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Arithmetic Progressions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsArithmetic Progressions5 Marks
Question
Show that $\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2$ and $\text{y}^2+\text{yz}+\text{z}^2$ are consecutive terms of an A.P., if x, y and z are in A.P.
✓

Answer

x, y and z are in A.P.
Let d be the common difference then,
$\text{y}=\text{x}+\text{d}$ and $\text{x}=\text{x}+2\text{d}$
To show $\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2+\text{zx}+\text{x}^2$ and consecutive terms of an A.P., it is enough to show that,
$(\text{z}^2+\text{zx}+\text{x}^2)-(\text{x}^2+\text{xy}+\text{y}^2)\\=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$\text{LHS}=(\text{z}^2+\text{zx}+\text{x}^2)-(\text{x}^2+\text{xy}+\text{y}^2)$
$(\text{z}^2+\text{zx}-\text{zy}-\text{y}^2)$
$=(\text{x}2\text{d})^2+(\text{x}+2\text{d})\text{x}-\text{x}(\text{x}+\text{d})-(\text{x}+\text{d})^2$
$=\text{x}^2+4\text{xd}+4\text{d}^2+\text{x}^2+2\text{xd}\\-\text{x}^2-\text{xd}-\text{x}^2-2\text{xd}-\text{d}^2$
$=3\text{xd}+\text{d}3^2$
$\text{RHS}=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$=(\text{x}+\text{d})^2+(\text{x}+\text{d})(\text{x}+2\text{d})-(\text{x}+2\text{d})\text{x}=\text{x}^2$
$=\text{x}^2+2\text{d}\text{x}+\text{d}^2+\text{x}^2+2\text{dx}\\+\text{xd}+2\text{d}^2-\text{x}^2-2\text{dx}-\text{x}^2$
$=3\text{xd}+3\text{d}^2$
$\therefore\text{LHS}=\text{RHS}$
$\therefore\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2$ and $\text{y}^2+\text{yz}+\text{z}^2$ are consecutive terms of an A.P.

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