Q. 665.0( 1 Vote )

The sum of

Answer :

Let the consecutive multiples of 7 be ‘a’ and a + 7


a2 + (a + 7)2 = 637


2a2 + 14a – 588 = 0


2a2 + 42a – 28a – 588 = 0


2a(a + 21) – 28(a + 21) = 0


(2a – 28)(a + 21) = 0


Thus, a = 14


Consecutive multiples of 7 are 14, 21

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