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)

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

Approved

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