Question

Q.5

Answer 1

(1)`(B+C)/a,(c+a)/b and (a+b)/C "are in A.P"` 

`⇒ (c+a)/b-(b+c)/a=(a+b)/c-(c+a)/b` 

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

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

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

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

`⇒(a-b)/(ab)=(b-c)/(bc)` 

`⇒1/b-1/a=1/c-1/b`

`⇒1/b+1/b=1/a+1/c`

`⇒2/b=1/a+1/c`

⇒1/a,1/b and 1/c are in A.P.

(2) `b+c/a, c+a/b and a+b/c "are in A.P."`

`⇒ (c+a)/b-(b+c)/a=(a+b)/c -(c+a)/b`

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

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

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

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

`⇒(a-b)/(ab)=(b-c)/(bc)` 

`⇒( a-b)/a=(b-c)/c`

`⇒ ca-bc=ab-ca`

`⇒2ca=ab+bc`

⇒ bc, ca and ab are in A.P