Question

$\huge\frak\purple{Question}$
The sum of four consecutive multiples of 7 is 70 find the multiples?
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Answer 1

Answer:

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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

Answer 2

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...!!!

Heya bro..

Let’s first understand what the s consecutivety.

Consecutive no. Means—->> no. Without any integral gaps for example.

Consecutive multiples of 3 are—-> 3..6..9...12..etc

Therefore similarly it’s ask about the sum of consecutive 4 multiples of 7.

So here we go now...

Let the first no be x.—1

S cons will be automatically 7+x—-2

3rd willbe x+14—3

And 4th is x+21—4

There fire re let’s make it all equations.

So now..let’s add all of them according to the question.

It will be

X+x+7+x+14 x+21=70—-given

So now..

Adding all—-

4x+42=70

So x =7.

Thereofore., other all no. Are

7,14,21,28..these are all the terms..

Hope it helps uh.

Else ask any doubt

Thanks..