If a,b,c are in A.P., then prove that the following are also in A.P
(i) `a^(2)(b+c),b^(2)(c+a),c^(2)(a+b)` ltbr gt(ii) `1/(sqrtb+sqrtc),1/(sqrtc+sqrta),1/(sqrta+sqrtb)`
(iii) `a(1/b+1/c),b(1/c+1/a),c(1/a+1/b)`

Explanation

Let `a^(2)(b+c),b^(2)(c+a),c^(2)(a+b)` are in A.P. Then,
`b^(2)(c+a)-a^(2)(b+c)=c^(2)(a+b)-b^(2)(c+a)`
or `c(b^(2)-a^(2))+ab(b-a)=a(c^(2)-b^(2))+bc(c-b)`
or `(b-a)(ab+bc+ca)=(c-b)(ab+bc+ca)`
or b-a=c-b
or 2b=a+c
Thus, a,b,c are in A.P., which is given.
(ii) Let `1/(sqrtb+sqrtc),1/(sqrtc+sqrta),1/(sqrta+sqrtb)` are in A.P.then,
`1/(sqrtc+sqrta)-1/(sqrtb+sqrtc)=1/(sqrta+sqrtb)-1/(sqrtc+sqrta)`
or `(sqrtb-sqrta)/((sqrtc+sqrta)(sqrtb+sqrtc))=((sqrtc-sqrtb))/((sqrta+sqrtb)(sqrtc+sqrta))`
or `(sqrtb-sqrta)/((sqrtc+sqrta)(sqrtb+sqrtc))=((sqrtc-sqrtb))/((sqrta+sqrtb)(sqrtc+sqrta))`
or `(sqrtb-sqrta)/(sqrtb+sqrtc)=(sqrtc-sqrtb)/(sqrta+sqrtb)`
or b-a=c-b
or 2b=a+c
Thus, a,b,c are in A.P., which is given.
Hence, `1/(sqrtb+sqrtc),1/(sqrtc+sqrta),1/(sqrta+sqrtb)` are in A.P.
(iii) a,b,c are in A.P. Then,
`a/(abc),b/(abc),c/(abc)` are in A.P.
[On dividing each term by abc]
`rArr1/(bc),1/(ca),1/(ab)` are in A.P.
`rArr(ab+bc+ca)/(bc),(ab+bc+ca)/(ca),(ab+bc+ca)/(ab)` are in A.P.
[On multiplying each term by ab+bc+ca]
`rArr(ab+bc+ca)/(bc)-1,(ab+bc+ca)/(ca)-1,(ab+bc+ca)/(ab)-1`
are in A.P. [On adding -1 to each term]
`rArr(ab+ac)/(bc),(ab+bc)/(ca),(bc+ca)/(ab)` are in A.P.
`rArra(1/b+1/c),b(1/c+1/a),c(1/a+1/b)` are in A.P.