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Question

If $a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma horizontal ellipsis comma a subscript 2 n plus 1 end subscript$ are in A.P., then $fraction numerator a subscript 2 n plus 1 end subscript minus a subscript 1 end subscript over denominator a subscript 2 n plus 1 end subscript plus a subscript 1 end subscript end fraction plus fraction numerator a subscript 2 n end subscript minus a subscript 2 end subscript over denominator a subscript 2 n end subscript plus a subscript 2 end subscript end fraction plus horizontal ellipsis plus fraction numerator a subscript n plus 2 end subscript minus a subscript n end subscript over denominator a subscript n plus 2 end subscript plus a subscript n end subscript end fraction$ is equal to

$\frac{n (n + 1)}{2} \times \frac{a_{2} - a_{1}}{a_{n + 1}}$
$\frac{n (n + 1)}{2}$
$(n + 1) (a_{2} - a_{1})$
None of these

The correct answer is: $fraction numerator n left parenthesis n plus 1 right parenthesis over denominator 2 end fraction cross times fraction numerator a subscript 2 end subscript minus a subscript 1 end subscript over denominator a subscript n plus 1 end subscript end fraction$

The general term can be given by
$t subscript r plus 1 end subscript equals fraction numerator a subscript 2 n plus 1 minus r end subscript minus a subscript r plus 1 end subscript over denominator a subscript 2 n plus 1 minus r end subscript plus a subscript r plus 1 end subscript end fraction comma blank r equals 0 comma blank 1 comma blank 2 comma blank horizontal ellipsis comma n minus 1$
$equals fraction numerator a subscript 1 end subscript plus open parentheses 2 n minus r close parentheses d minus left curly bracket a subscript 1 end subscript plus r d right curly bracket over denominator a subscript 1 end subscript plus open parentheses 2 n minus r close parentheses d plus left curly bracket a subscript 1 end subscript plus r d right curly bracket end fraction$
$equals fraction numerator left parenthesis n minus r right parenthesis d over denominator a subscript 1 end subscript plus n d end fraction$
Therefore, the required sum is
$S subscript n end subscript equals not stretchy sum from r equals 0 to n minus 1 of t subscript r plus 1 end subscript$
$equals not stretchy sum from r equals 0 to n minus 1 of fraction numerator left parenthesis n minus r right parenthesis d over denominator a subscript 1 end subscript plus n d end fraction$
$equals open square brackets fraction numerator n plus open parentheses n minus 1 close parentheses plus open parentheses n minus 2 close parentheses plus horizontal ellipsis plus 1 over denominator a subscript 1 end subscript plus n d end fraction close square brackets d$
$equals fraction numerator n left parenthesis n plus 1 right parenthesis d over denominator 2 a subscript n plus 1 end subscript end fraction$
$equals fraction numerator n left parenthesis n plus 1 right parenthesis over denominator 2 end fraction blank fraction numerator a subscript 2 end subscript minus a subscript 1 end subscript over denominator a subscript n plus 1 end subscript end fraction blank left square bracket because d equals a subscript 2 end subscript minus a subscript 1 end subscript right square bracket$

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