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Maths-

General

Easy

Question

Consider the ten numbers $a r comma blank a r to the power of 2 end exponent comma a r to the power of 3 end exponent comma horizontal ellipsis comma a r to the power of 10 end exponent$ . If their sum is 18 and the sum of their reciprocals is 6 then the product of these ten numbers, is

81
243
343
324

The correct answer is: 243

Given $fraction numerator a r left parenthesis r to the power of 10 end exponent minus 1 right parenthesis over denominator r minus 1 end fraction equals 18$ (1)
Also $fraction numerator fraction numerator 1 over denominator a r end fraction open parentheses 1 minus fraction numerator 1 over denominator r to the power of 10 end exponent end fraction close parentheses over denominator 1 minus fraction numerator 1 over denominator r end fraction end fraction equals 6$
$rightwards double arrow fraction numerator 1 over denominator a r to the power of 11 end exponent end fraction bullet fraction numerator left parenthesis r to the power of 10 end exponent minus 1 right parenthesis r over denominator r minus 1 end fraction equals 6$
$rightwards double arrow fraction numerator 1 over denominator a to the power of 2 end exponent blank r to the power of 11 end exponent end fraction bullet fraction numerator a r left parenthesis r to the power of 10 end exponent minus 1 right parenthesis over denominator r minus 1 end fraction equals 6$ (2)
From (1) and (2),
$fraction numerator 1 over denominator a to the power of 2 end exponent blank r to the power of 11 end exponent end fraction bullet 18 equals 6$
$rightwards double arrow a to the power of 2 end exponent r to the power of 11 end exponent equals 3$
Now $P equals a to the power of 10 end exponent r to the power of 55 end exponent equals open parentheses a to the power of 2 end exponent r to the power of 11 end exponent close parentheses to the power of 5 end exponent equals 3 to the power of 5 end exponent equals 243$

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