Question

i want to know that if they form an ap

Answer 1

Are if Roots are equal then apply Roots sum and products relation

➡(a-b)x² +(b-c)x + (c-a) = 0

Top proove c,a,b are in A.P

it must satisfy 2b = a+c

so, By hit and trial method

put x = 1

➡a-b)x² +(b-c)x + (c-a) = 0

▶a-b) +(b-c) + (c-a) = 0

so , 1 is the root of given equation

then , then other root is also 1

▶
$1 + 1 = \frac{ - b}{2a} \\ \\ \frac{ - (b - c)}{2(a - b)} = 2 \\ \\ - b + c = 4a - 4b \\ \\ - b + 4b = 4a - c \\ \\ 3b = 4a - c$


clearly , the relation is not of the form 2b=a+c for being in A.P

so its not in A.P