Question

if a,b,c are in A.P then prove that 1/√b+√c ,1/√c+√a ,1/√a+√b are in A.P​

Answer 1

Given : a,b,c are in A.P

To find : prove that 1/√b+√c ,1/√c+√a ,1/√a+√b are in A.P

Solution:

a,b,c are in A.P

=> b - a  = c - b   = d

also   c - b  + b -a  = c - a

=> c - a = 2(b-a) = 2(c - a)  = 2d

=> d = (c- a)/2

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

iff

(1/(√b+√c)  +  1/(√a+√b)  = 2/(√c+√a)

LHS = (1/(√b+√c)  +  1/(√a+√b)

1/(√b+√c)  = √c - √b / (c - b)   = (√c - √b)/d

1/(√a+√b)  =    √b - √a / (b - a) = (√b - √a)/d

= (√c - √b)/d +  (√b - √a)/d

=  (√c - √b + √b - √a )/d

=  (√c -   √a )/d

=  (√c -   √a )/ (c - a)/2

= 2  (√c -   √a )/ (c - a)

= 2  (√c -   √a )/ (√c +   √a )(√c - √a)

=  2  /(√c +   √a )

= RHS

QED

Hence proved

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