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# The Sum of the Squares of Two Consecutive Multiples of 7 is 637. Find the Multiples. - Mathematics

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#### Question

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.

#### Solution

Let one of the number be 7x then the other number be 7(x + 1).

Then according to question,

$\left( 7x \right)^2 + \left[ 7\left( x + 1 \right) \right]^2 = 637$

$\Rightarrow 49 x^2 + 49( x^2 + 2x + 1) = 637$

$\Rightarrow 49 x^2 + 49 x^2 + 98x + 49 - 637 = 0$

$\Rightarrow 98 x^2 + 98x - 588 = 0$

$\Rightarrow x^2 + x - 6 = 0$

$\Rightarrow x^2 + 3x - 2x - 6 = 0$

$\Rightarrow x(x + 3) - 2(x + 3) = 0$

$\Rightarrow (x - 2)(x + 3) = 0$

$\Rightarrow x - 2 = 0 \text { or } x + 3 = 0$

$\Rightarrow x = 2 \text { or } x = - 3$

Since, the numbers are multiples of 7,

Therefore, one number = 7 × 2 =14.

Then another number will be $7(x + 1) = 7 \times 3 = 21$

Thus, the two consecutive multiples of 7 are 14 and 21.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Solution for Class 10 Maths (2018 (Latest))