If (b-c)(b-c), (c-a)(c-a), (a-b)(a-b) are in AP then show that 1/(b-c), 1/(c-a), 1/(a-b) are in AP. {USE :ADD ab+bc+ca-a^2-b^2-c^2 to each term} 

Asked by nk.pandit02 | 12th Mar, 2017, 01:46: PM

Expert Answer:

begin mathsize 16px style To space prove colon space fraction numerator 1 over denominator straight b minus straight c end fraction comma space fraction numerator 1 over denominator straight c minus straight a end fraction comma space fraction numerator 1 over denominator straight a minus straight b end fraction space are space in space AP
rightwards double arrow To space prove colon fraction numerator 1 over denominator straight c minus straight a end fraction minus space fraction numerator 1 over denominator straight b minus straight c end fraction equals fraction numerator 1 over denominator straight a minus straight b end fraction minus fraction numerator 1 over denominator straight c minus straight a end fraction
rightwards double arrow To space prove colon fraction numerator 1 over denominator straight a minus straight b end fraction plus space fraction numerator 1 over denominator straight b minus straight c end fraction equals fraction numerator 2 over denominator straight c minus straight a end fraction space
rightwards double arrow To space prove colon left parenthesis straight b minus straight c plus straight a minus straight b right parenthesis left parenthesis straight c minus straight a right parenthesis equals 2 left parenthesis straight a minus straight b right parenthesis left parenthesis straight b minus straight c right parenthesis
rightwards double arrow To space prove colon left parenthesis straight a minus straight c right parenthesis left parenthesis straight c minus straight a right parenthesis equals 2 left parenthesis straight a minus straight b right parenthesis left parenthesis straight b minus straight c right parenthesis
rightwards double arrow To space prove colon ac minus straight a squared minus straight c squared plus ac equals 2 left square bracket ab minus ac minus straight b squared plus bc right square bracket
rightwards double arrow To space prove colon ac minus straight a squared minus straight c squared plus ac equals 2 ab minus 2 ac minus 2 straight b squared plus 2 bc
rightwards double arrow To space prove colon 2 ab minus 4 ac minus 2 straight b squared plus 2 bc plus straight a squared plus straight c squared equals 0 space space space..... left parenthesis straight i right parenthesis

Given space that space left parenthesis straight b minus straight c right parenthesis left parenthesis straight b minus straight c right parenthesis comma space left parenthesis straight c minus straight a right parenthesis left parenthesis straight c minus straight a right parenthesis comma space left parenthesis straight a minus straight b right parenthesis left parenthesis straight a minus straight b right parenthesis space are space in space AP
rightwards double arrow left parenthesis straight c minus straight a right parenthesis left parenthesis straight c minus straight a right parenthesis minus left parenthesis straight b minus straight c right parenthesis left parenthesis straight b minus straight c right parenthesis equals left parenthesis straight a minus straight b right parenthesis left parenthesis straight a minus straight b right parenthesis minus left parenthesis straight c minus straight a right parenthesis left parenthesis straight c minus straight a right parenthesis
rightwards double arrow left parenthesis straight c minus straight a right parenthesis squared negative left parenthesis straight b minus straight c right parenthesis squared equals left parenthesis straight a minus straight b right parenthesis squared minus left parenthesis straight c minus straight a right parenthesis squared
rightwards double arrow left parenthesis straight a minus straight b right parenthesis squared plus left parenthesis straight b minus straight c right parenthesis squared minus 2 left parenthesis straight c minus straight a right parenthesis squared equals 0
rightwards double arrow straight a squared minus 2 ab plus straight b squared plus straight b squared minus 2 bc plus straight c squared minus 2 straight c squared plus 4 ac minus 2 straight a squared equals 0
rightwards double arrow negative 2 ab plus 4 ac plus 2 straight b squared minus 2 bc minus straight a squared minus straight c squared equals 0
that space is comma space 2 ab minus 4 ac minus 2 straight b squared plus 2 bc plus straight a squared plus straight c squared equals 0
So comma space space from space left parenthesis straight i right parenthesis comma space we space get
fraction numerator 1 over denominator straight b minus straight c end fraction comma space fraction numerator 1 over denominator straight c minus straight a end fraction comma space fraction numerator 1 over denominator straight a minus straight b end fraction space are space in space AP end style
 
This is the method for the prove. I am not clear as in why you want this: {USE :ADD ab+bc+ca-a^2-b^2-c^2 to each term}. But I have got the proof as shown above. Hope this method is clear. Thank you.

Answered by Rebecca Fernandes | 12th Mar, 2017, 02:10: PM