Let $a_{n}$ be the $n^{t h}$ term of an $A P$ .Show that
${a_{1}}^{2} - {a_{2}}^{2} + {a_{3}}^{2} - {a_{4}}^{2} + . . . . . . . . . . + {a^{2}}_{2 n - 1} - {a^{2}}_{2 n} = \frac{n}{2 n - 1} ({a_{1}}^{2} - {a_{2 n}}^{2})$

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Solution

${a_{1}}^{2} - {a_{2}}^{2} + {a_{3}}^{2} - {a_{4}}^{2} + {a_{5}}^{2} - {a_{6}}^{2} + . . . . . . . . {a^{2}}_{2 n - 1} - {a^{2}}_{2 n}$
$(a_{1} - a_{2}) (a_{1} + a_{2}) + (a_{3} - a_{4}) (a_{3} + a_{4}) . . . . .$
$= - d [a_{1} + a_{2} + a_{3} . . . . . . . . a_{2 n}]$
$= - d \frac{[n (a_{1} + a_{2 n})] (a_{1} - a_{2 n})}{(a_{1} - a_{2 n})}$
$= \frac{- d n [{a_{1}}^{2} - {a^{2}}_{2 n}]}{- d (2 n - 1)} = \frac{n}{2 n - 1} (a_{1}^{2} - a_{2 n}^{2})$

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