Question

19. If a b c , ,2 2 2 are in A.P. then show that

Answer 1

If a^2,b^2,c^2 are in AP
b^2-a^2=c^2-b^2 (b-a)=(c-b)(c+b)/(b+a)
1/(c+a)-1/(c+b)=[(c+b)-(c+a)]/(c+a)(c+b)
=(b-a)/(c+a)(c+b)
=(c-b)(c+b)/(c+a)(c+b)(b+a)
=(c-b)/(c+a)(b+a) (1)
1/(b+a)-1/(c+a)
=[(c+a)-(b+a)]/(b+a)(c+a)
=(c-b)/(b+a)(c+a) (2)
from eq 1 &2
they r in AP

Answer 2

Solution :

It is given that ,

a² , b² , c² are in A.P

=> b² - a² = c² - b² [ common difference ]

=> (b+a)(b-a)= (c+b)(c-a)

=> (b-a)/(c+b) = (c-b)/(b+a)

=>[b+c-c-a]/[(c+b)(c+a)]=[c+a-a-b]/[(b+a)(c+a)]

=>[(b+c)-(c+a)]/[(c+b)(c+a)]

= [(c+a)-(a+b)]/[(b+a)(c+a)]

=> (b+c)/[(c+b)(c+a)] -(c+a)/[(c+b)(c+a)]

= (c+a)/[(b+a)(c+a)] - (a+b)/[(b+a)(c+a)]

=> 1/(c+a) - 1/(c+b) = 1/(b+a) - 1/(c+a)

Therefore

1/(b+c) , 1/(c+a) , 1/(a+b) are in A.P

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