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].
