Let the sequence {a_1},{a_2},{a_3},.......... | vedclass

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Question

(a) Since ${a_1},\;{a_2},\,{a_3},...........,{a_n}$ form an $A.P.$

therefore, ${a_2} - {a_1} = {a_4} - {a_3} = ....... = {a_{2n}} - {a_{2n - 1}} =

Vedclass · Accepted Answer

$\frac{n}{{2n - 1}}(a_1^2 - a_{2n}^2)$

Vedclass · Answer

(a) Since ${a_1},\;{a_2},\,{a_3},...........,{a_n}$ form an $A.P.$

therefore, ${a_2} - {a_1} = {a_4} - {a_3} = ....... = {a_{2n}} - {a_{2n - 1}} = d$

Here $a_1^2 - a_2^2 + a_3^2 - a_4^2 +$$ ....... + a_{2n - 1}^2 - a_{2n}^2$

$ = ({a_1} - {a_2})({a_1} + {a_2}) + ({a_3} - {a_4})({a_3} + {a_4}) +$$ ...... + ({a_{2n - 1}} - {a_{2n}})({a_{2n - 1}} + {a_{2n}})$

$ = - d({a_1} + {a_2} + ....... + {a_{2n}}) = - d\left\{ {\frac{{2n}}{2}({a_1} + {a_{2n}})} ight\}$

Also we know ${a_{2n}} = {a_1} + (2n - 1)d$$ \Rightarrow $$d = \frac{{{a_{2n}} - {a_1}}}{{2n - 1}}$

$ \Rightarrow $ $ - d = \frac{{{a_1} - {a_{2n}}}}{{2n - 1}}$.

$	herefore $ Therefore the sum is

= $\frac{{n({a_1} - {a_{2n}}).({a_1} + {a_{2n}})}}{{2n - 1}} = \frac{n}{{2n - 1}}(a_1^2 - a_{2n}^2)$.

Let the sequence ${a_1},{a_2},{a_3},.............{a_{2n}}$ form an $A.P. $ Then $a_1^2 - a_2^2 + a_3^3 - ......... + a_{2n - 1}^2 - a_{2n}^2 = $

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