I am looking for 3 consecutive multiples of 7. The sum of the squares of the 2 smallest is 147 less than the square of the largest. The first multiple of 7 is (x−7), the 2nd multiple of 7 is (x), and the 3rd multiple of 7(x+7). Find the numbers.

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Step 1   I am looking for 3 consecutive multiples of 7. The sum of the squares of the 2 smallest is 147 less than the square of the largest. The first multiple of 7 is (x−7), the 2nd multiple of 7 is (x), and the 3rd multiple of 7(x+7). Find the numbers.   According to question,  (x−7)2+x2=(x+7)2−147  ⇒x2+49−14x+x2=x2+49+14x−147  ⇒x2−14x=14x−147  ⇒x2−28x+147=0  ⇒x2−(21+7)x+147=0  ⇒x2−21x−7x+147=0  ⇒x(x−21)−7(x−21)=0  ⇒(x−21)−(x−7)=0  x−21=0,x−7=0  ⇒x=21,x=7  Step 2  when x=21  the consecutive multiples of 7 is  (x−7)=21−7=14  x=21  x+7=21+7=28  when x=7  the consecutive multiples of 7 is  (x−7)=7−7=0  x=7  x+7=7+7=14

I am looking for 3 consecutive multiples of 7. The first multiple of 7 is (x-7), the 2nd multiple of 7 is (x), and the 3rd multiple of 7 (x+7). Find the numbers.

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