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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
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