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)

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

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