The sum of four consecutive multiples of 7 is 70 find the numbers
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Hey!
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- 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...!!!
Have a great day ahead !
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