Question

Three consecutive odd numbers are such that three times the largest is one less than double the sum of other two. Find the numbers.

Answer 1

Final Answer:

Three consecutive odd numbers are 9, 11 and 13, where three times the largest is one less than double the sum of other two.

Given:

Three consecutive odd numbers are such that three times the largest is one less than double the sum of other two.

To Find:

The three consecutive odd numbers.

Explanation:

The following points are important.

• The consecutive numbers can be any of those sequential numbers, placed in the number line, such that differ by one from the consecutive numbers.
• The consecutive odd numbers are those sequential numbers, indicated in the number line, such that differ by two from the consecutive odd numbers.

Step 1 of 4

Assume that three consecutive odd numbers are $(2n+1),(2n+3),(2n+5)$, of which the largest one is $(2n+5)$.

The expression for three times the largest odd number is

$=3(2n+5)$

Step 2 of 4

The expression for double the sum of other two odd numbers is

$2\times [(2n+1)+(2n+3)]\\=2(2n+1+2n+3)\\=2(4n+4)\\=2\times4(n+1)\\=8(n+1)$

Step 3 of 4

Write the following equation as per the statement and solve it.

$3(2n+5)=8(n+1)-1\\6n+15=8n+8-1\\8n-6n=15-8+1\\2n=8\\n=4$

Step 4 of 4

So, these three consecutive odd numbers are

$(2n+1)=(2\times4+1)=9\\(2n+3)=(2\times4+3)=11\\(2n+5)=(2\times8+5)=13$

Therefore, the required three consecutive odd numbers are 9, 11 and 13,where three times the largest is one less than double the sum of other two.

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