Question

11. The sum of four consecutive multiples of 7 is 70. Find these multiples​

Answer 1

Answer:

Let 4 consecutive numbers be- n, n+1, n+2, n+3

As they are multiples of 7,

Consecutive multiples are - 7n, 7(n+1) , 7(n+2) , 7 (n+3)

Sum of these 4 consecutive multiples = 70

7n+7(n+1)+7(n+2)+7(n+3)= 70

7n+7n+7+7n+14+7n+21=70

28n+42=70

28n= 70-42

28n=28

n= 1

Now,

One number = 7×1=7

Second number = 7(1+1) , 7×2=14

Third number= 7(1+2) , 7×3=21

Fourth number = 7(1+3), 7×4=28

CHECK -:

7+14+21+28=70 , as told in the Question.

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Hope it helps...!!!

Step-by-step explanation:

Answer 2

Answer:

Hey! 7+14+21+28=70 , as told in the Question

Step-by-step explanation:

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