Question

The sum of four consecutive multiples of 7 is 322. Find the smallest multipleinvolved.​

Answer 1

SOLUTION

GIVEN

The sum of four consecutive multiples of 7 is 322.

TO DETERMINE

The smallest multiple involved.

EVALUATION

Let four consecutive multiples of 7 are

7x , 7x + 7 , 7x + 14 , 7x + 21

So by the given condition

7x + 7x + 7 + 7x + 14 + 7x + 21 = 322

$\implies \sf{28x + 42 = 322}$

$\implies \sf{28x = 280}$

$\implies \sf{x = 10}$

So the numbers are 70 , 77 , 84 , 91

FINAL ANSWER

The smallest multiple involved = 70

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Answer 2

Given : The sum of four consecutive multiples of 7 is 322.

To Find :  the smallest multiple involved.​

Solution:

four consecutive multiples of 7

7k , 7(k + 1) , 7(k + 2) , 7(k + 3)

Sum = 322

=> 7k + 7(k + 1) + 7(k + 2) +  7(k + 3) = 322

=> 7(k + k + 1 + k + 2 + k + 3) = 322

=> 7(4k + 6) = 322

=> 4k + 6  = 46

=> 4k = 40

=> k = 10

70 , 77 , 84 , 91

Smallest multiple involved = 70

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