Question

Three consecutive integers are such that when they are taken in increasing order multiplied by 2 3 and 4 respectively they add up to 74 find these numbers. pls answer quick​

Answer 1

Given :

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

To find :

• Those numbers =?

Step-by-step explanation :

Let, the first consecutive integer be, x.

Then, the second consecutive integer be, x + 1

And, the third consecutive integer be, x + 2

It is Given that,

They, are taken in increasing order and multiplied by 2,3 and 4 respectively.

So,

x × 2 = 2x

3(x+1) = 3(x+1)

4(x+2) = 4(x+2)

According to the question,

2x + 3(x + 1) + 4(x + 2) = 74

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

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

9x + 11 = 74

9x = 74 - 11

9x = 63

x = 63/9

x = 7.

Therefore, We got the value of, x = 7.

Hence,

The first consecutive integer, x = 7

Then, the second consecutive integer, x + 1 = 8

And, the third consecutive integer, x + 2 = 9

Answer 2

${ \huge{ \bold{ \underline{ \underline{ \blue{Question:-}}}}}}$

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

_______________

${ \huge{ \bold{ \underline{ \underline{ \red{Answer:-}}}}}}$

✒Given : -

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

✒To Find : -

• The numbers ..

✒Let ,

• First Consecutive Integer be = y
• Second Consecutive Integer be = y + 1
• Third Consecutive Integer be = y + 2

$\dashrightarrow\sf{y\times{2}=2y}$

$\dashrightarrow\sf{3(y+1)=3(y+1)}$

$\dashrightarrow\sf{4(y+2)=4(y+2)}$

✒Now ,

$\leadsto\sf{{ \large{ \bold{ \underline{ \underline{ \purple{According\:to\:the\:Question:-}}}}}}}$

$\dashrightarrow\sf{2y+3(y+1)+4(y+2)=74}$

$\dashrightarrow\sf{2y+3y+3+4y+8=74}$

$\dashrightarrow\sf{2y+3y+4y+3+8=74}$

$\dashrightarrow\sf{9y+11=74}$

$\dashrightarrow\sf{9y=74-11}$

$\dashrightarrow\sf{9y=63}$

$\dashrightarrow\sf{y=\cancel\dfrac{63}{9}}$

$\leadsto\sf{{ \large{ \boxed{ \bold{ \bold{ \green{y=7}}}}}}}$

✒Hence , Value of y is 7 ..

✒Therefore ,

$\dashrightarrow\sf{First\:consecutive\:Integer=(y)}$

$\dashrightarrow\sf\bold{7}$

$\dashrightarrow\sf{Second\:consecutive\:Integer=(y+1)}$

$\dashrightarrow\sf{7+1}$

$\leadsto\sf\bold{8}$

$\dashrightarrow\sf{Third\:consecutive\:Integer=(y+2)}$

$\dashrightarrow\sf{7+2}$

$\leadsto\sf\bold{9}$