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Maths-

General

Easy

Question

If $open parentheses 1 plus 3 plus 5 plus horizontal ellipsis plus p close parentheses plus open parentheses 1 plus 3 plus 5 plus horizontal ellipsis plus q close parentheses equals left parenthesis 1 plus 3 plus 5 plus horizontal ellipsis plus r right parenthesis$ where each set of parentheses contains the sum of of $p plus q plus r$ (where $p greater than 6$ ) is

12
21
45
54

The correct answer is: 21

We know that $1 plus 3 plus 5 plus horizontal ellipsis plus open parentheses 2 k minus 1 close parentheses equals k to the power of 2 end exponent$ . Thus, the given equation can be written as
$open parentheses fraction numerator p plus 1 over denominator 2 end fraction close parentheses to the power of 2 end exponent plus open parentheses fraction numerator q plus 1 over denominator 2 end fraction close parentheses to the power of 2 end exponent equals open parentheses fraction numerator r plus 1 over denominator 2 end fraction close parentheses to the power of 2 end exponent$
$rightwards double arrow open parentheses p plus 1 close parentheses to the power of 2 end exponent plus open parentheses q plus 1 close parentheses to the power of 2 end exponent equals open parentheses r plus 1 close parentheses to the power of 2 end exponent$
As $p greater than 6 comma blank p plus 1 greater than 7$ , we may take $p plus 1 equals 8 comma blank q plus 1 equals 6 comma blank r plus 1 equals 10$
Hence, $p plus q plus r equals 21$

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