To find the limit of the sequence [tex]\( a_n \)[/tex] defined as [tex]\( a_n = \frac{15 + 25 + 35 + \ldots + (10n - 5)}{n^2} \)[/tex], we need to examine the behavior of this sequence as [tex]\( n \)[/tex] approaches infinity. Let's break down the expression step-by-step.
First, let's rewrite the numerator in a more generalized form:
[tex]\[ S_n = 15 + 25 + 35 + \ldots + (10n - 5) \][/tex]
This sequence can be described in terms of a sum involving [tex]\( k \)[/tex]:
[tex]\[ S_n = \sum_{k=1}^n (10k - 5) \][/tex]
We can split this sum into two separate sums:
[tex]\[ S_n = \sum_{k=1}^n 10k - \sum_{k=1}^n 5 \][/tex]
The first sum [tex]\( \sum_{k=1}^n 10k \)[/tex] is a simple arithmetic series:
[tex]\[ \sum_{k=1}^n 10k = 10 \sum_{k=1}^n k = 10 \left( \frac{n(n+1)}{2} \right) = 5n(n+1) \][/tex]
The second sum [tex]\( \sum_{k=1}^n 5 \)[/tex] is simply [tex]\( 5 \)[/tex] added [tex]\( n \)[/tex] times:
[tex]\[ \sum_{k=1}^n 5 = 5n \][/tex]
Therefore, the sum [tex]\( S_n \)[/tex] is:
[tex]\[ S_n = 5n(n+1) - 5n = 5n^2 + 5n - 5n = 5n^2 \][/tex]
Now we substitute [tex]\( S_n \)[/tex] back into the sequence definition:
[tex]\[ a_n = \frac{S_n}{n^2} = \frac{5n^2}{n^2} = 5 \][/tex]
As you can see, the expression simplifies to a constant value. Therefore, the limit of the sequence as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} 5 = 5 \][/tex]
So, the limit of the sequence [tex]\( a_n \)[/tex] is:
[tex]\[ \boxed{5} \][/tex]
