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Maths-

General

Easy

Question

If, for a positive integer n, the quadratic equation $x left parenthesis x plus 1 right parenthesis plus left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis plus horizontal ellipsis plus left parenthesis x plus stack n minus 1 with bar on top right parenthesis left parenthesis x plus n right parenthesis equals 10 n$ has two consecutive integral solutions, then n is equal to

9
10
11
12

The correct answer is: 11

The given quadratic equation is
$x left parenthesis x plus 1 right parenthesis plus left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis plus midline horizontal ellipsis plus left square bracket x plus left parenthesis n minus 1 right parenthesis right square bracket left parenthesis x plus n right parenthesis equals 10 n$
After simplifying, we get
$n x to the power of 2 end exponent plus left curly bracket 1 plus 3 plus 5 plus 7 plus midline horizontal ellipsis plus left parenthesis 2 n minus 1 right parenthesis x plus left square bracket left parenthesis 0 times 1 right parenthesis plus left parenthesis 1 times 2 right parenthesis$
$plus left parenthesis 2 times 3 right parenthesis plus midline horizontal ellipsis plus left parenthesis n minus 1 right parenthesis n right curly bracket equals 10 n$
$table row cell n x to the power of 2 end exponent plus n to the power of 2 end exponent x plus open square brackets fraction numerator n open parentheses n to the power of 2 end exponent minus 1 close parentheses over denominator 3 end fraction close square brackets minus 10 n equals 0 end cell row cell x to the power of 2 end exponent plus n x plus open parentheses fraction numerator n to the power of 2 end exponent minus 1 minus 30 over denominator 3 end fraction close parentheses equals 0 end cell row cell x to the power of 2 end exponent plus n x plus open parentheses fraction numerator n to the power of 2 end exponent minus 31 over denominator 3 end fraction close parentheses equals 0 end cell end table$
Using n = 11 (where n∈I), we get
$x to the power of 2 end exponent plus 11 x plus open parentheses fraction numerator 121 minus 31 over denominator 3 end fraction close parentheses equals 0$
$table row cell x to the power of 2 end exponent plus 11 x plus 30 equals 0 end cell row cell left parenthesis x plus 6 right parenthesis left parenthesis x plus 5 right parenthesis equals 0 end cell end table$
Therefore, x = -5, -6 (i.e., two consecutive integral solutions). Thus, n = 11.

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