To approximate \( 11\sqrt{31} \) by using a sequence of values that converges to \( \sqrt{31} \), we need to first find a good approximation for \( \sqrt{31} \).
We know that \( \sqrt{25} = 5 \) and \( \sqrt{36} = 6 \), so \( \sqrt{31} \) is between 5 and 6.
We can refine this approximation by checking the squares of numbers between 5 and 6. Let's try 5.5:
\[ 5.5^2 = 30.25 \]
Next, let's check 5.6:
\[ 5.6^2 = 31.36 \]
Now we see that \( \sqrt{31} \) is between 5.5 and 5.6. Let's average these to refine our guess further:
\[ \text{Midpoint} = \frac{5.5 + 5.6}{2} = 5.55 \]
Now we calculate \( 5.55^2 \):
\[ 5.55^2 = 30.8025 \]
Next, check 5.57:
\[ 5.57^2 = 31.0449 \]
Now we know \( \sqrt{31} \) is between 5.55 and 5.57. Let's try 5.56:
\[ 5.56^2 = 30.9136 \]
Now we know that \( \sqrt{31} \) is between 5.56 and 5.57. We can conclude that:
\[ \sqrt{31} \approx 5.57 \]
Now, multiply this approximation by 11 to find \( 11\sqrt{31} \):
\[ 11\sqrt{31} \approx 11 \times 5.57 = 61.27 \]
Rounding to one decimal place, we get:
\[ \boxed{61.3} \]
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