Question

Show that${(\mathrm{a}-\mathrm{b})}^2$, $({\mathrm{a}}^2+{\mathrm{b}}^2)$ and ${(\mathrm{a}+\mathrm{b})}^2$ are In A.P.

Answer 1

To prove : ${(\mathrm{a}-\mathrm{b})}^2$, $({\mathrm{a}}^2+{\mathrm{b}}^2)$ and ${(\mathrm{a}+\mathrm{b})}^2$ are in A.P

Note: A sequence is said to be in A.P when the difference between its all the successive and preceding terms terms are equal. This difference is known as “common difference”.

Obtain the difference between two successive terms of the sequence given above, we get

Common difference between $({\mathrm{a}}^2+{\mathrm{b}}^2)$ and ${(\mathrm{a}-\mathrm{b})}^2$$=$$({\mathrm{a}}^2+{\mathrm{b}}^2)-({\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab})$

$=$ ${\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{ab}$

$=2\mathrm{ab}$ (like terms with opposite sign cancel each other)

Common difference between ${(\mathrm{a}+\mathrm{b})}^2$ and ${({\mathrm{a}}^2+\mathrm{b}}^2)$$=$$({\mathrm{a}}^2+{\mathrm{b}}^2)-({\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab})$

$=$ ${\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{a}}^2-{\mathrm{b}}^2+2\mathrm{ab}$

$=2\mathrm{ab}$ (like terms with opposite sign cancel each other)

Common difference between all the terms of the given sequence are equal.

Thus, ${(\mathrm{a}-\mathrm{b})}^2$,$({\mathrm{a}}^2+{\mathrm{b}}^2)$and ${(\mathrm{a}+\mathrm{b})}^2$ are in A.P