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Chanchal Kumar Grade: 11
        

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


(i) 2 (a-b) = a-c = 2 (b-c)


(ii)(a-c)^2 = 4(b^2-ac)

7 years ago

Answers : (2)

Askiitians_Expert Yagyadutt
askIITians Faculty
74 Points
										

Hiii chanchal....it is very easy.....just a small trick you have to apply ..


Let   a = p -d


b = p


c = p + d


 


now  a b c are in AP with common difference ....d ..


1) 2 (a-b) = a-c = 2 (b-c)


 


a-b = - d 


a - c = -2d


b -c = - d


 


so it proved from their value ...that  2(a-b) = a-c = 2(b-c)


 


2)(a-c)^2 = 4(b^2-ac)


 


(a-c) = -2d  => (a-c)^2 = 4d^2


ac = ( p-d)(p+d) = p^2 -d^2


b^2 = p^2


b^2  - ac =  d^2


4(b^2-ac) = 4d^2


 


hence ....4(b^2-4ac) = (a-c)^2 ....proved ..


 


 


Let me know if solution is insufficient for understanding....i will describe it more ..


 


Regards


Yagya


askiitians_expert

7 years ago
vikas askiitian expert
510 Points
										

i) 2(a-b)= (a-c) = 29b-c)


 


a,b,c are in ap so [b-a =c-b]..........1




2(a-b) =2(a - (a+c/2))=(a-c )        (by putting b=a+c/2)


2(a-b)=2((2b-c) -b)=2(b-c)             (by putting a=2b-c )




2)  (a-c)2 = 4(b2 - 4ac)


(b2 - 4ac) = 4( (a+c/2)2 -ac )         (by putting b=a+c/2)


                   =(a2 + c2 -2ac)


                   =(a-c)2

7 years ago
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