) + \Big( \frac{a+c}{2} \Big) c+ca \Big]$

$\displaystyle = 2 \Big[ \frac{a^2+ac+ac+c^2+2ac}{2} \Big]$

$\displaystyle = a^2 + 4ac + c^2 = \text{ LHS. Hence proved. }$

$\displaystyle \text{iii) } \text{Given } a, b, c \text{ are in A.P. }$

$\displaystyle \therefore 2b = a+c \Rightarrow b = \frac{a+c}{2}$

$\displaystyle \text{To prove: } a^3 + c^3 + 6abc = 8b^3$

$\displaystyle \text{RHS } = 8b^3 = 8 \Big( \frac{a+c}{2} \Big)^3$

$\displaystyle = a^3 + c^3 + 3ac(a+c)$

$\displaystyle = a^3 + c^3 + 3ac(2b)$

$\displaystyle = a^3 + c^3 +6abc = \text{ LHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 6: If } a \Big( \frac{1}{b}+\frac{1}{c} \Big) , \ b \Big( \frac{1}{c}+\frac{1}{a} \Big) , \ c \Big( \frac{1}{a}+\frac{1}{b} \Big) \text{ are in A.P. , prove that } a, b, c \text{ are in A.P. }$

$\displaystyle \text{Given } a \Big( \frac{1}{b}+\frac{1}{c} \Big) , \ b \Big( \frac{1}{c}+\frac{1}{a} \Big) , \ c \Big( \frac{1}{a}+\frac{1}{b} \Big) \text{ are in A.P. }$

$\displaystyle \Rightarrow a \Big( \frac{1}{b}+\frac{1}{c} \Big) + 1, \ b \Big( \frac{1}{c}+\frac{1}{a} \Big) + 1, \ c \Big( \frac{1}{a}+\frac{1}{b} \Big) + 1 \text{ are in A.P. }$

$\displaystyle \Rightarrow a \Big( \frac{1}{b}+\frac{1}{c} +\frac{1}{a} \Big) , \ b \Big( \frac{1}{c}+\frac{1}{a} + \frac{1}{b} \Big) , \ c \Big( \frac{1}{a}+\frac{1}{b} + \frac{1}{c} \Big) \text{ are in A.P. }$

$\displaystyle \text{Dividing by } \Big( \frac{1}{a}+\frac{1}{b} + \frac{1}{c} \Big)$ we get

$\displaystyle a, b, c \text{ are in A.P. } . \text{ Hence proved.}$

$\displaystyle \\$

Question 7: Show that $\displaystyle x^2 + xy + y^2, z^2 + zx + x^2$ and $\displaystyle y^2 + yz + z^2$ are consecutive terms of an A.P., if $\displaystyle x, y, z \text{ are in A.P. }$

$\displaystyle \text{Given } x, y, z \text{ are in A.P. }$.

$\displaystyle \text{Therefore } 2y = x+ z \Rightarrow y = \frac{x+z}{2}$

$\displaystyle \text{To prove: } (x^2 + xy + y^2) , (z^2 + zx + x^2 ) , (y^2 + yz + z^2) \text{ are in A.P. }$

$\displaystyle \Rightarrow (x^2 + xy + y^2) + (y^2 + yz + z^2) = 2(z^2 + zx + x^2 )$

$\displaystyle \text{LHS } = (x^2 + xy + y^2) + (y^2 + yz + z^2)$

$\displaystyle = x^2 + z^2 + 2y^2 + xy +yz$

$\displaystyle = x^2 + z^2 + 2 \Big( \frac{x+z}{2} \Big)^2 + (x+z) \Big( \frac{x+z}{2} \Big )$

$\displaystyle = x^2 + z^2 + 2 \Big[ \frac{(x+z)^2}{4} \Big ] + \frac{(x+z)^2}{2}$

$\displaystyle = x^2 + z^2 + ( x+z)^2$

$\displaystyle = x^2 + z^2 + x^2 + 2xy + z^2$

$\displaystyle = 2x^2 + 2z^2 + 2xy$

$\displaystyle = 2 (x^2 + z^2 + xy)$

$\displaystyle \text{Therefore } x^2 + xy + y^2, z^2 + zx + x^2$ and $\displaystyle y^2 + yz + z^2$ are consecutive terms of an A.P.