Class 11: Arithmetic Progressions

$\displaystyle \textbf{Question 1: }\text{If }\frac{1}{a},\frac{1}{b},\frac{1}{c}\text{ are in A.P., prove that:}$
$\displaystyle \text{i) }\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}\text{ are in A.P.}$
$\displaystyle \text{ii) }a(b+c),b(c+a),c(a+b)\text{ are in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{1}{a},\frac{1}{b},\frac{1}{c}\text{ are in A.P.}$
$\displaystyle \therefore \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
$\displaystyle \frac{2}{b}=\frac{a+c}{ac}$
$\displaystyle \therefore b(a+c)=2ac\qquad\ldots\text{(i)}$
$\displaystyle \text{i) We have to prove that}$
$\displaystyle 2\left(\frac{a+c}{b}\right)=\frac{b+c}{a}+\frac{a+b}{c}$
$\displaystyle \frac{b+c}{a}+\frac{a+b}{c}-2\left(\frac{a+c}{b}\right)$
$\displaystyle =\frac{bc(b+c)+ab(a+b)-2ac(a+c)}{abc}$
$\displaystyle =\frac{b^2c+bc^2+a^2b+ab^2-2ac(a+c)}{abc}$
$\displaystyle =\frac{b^2(a+c)+b(a^2+c^2)-2ac(a+c)}{abc}$
$\displaystyle =\frac{b(a+c)^2-2ac(a+c)}{abc}$
$\displaystyle =\frac{(a+c)\left[b(a+c)-2ac\right]}{abc}$
$\displaystyle =0\qquad\text{Using (i)}$
$\displaystyle \therefore 2\left(\frac{a+c}{b}\right)=\frac{b+c}{a}+\frac{a+b}{c}$
$\displaystyle \therefore \frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}\text{ are in A.P.}$
$\displaystyle \text{ii) We have to prove that}$
$\displaystyle 2b(c+a)=a(b+c)+c(a+b)$
$\displaystyle \text{LHS}=2b(a+c)$
$\displaystyle =2(ab+bc)$
$\displaystyle \text{RHS}=a(b+c)+c(a+b)$
$\displaystyle =ab+ac+ac+bc$
$\displaystyle =ab+bc+2ac$
$\displaystyle =ab+bc+b(a+c)\qquad\text{Using (i)}$
$\displaystyle =ab+bc+ab+bc$
$\displaystyle =2(ab+bc)$
$\displaystyle \therefore \text{LHS}=\text{RHS}$
$\displaystyle \therefore a(b+c),b(c+a),c(a+b)\text{ are in A.P.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{If }a^2,b^2,c^2\text{ are in A.P., prove that }\frac{a}{b+c},\frac{b}{c+a},$
$\displaystyle \frac{c}{a+b}\text{ are in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a^2,b^2,c^2\text{ are in A.P.}$
$\displaystyle \therefore 2b^2=a^2+c^2$
$\displaystyle b^2-a^2=c^2-b^2$
$\displaystyle (b-a)(b+a)=(c-b)(c+b)$
$\displaystyle \frac{b-a}{b+c}=\frac{c-b}{a+b}$
$\displaystyle \frac{b-a}{(b+c)(a+c)}=\frac{c-b}{(a+b)(a+c)}$
$\displaystyle \frac{1}{a+c}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{a+c}$
$\displaystyle \therefore \frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b}\text{ are in A.P.}$
$\displaystyle \text{Multiplying each term by }a+b+c,$
$\displaystyle \frac{a+b+c}{b+c},\frac{a+b+c}{a+c},\frac{a+b+c}{a+b}\text{ are in A.P.}$
$\displaystyle \therefore 1+\frac{a}{b+c},1+\frac{b}{a+c},1+\frac{c}{a+b}\text{ are in A.P.}$
$\displaystyle \text{Subtracting }1\text{ from each term,}$
$\displaystyle \therefore \frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}\text{ are in A.P.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{If }a,b,c\text{ are in A.P., show that:}$
$\displaystyle \text{i) }a^2(b+c),b^2(c+a),c^2(a+b)\text{ are also in A.P.}$
$\displaystyle \text{ii) }b+c-a,c+a-b,a+b-c\text{ are in A.P.}$
$\displaystyle \text{iii) }bc-a^2,ca-b^2,ab-c^2\text{ are in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a,b,c\text{ are in A.P.}$
$\displaystyle \therefore 2b=a+c\qquad\ldots\text{(1)}$
$\displaystyle \text{i) We have to prove that}$
$\displaystyle 2b^2(c+a)=a^2(b+c)+c^2(a+b)$
$\displaystyle \text{LHS}=2b^2(a+c)$
$\displaystyle =2b^2(2b)$
$\displaystyle =4b^3$
$\displaystyle \text{RHS}=a^2(b+c)+c^2(a+b)$
$\displaystyle =a^2b+a^2c+ac^2+bc^2$
$\displaystyle =b(a^2+c^2)+ac(a+c)$
$\displaystyle =b\left[(a+c)^2-2ac\right]+ac(a+c)$
$\displaystyle =b\left[(2b)^2-2ac\right]+2abc$
$\displaystyle =4b^3-2abc+2abc$
$\displaystyle =4b^3$
$\displaystyle \therefore \text{LHS}=\text{RHS}$
$\displaystyle \therefore a^2(b+c),b^2(c+a),c^2(a+b)\text{ are in A.P.}$
$\displaystyle \text{ii) We have to prove that}$
$\displaystyle 2(c+a-b)=(b+c-a)+(a+b-c)$
$\displaystyle \text{LHS}=2(a+c-b)$
$\displaystyle =2(2b-b)$
$\displaystyle =2b$
$\displaystyle \text{RHS}=b+c-a+a+b-c$
$\displaystyle =2b$
$\displaystyle \therefore \text{LHS}=\text{RHS}$
$\displaystyle \therefore b+c-a,c+a-b,a+b-c\text{ are in A.P.}$
$\displaystyle \text{iii) We have to prove that}$
$\displaystyle 2(ca-b^2)=(bc-a^2)+(ab-c^2)$
$\displaystyle \text{RHS}=bc-a^2+ab-c^2$
$\displaystyle =c(b-c)+a(b-a)$
$\displaystyle =c\left(\frac{a+c}{2}-c\right)+a\left(\frac{a+c}{2}-a\right)$
$\displaystyle =\frac{c(a-c)+a(c-a)}{2}$
$\displaystyle =\frac{2ac-a^2-c^2}{2}$
$\displaystyle =ac-\frac{1}{2}(a^2+c^2)$
$\displaystyle =ac-\frac{1}{2}\left[(a+c)^2-2ac\right]$
$\displaystyle =ac-\frac{1}{2}(4b^2-2ac)$
$\displaystyle =2ac-2b^2$
$\displaystyle =2(ca-b^2)$
$\displaystyle =\text{LHS}$
$\displaystyle \therefore bc-a^2,ca-b^2,ab-c^2\text{ are in A.P.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}\text{ are in A.P. and }a+b+c\neq0,$
$\displaystyle \text{prove that:}$
$\displaystyle \text{i) }\frac{1}{a},\frac{1}{b},\frac{1}{c}\text{ are in A.P.}$
$\displaystyle \text{ii) }bc,ca,ab\text{ are in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Given, }\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}\text{ are in A.P.}$
$\displaystyle \therefore \frac{c+a}{b}-\frac{b+c}{a}=\frac{a+b}{c}-\frac{c+a}{b}$
$\displaystyle \frac{a(a+c)-b(b+c)}{ab}=\frac{b(a+b)-c(a+c)}{bc}$
$\displaystyle \frac{(a-b)(a+b+c)}{ab}=\frac{(b-c)(a+b+c)}{bc}$
$\displaystyle \text{Since }a+b+c\neq0,$
$\displaystyle \frac{a-b}{ab}=\frac{b-c}{bc}$
$\displaystyle \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$
$\displaystyle \therefore \frac{1}{a},\frac{1}{b},\frac{1}{c}\text{ are in A.P.}$
$\displaystyle \text{ii) From part (i),}$
$\displaystyle \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
$\displaystyle \text{Multiplying both sides by }abc,$
$\displaystyle 2ac=bc+ab$
$\displaystyle \therefore 2(ca)=bc+ab$
$\displaystyle \therefore bc,ca,ab\text{ are in A.P.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }a,b,c\text{ are in A.P., prove that:}$
$\displaystyle \text{i) }(a-c)^2=4(a-b)(b-c)$
$\displaystyle \text{ii) }a^2+c^2+4ac=2(ab+bc+ca)$
$\displaystyle \text{iii) }a^3+c^3+6abc=8b^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a,b,c\text{ are in A.P.}$
$\displaystyle \therefore 2b=a+c$
$\displaystyle \therefore b=\frac{a+c}{2}\qquad\ldots\text{(1)}$
$\displaystyle \text{i) We have to prove that }(a-c)^2=4(a-b)(b-c).$
$\displaystyle \text{RHS}=4(a-b)(b-c)$
$\displaystyle =4\left(a-\frac{a+c}{2}\right)\left(\frac{a+c}{2}-c\right)$
$\displaystyle =4\left(\frac{a-c}{2}\right)\left(\frac{a-c}{2}\right)$
$\displaystyle =(a-c)^2$
$\displaystyle =\text{LHS}$
$\displaystyle \text{ii) We have to prove that }a^2+c^2+4ac=2(ab+bc+ca).$
$\displaystyle \text{RHS}=2(ab+bc+ca)$
$\displaystyle =2\left[a\left(\frac{a+c}{2}\right)+\left(\frac{a+c}{2}\right)c+ca\right]$
$\displaystyle =2\left(\frac{a^2+ac+ac+c^2+2ac}{2}\right)$
$\displaystyle =a^2+c^2+4ac$
$\displaystyle =\text{LHS}$
$\displaystyle \text{iii) We have to prove that }a^3+c^3+6abc=8b^3.$
$\displaystyle \text{RHS}=8b^3$
$\displaystyle =8\left(\frac{a+c}{2}\right)^3$
$\displaystyle =(a+c)^3$
$\displaystyle =a^3+c^3+3ac(a+c)$
$\displaystyle =a^3+c^3+3ac(2b)$
$\displaystyle =a^3+c^3+6abc$
$\displaystyle =\text{LHS}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }a\left(\frac{1}{b}+\frac{1}{c}\right),\ b\left(\frac{1}{c}+\frac{1}{a}\right),$
$\displaystyle c\left(\frac{1}{a}+\frac{1}{b}\right)\text{ are in A.P. and }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\neq0,$
$\displaystyle \text{prove that }a,b,c\text{ are in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a\left(\frac{1}{b}+\frac{1}{c}\right),\ b\left(\frac{1}{c}+\frac{1}{a}\right),$
$\displaystyle c\left(\frac{1}{a}+\frac{1}{b}\right)\text{ are in A.P.}$
$\displaystyle \text{Adding }1\text{ to each term,}$
$\displaystyle a\left(\frac{1}{b}+\frac{1}{c}\right)+1,\ b\left(\frac{1}{c}+\frac{1}{a}\right)+1,$
$\displaystyle c\left(\frac{1}{a}+\frac{1}{b}\right)+1\text{ are in A.P.}$
$\displaystyle \therefore a\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right),$
$\displaystyle b\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right),$
$\displaystyle c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\text{ are in A.P.}$
$\displaystyle \text{Since }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\neq0,\text{ dividing each term by this common factor,}$
$\displaystyle \therefore a,b,c\text{ are in A.P.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Show that }x^2+xy+y^2,\ z^2+zx+x^2\text{ and }y^2+yz+z^2\text{ are}$
$\displaystyle \text{consecutive terms of an A.P., if }x,y,z\text{ are in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }x,y,z\text{ are in A.P.}$
$\displaystyle \therefore 2y=x+z$
$\displaystyle \therefore y=\frac{x+z}{2}$
$\displaystyle \text{We have to prove that}$
$\displaystyle (x^2+xy+y^2)+(y^2+yz+z^2)=2(z^2+zx+x^2).$
$\displaystyle \text{LHS}=(x^2+xy+y^2)+(y^2+yz+z^2)$
$\displaystyle =x^2+z^2+2y^2+y(x+z)$
$\displaystyle =x^2+z^2+2\left(\frac{x+z}{2}\right)^2+\frac{x+z}{2}(x+z)$
$\displaystyle =x^2+z^2+\frac{(x+z)^2}{2}+\frac{(x+z)^2}{2}$
$\displaystyle =x^2+z^2+(x+z)^2$
$\displaystyle =x^2+z^2+x^2+2xz+z^2$
$\displaystyle =2(x^2+xz+z^2)$
$\displaystyle =2(z^2+zx+x^2)$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore x^2+xy+y^2,\ z^2+zx+x^2,\ y^2+yz+z^2\text{ are in A.P.}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$