Question

please help me with my friends​

Answer 1

   

Oh, again!!! Answering it fifth time!!!

Roots are equal. So discriminant is zero.  

               

Therefore,  

$(b-c)^2-(4(a-b)(c-a)) = 0 \\ \\ (b^2-2bc+c^2)-(4(ac-a^2-bc+ab)) = 0 \\ \\ (b^2-2bc+c^2)-(4(ab-bc+ac-a^2)) = 0 \\ \\ (b^2-2bc+c^2)-(4ab-4bc+4ac-4a^2)=0 \\ \\ b^2-2bc+c^2-4ab+4bc-4ac+4a^2=0 \\ \\ 4a^2+b^2+c^2-4ab+2bc-4ac=0 \\ \\ (2a-b-c)^2=0 \\ \\ 2a-b-c=0 \\ \\ 2a=b+c \\ \\ 2a=c+b$

From this, we can say that c, a, b are in AP because, when we consider three consecutive terms in an AP, the sum of first and third terms is equal to double the second term.

Let me show you.

$Let \ \ c=T_1,\ a,=T_2,\ \ \ \&\ \ \ b=T_3 \\ \\ \therefore\ T_2=T_1+d\ \ \ \& \ \ \ T_3=T_1+2d \\ \\ \\ \\ c+b \\ \\ T_1+T_3 \\ \\ T_1+T_1+2d \\ \\ T_1+T_1+d+d \\ \\ T_1+d+T_1+d \\ \\ 2(T_1+d) \\ \\ 2 \times T_2 \\ \\ 2a \\ \\ \\ \therefore\ 2a=c+b$

∴ c, a, b are in AP.

Hence showed.

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Thank you. ^_^

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