Math, asked by Sidhi38251, 3 months ago

Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

Answers

Answered by amitsharma777222999
2

Step-by-step explanation:

numbers be n-1, n, n+1

2(n-1)+3n+4(n+1)=74

2n+3n+4n+2=74

9n=72

n=8

numbers are 7,8,9

Answered by Rocksteady
1

Answer:

The three consecutive integers are 7, 8 and 9.

Step-by-step explanation:

Let the first integer be x.

Then, second integer will be x + 1.

And, the third integer will be x + 2.

After taking in increasing order,

x × 2 = 2x (I)

(x + 1) × 3 = 3x + 3 (II)

(x + 2) × 4 = 4x + 8 (III)

Now, adding equations (I), (II) and (III),

2x + 3x + 3 + 4x + 8 = 74

9x + 11 = 74

9x = 74 - 11

9x = 63

x = 63 ÷ 9

•°• x = 7

Finally,

First integer = x = 7

Second integer = x + 1 = 7 + 1 = 8

Third integer = x + 2 = 7 + 2 = 9

Hence, the three consecutive integers are 7, 8 and 9.

Adiós! Hope this helps you.

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