Question

Q.6

Answer 1

`1/a,1/b and 1/c "are in A.P"` 

`⇒ 2/b=1/a+1/c`

`⇒1/b+1/b=1/a+1/c`

`⇒ 1/b-1/a=1/c-1/b`

`⇒(a-b)/(ab)=(b-c)/(bc)`

`⇒((a-b)(a+b+c))/(ab)=((b-c)(a+b+c))/(bc)`

`⇒((a-b)[(a+b)+c])/(ab)=((b-c)[a+(b+c)])/(bc)`

`⇒ ((a-b)(a+b)+(a-b)c)/(ab)=((b-c)a+(b^2-c^2))/(bc)`

`⇒((a^2-b^2)+(a-b)c)/(ab)=((b-c)a+(b^2-c^2))/(bc)`

`⇒(a^2-b^2+ac-bc)/(ab)=(ab-ca+b^2-c^2)/(bc)`

`⇒(a^2+ac-b^2-bc)/(ab)=(ab+b^2-c^2-ca)/(bc)`

`⇒(a(a+c)-b(b+c))/(ab)=(b(a+b)-c(c+a))/(bc)`

`⇒(c+a)/b-(b+c)/a=(a+b)/c-(c+a)/b`

`⇒(c+a)/b+(c+a)/b=(a+b)/c+(b+c)/a`

`⇒2((c+a)/b)=(a+b)/c+(b+c)/a`

`⇒(b+c)/a,(c+a)/b and (a+b)/c "are also n A.P"`