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What is the smallest sum of six consecutive natural numbers, given that the last one is twice the first number?

Sagot :

To solve the problem of finding the smallest sum of six consecutive natural numbers, given that the last one is twice the first number, follow these steps:

1. Denote the first number as [tex]\( n \)[/tex]:
- Let's assume the first number in the sequence of six consecutive natural numbers is [tex]\( n \)[/tex].

2. Express the six consecutive numbers:
- The six consecutive numbers starting from [tex]\( n \)[/tex] are [tex]\( n \)[/tex], [tex]\( n+1 \)[/tex], [tex]\( n+2 \)[/tex], [tex]\( n+3 \)[/tex], [tex]\( n+4 \)[/tex], and [tex]\( n+5 \)[/tex].

3. Set up the given condition:
- According to the problem, the last number [tex]\( n+5 \)[/tex] is twice the first number [tex]\( n \)[/tex]. This gives us the equation:
[tex]\[ n + 5 = 2n \][/tex]

4. Solve for [tex]\( n \)[/tex]:
- Rearrange the equation to isolate [tex]\( n \)[/tex]:
[tex]\[ n + 5 = 2n \][/tex]
[tex]\[ 5 = 2n - n \][/tex]
[tex]\[ n = 5 \][/tex]

5. Identify the six consecutive numbers:
- Since [tex]\( n = 5 \)[/tex], the six consecutive numbers are:
[tex]\[ 5, 6, 7, 8, 9, 10 \][/tex]

6. Calculate the sum of these numbers:
- Sum the numbers [tex]\( 5, 6, 7, 8, 9, 10 \)[/tex]:
[tex]\[ 5 + 6 + 7 + 8 + 9 + 10 = 45 \][/tex]

Therefore, the smallest sum of the six consecutive natural numbers, given that the last one is twice the first number, is [tex]\( 45 \)[/tex].