Question

The addition of three successive multiple

Answer 1

$\huge \mathfrak{answer}$

Given:

Sum of 3 successive multiples of 9 is 81.

To Find

The 3 successive multiples

Answer

Let the multiple of 9 be ⇒ \sf{ 9y, \ 9(y+1), \ 9(y+2)}9y, 9(y+1), 9(y+2)

We know that they will sum up to 81 so our new equation to find the value of 'x' would be ⇒

\sf {9y+9(y+1)+9(y+2)=81}9y+9(y+1)+9(y+2)=81

Step 1: Simplify the equation.

{9y+9y+9+9y+18}=819y+9y+9+9y+18=81

Step 2: Combine Like Terms.

(9y+9y+9y)+(9+18)=81(9y+9y+9y)+(9+18)=81

27y+27=8127y+27=81

Step 3: Subtract 27 from both sides.

27y+27-27=81-2727y+27−27=81−27

27y=5427y=54

Step 4: Divide both sides by 27.

\frac{27y}{27}=\frac{54}{27}

27

27y

=

27

54

y=2y=2

Now we know that y=2y=2 . Now we will give 'x' it's value and answer.

\sf{9y=9\times 2=18 }9y=9×2=18

\sf 9(y+1)=9(2+1) = 9\times3 = 279(y+1)=9(2+1)=9×3=27

\sf 9(y+2)=9(2+2)=9\times 4 =369(y+2)=9(2+2)=9×4=36

∴ The three successive multiples of 9 are 18, 27 and 36 which sum up to 81.