Question

If a, b, c is in A.P., then show that:

bc − a², ca − b², ab − c² are in A.P.

Answer 1

\[\text { Since a, b, c are in A . P . , we have: } \]

\[2b = a + c\]

\[\text {  We have to prove the following: } \]

\[2(ca - b^2 ) = \left( bc - a^2 + ab - c^2 \right)\]

\[\text { RHS }: bc - a^2 + ab - c^2 \]

\[ = c(b - c) + a(b - a)\]

\[ = c\left( \frac{a + c}{2} - c \right) + a\left( \frac{a + c}{2} - a \right) \left( \because 2b = a + c \right)\]

\[ = c\left( \frac{a + c - 2c}{2} \right) + a\left( \frac{a + c - 2a}{2} \right)\]

\[ = \frac{c\left( a - c \right)}{2} + a\left( \frac{c - a}{2} \right)\]

\[ = \frac{ca}{2} - \frac{c^2}{2} + \frac{ac}{2} - \frac{a^2}{2}\]

\[ = ac - \frac{1}{2}\left( c^2 + a^2 \right)\]

\[ = ac - \frac{1}{2}\left( 4 b^2 - 2ac \right) \left( \because a^2 + c^2 + 2ac = 4 b^2 \Rightarrow a^2 + c^2 = 4 b^2 - 2ac \right)\]

\[ = ac - 2 b^2 + ac\]

\[ = 2ac - 2 b^2 \]

\[ = 2\left( ac - b^2 \right)\]

\[ =\text {  LHS }\]

\[\text { Hence, proved } . \]