Question

the sum of three consecutive multiples of 9 is 999. find the numbers . plz fast its urgentplz its my paper
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Answer 1

Answer:

Given :-

• The sum of three consecutive multiples of 9 is 999.

To Find :-

• What are the numbers.

Solution :-

Let,

➢ First consecutive multiples = 9a

➢ Second consecutive multiples = 9(a + 1)

➢ Third consecutive multiples = 9(a + 2)

According to the question,

⇒ 9a + 9(a + 1) + 9(a + 2) = 999

⇒ 9a + 9a + 9 + 9a + 18 = 999

⇒ 9a + 9a + 9a + 9 + 18 = 999

⇒ 27a + 27 = 999

⇒ 27a = 999 - 27

⇒ 27a = 972

⇒ a = 972/27

➲ a = 36

Hence, the required numbers are :

❐ First Consecutive Multiples :

➟ 9a

➟ 9 × 36

➠ 324

❐ Second Consecutive Multiples :

➟ 9(a + 1)

➟ 9(36 + 1)

➠ 333

❐ Third Consecutive Multiples :

➟ 9(a + 2)

➟ 9(36 + 2)

➠ 342

∴ The numbers are 324, 333 and 342.

Answer 2

Answer:

• The consecutive multiples of 9 are 324, 333 and 342.

Step-by-step explanation:

Given,

• The sum of three consecutive multiples of 9 is 999.

To Find,

• The numbers.

Solution,

Let's,

• The first multiple of 9 = 9x

Then,

• The second multiple of 9 = 9x + 9

Then,

• The third multiple of 9 = 9x + 18

The sum of three consecutive multiples of 9 = 999

• → (9x) + (9x + 9) + (9x + 18) = 999
• → 9x + 9x + 9 + 9x + 18 = 999
• → 27x + 27 = 999
• → 27(x + 1) = 999
• → x + 1 = 37
• → x = 36

The value of x is 36.

The first multiple of 9 = 9x

• The first multiple of 9 = 324

The second multiple of 9 = 9x + 9

• The second multiple of 9 = 333

The third multiple of 9 = 9x + 18

• The third multiple of 9 = 342

Required Answer,

• The consecutive multiples of 9 are 324, 333 and 342.