Question

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

Answer 1

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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Answer 2

Answer:

Smallest multiple involved is 70.

Step-by-step explanation:

Given :

• Sum of four consecutive multiples of 7 is 322.

To find :

Smallest multiple involved.

Solution :-

Let the four multiples be :

• (n), (n+1), (n+2), (n+3).

Then, the multiples of 7 will be :

• (7n), 7(n+1), 7(n+2), 7(n+3).

Their sum is 322.

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

⇒ 28n + 42 = 322

⇒ 28n = 280

⇒ n = 280/28

⇒ n = 10

Value of n is 10.

Multiples are :

• (7n) = 70

• (7n + 7) = 77

• (7n + 14) = 84

• (7n + 21) = 91

Clearly, the smallest multiple involved is 70.

Verification :

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

⟹ 70+77+84+91 = 322

⟹ 322 = 322

⟹ L.H.S = R.H.S