if (b-c)^2, (c-a)^2, (a-b)^2 are in AP prove that 1/(b-c) , 1/(c-a) , 1/(a-b) are in AP - Maths - Sequences and Series

Question

if (b-c)^2, (c-a)^2, (a-b)^2 are in AP prove that 1/(b-c) ,
1/(c-a) , 1/(a-b) are in AP

ramandeep.singh · Accepted Answer

$(b - c)^{2}, (c - a)^{2} a n d (a - b)^{2}$ are in A.P.

$∴ (c - a)^{2} = \frac{(b - c)^{2} + (a - b)^{2}}{2} \Rightarrow 2 c^{2} + 2 a^{2} - 4 a c = 2 b^{2} + a^{2} + c^{2} - 2 b c - 2 a b \Rightarrow a^{2} - 2 a c + c^{2} = 2 b^{2} - 2 b c - 2 a b + 2 a c \Rightarrow (a - c)^{2} = 2 (b - a) (b - c) \Rightarrow \frac{a - c}{2 (a - b) (b - c)} = \frac{1}{c - a} \Rightarrow \frac{(a - b) + (b - c)}{2 (a - b) (b - c)} = \frac{1}{c - a} \Rightarrow \frac{1}{b - c} + \frac{1}{a - b} = 2 (\frac{1}{c - a}) \Rightarrow \frac{1}{c - a} - \frac{1}{b - c} = \frac{1}{a - b} - \frac{1}{c - a}$

Therefore, $\frac{1}{b - c}, \frac{1}{c - a} a n d \frac{1}{a - b}$ are in A.P.

if (b-c)^2, (c-a)^2, (a-b)^2 are in AP prove that 1/(b-c) , 1/(c-a) , 1/(a-b) are in AP.