Question

if a,b,c,are in a.p prove that.

Answer 1

a, b, c ... in AP. 

∴ 2b = a + c .......... (1) 

∴ To Prove : 1/(√b + √c), 1/(√c + √a), 1/(√a + √b) ... in AP 

⇒ To Prove : [ 1/(√c+√a) ] - [ 1/(√b+√c) ] = [ 1/(√a+√b) ] - [ 1/(√c+√a) ] 

⇒ To Prove : [ (√b-√a) / (√b+√c) ] = [ (√c-√b) / (√a+√b) ] 

⇒ To Prove : (√b-√a)·(√b+√a) = (√c-√b)·(√c+√b) 

⇒ To Prove : b - a = c - b 

⇒ To Prove : 2b = a + c, which is True ... from (1) 

Hence, the result.

Answer 2

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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