Class 11: Arithmetic Progressions – Exercise 19.5 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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: $2b(c+a) = a( b+c) + c( a+b)$

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

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

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

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

$\\$

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.

$\\$

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.

$\\$

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.

$\\$

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.

$\\$

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.