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

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

ii) a(b+c), b(c+a), c(a+b) are in A.P.

Answer:

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

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

\Rightarrow 2ac = ab + bc

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

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

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

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

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

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

Hence proved.

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

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

\Rightarrow 2ac = ab + bc

To prove: 2b(c+a) = a( b+c) + c( a+b)

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

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

\Rightarrow 2bc + 2ab = ab + ab + bc + bc

\Rightarrow 2ab + 2bc = 2ab + 2bc

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

Answer:

Given a^2, b^2, c^2 are in A.P.

2b^2 = a^2 + c^2

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

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

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

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

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

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

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

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

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

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Question 3: If a, b, c are in A.P., then show that:

i) a^2(b+c), b^2(c+a), c^2(a+b) are also in A.P.

ii) b+c-a, c+a-b, a+b-c are in A.P.

iii) bc-a^2, ca-b^2, ab-c^2 are in A.P.

Answer:

i)      Given a, b, c are in A.P. Therefore 2b = a+ c

To prove: 2b^2(a+c) = a^2(b+c) + c^2(a+b)

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

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

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

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

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

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

ii)      Given a, b, c are in A.P. Therefore 2b = a+ c

To prove: 2(c+a-b) = ( b+c-a) + ( a + b - c)

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

\Rightarrow 2b = 2b      Hence proved.

iii)     Given a, b, c are in A.P. Therefore 2b = a+ c

To prove: 2 ( ca-b^2) = ( bc-a^2) + ( ab - c^2)

RHS = ( bc-a^2) + ( ab - c^2)

= c ( b-c) + a(b-a)

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

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

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

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

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

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

= ac - 2b^2 +ac

= 2ac - 2b^2 = LHS. Hence proved.

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

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

ii) bc, ca, ab are in A.P.

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\Rightarrow ac-bc = ab-ac .    Hence proved.

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Question 5: If a, b, c are in A.P., then prove that:

i) (a-c)^2 = 4 ( a-b)(b-c)

ii) a^2 + c^2 +4ac = 2 ( ab +bc + ca)

iii) a^3 + c^3 + 6abc = 8b^3

Answer:

i)       Given a, b, c are in A.P.

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

To prove: (a-c)^2 = 4(a-b)(b-c)

RHS = 4(a-b)(b-c)

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

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

= (a-c)^2 = LHS      Hence proved.

ii)      Given a, b, c are in A.P.

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

To prove: a^2 +c^2+4ac = 2(ab+bc+ca)

RHS = 2(ab+bc+ca)

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

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

= a^2 + 4ac + c^2 = LHS. Hence proved.

iii)     Given a, b, c are in A.P.

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

To prove: a^3 + c^3 + 6abc = 8b^3

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

= a^3 + c^3 + 3ac(a+c)

= a^3 + c^3 + 3ac(2b)

= a^3 + c^3 +6abc = LHS. Hence proved.

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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) are in A.P., prove that a, b, c are in A.P. 

Answer:

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) are in A.P.

\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 are in A.P.

\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) are in A.P.

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

a, b, c are in A.P.. Hence proved.

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

Answer:

Given x, y, z are in A.P..

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

To prove: (x^2 + xy + y^2) , (z^2 + zx + x^2 ) ,  (y^2 + yz + z^2) are in A.P.

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

LHS = (x^2 + xy + y^2) +  (y^2 + yz + z^2)

= x^2 + z^2 + 2y^2 + xy +yz

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

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

= x^2 + z^2 + ( x+z)^2

= x^2 + z^2 + x^2 + 2xy + z^2

= 2x^2 + 2z^2 + 2xy

= 2 (x^2 + z^2 + xy)

Therefore x^2 + xy + y^2, z^2 + zx + x^2 and y^2 + yz + z^2 are consecutive terms of an A.P.