if a^2,b^2,c^2 are in ap then show b+c,c+a,a+b are in h.p

here a^2, b^2,and c^2 are in ap

therefore  b^2-a^2=c^2-b^2

               2(b^2)=a^2+c^2....................1

now a^2, b^2,and c^2 are in ap

therefore 1/a^2, 1/b^2, 1/c^2 are in hp

                   1/(b^2)^2=1/(a^2)^2*1/(c^2)^2

                  b^4=a^2*c^2 (taking square root on both sides and reciprocle)

                  b^2=ac...............................2

from 1&2      2(ac)=a^2+c^2

                   a^2-2ac+c^2=0

                  (a-c)^2=0

                   a-c=0

                   a=c.............................3

from 3&2          b^2=c^2

                        b=c.........................4

 from 3&4           a=b=c

                       a+b=b+c=c+a

                       (b+c)^2=(a+b)*(c+a)

               and therefore 

Dear Student,
Please find below the solution to your problem.

given,a2,b2and c2are in AP
=>b2-a2=c2-b2
=>[b-a][b+a]=[c+b][c-b]
=>[b-a]/[c+b]=[c-b]/[b+a]
=>[b-a]/[b+c]=[c-b]/[a+b]
=>{ [b-a]/[b+c] } * {1/[c+a] } = { [c-b]/[a+b] } * {1/[c+a] }
=>{ [b-a] } / { [b+c][c+a] } = { [c-b] } / { [a+b][c+a] }
=>{ [b-a+c-c] } / { [b+c][c+a] } = { [c-b+a-a] } / { [a+b][c+a] }
=>{ [b+c-a-c] } / { [b+c][c+a] } = { [c+a-b-a] } / { [a+b][c+a] }
=>{ [b+c]-[a+c] } / { [b+c][c+a] } = { [c+a]-[b+a] } / { [c+a][a+b] }
=>{ [b+c] / [b+c][c+a] } – { [a+c] /[b+c][c+a] } = { [c+a] / [c+a][b+a] } – { [b+a] / [c+a][b+a] }
=>{ 1/[c+a] } – { 1/[b+c] } = { 1/[b+a] } – { 1/[c+a] }
=>2 * {1/[c+a] } = {1/[b+a] } + {1/[b+c] }
therefore ,1/[a+b] , 1/[c+a] and 1/[b+c] are in AP
=> [a+b] , [c+a] and [b+c] are in HP
=> [b+c] , [c+a] and [a+b] are in HP

Thanks and Regards