Question

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

Answer 1

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➺The sum of four consecutive multiples of 7 is 322. Find the smallest multiple involved.

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➺let's suppose four multiples are

=( n), (n+1),(n+2),(n+3).

➺than they are the multiple of 7 than =(7n), 7(n+1), 7(n+2), 7(n+3).

➺than, according to question,

(7n) +(7n+7)+(7n+14)+(7n+21)=322

28n+42=322

28n= 322-42

n= 280/28

hence, n= 10

n= 7×10 = 70

n+1= 7(10+1)= 77

n+2= 7(10+2) = 84

n+3= 7(10+3) = 91

➺CHECK OUT:-

CHECK OUT:-70+77+84+91 = 322

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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