An arithmetic progression is given as 600 + 650 + 700 + 750 + u. Find the sum of the first 30 terms. 3. Without using mathematical tables or a calculator, evaluate rac 125 + 64 sqrt(5) + sqrt(2).

Here, $n = 30$, $a = 600$, and $d = 50$.

The sum of the first 30 terms is:

3. Without using mathematical tables or a calculator, evaluate $\frac{\log 125 + \log 64}{\log \sqrt{5} + \log \sqrt{2}}$.

So, $\log 125 + \log 64 = \log (5^3 \times 2^6)$.

So, $\log \sqrt{5} + \log \sqrt{2} = \log (5^{1/2} \times 2^{1/2}) = \log ((5 \times 2)^{1/2}) = \log (10^{1/2})$.

For the numerator: $\log (5^3 \times 2^6) = \log 5^3 + \log 2^6 = 3 \log 5 + 6 \log 2$.
For the denominator: $\log (10^{1/2}) = \frac{1}{2} \log 10$.
Assuming base 10 logarithms (common when no base is specified), $\log 10 = 1$.
So the denominator becomes $\frac{1}{2} \times 1 = \frac{1}{2}$.

This can be rewritten as:

This approach seems to lead to a more complex expression rather than a single numerical value. Let's re-evaluate Step 1 and Step 2 using a different combination.

Alternative Step 1: Simplify the numerator.

Alternative Step 2: Simplify the denominator.

So the expression becomes:

Since $\log 10 = 1$:

This still doesn't yield a simple numerical value without a calculator for $\log 2$.

Let's try to make the bases match more directly.
Numerator: $\log 125 + \log 64 = \log (5^3) + \log (2^6) = 3 \log 5 + 6 \log 2$.
Denominator: $\log \sqrt{5} + \log \sqrt{2} = \log (5^{1/2}) + \log (2^{1/2}) = \frac{1}{2} \log 5 + \frac{1}{2} \log 2$.
So the expression is:

Factor out common terms:

This is not correct. The numerator is $3 \log 5 + 6 \log 2$.
Let's factor out 3 from the numerator: $3(\log 5 + 2 \log 2)$.
Let's factor out $\frac{1}{2}$ from the denominator: $\frac{1}{2}(\log 5 + \log 2)$.
The expression is $\frac{3(\log 5 + 2 \log 2)}{\frac{1}{2}(\log 5 + \log 2)}$.
This still doesn't simplify nicely.

Let's re-examine the product in the numerator and denominator.
Numerator: $\log 125 + \log 64 = \log (125 \times 64) = \log (8000)$.
Denominator: $\log \sqrt{5} + \log \sqrt{2} = \log (\sqrt{5} \times \sqrt{2}) = \log (\sqrt{10})$.
So the expression is $\frac{\log 8000}{\log \sqrt{10}}$.
We can write $8000 = 8 \times 1000 = 2^3 \times 10^3$.
And $\sqrt{10} = 10^{1/2}$.
So, $\frac{\log (2^3 \times 10^3)}{\log (10^{1/2})} = \frac{\log (2^3) + \log (10^3)}{\log (10^{1/2})} = \frac{3 \log 2 + 3 \log 10}{\frac{1}{2} \log 10}$.
If the base of the logarithm is 10, then $\log 10 = 1$.
So, $\frac{3 \log 2 + 3}{\frac{1}{2}} = 6 \log 2 + 6$. This is still not a single number.

There must be a way to make the terms cancel out.
Let's try to express everything in terms of $\log 2$ and $\log 5$.
Numerator: $\log 125 + \log 64 = \log (5^3) + \log (2^6) = 3 \log 5 + 6 \log 2$.
Denominator: $\log \sqrt{5} + \log \sqrt{2} = \log (5^{1/2}) + \log (2^{1/2}) = \frac{1}{2} \log 5 + \frac{1}{2} \log 2$.
The expression is:

Factor out 3 from the numerator: $3(\log 5 + 2 \log 2)$.
Factor out $\frac{1}{2}$ from the denominator: $\frac{1}{2}(\log 5 + \log 2)$.
This is $\frac{3(\log 5 + \log 2^2)}{\frac{1}{2}(\log 5 + \log 2)} = \frac{3(\log 5 + \log 4)}{\frac{1}{2}(\log 5 + \log 2)}$.
This is $\frac{3 \log (5 \times 4)}{\frac{1}{2} \log (5 \times 2)} = \frac{3 \log 20}{\frac{1}{2} \log 10}$.
If the base is 10, then $\log 10 = 1$.
So, $\frac{3 \log 20}{\frac{1}{2}} = 6 \log 20$. This is still not a single number.

Let's re-check the problem statement. "Without using mathematical tables or a calculator, evaluate". This implies the answer should be a simple integer or fraction.

Let's reconsider the numerator and denominator products:
Numerator: $\log 125 + \log 64 = \log (125 \times 64) = \log (8000)$.
Denominator: $\log \sqrt{5} + \log \sqrt{2} = \log (\sqrt{5} \times \sqrt{2}) = \log (\sqrt{10})$.

Now, let's try to relate 8000 and $\sqrt{10}$.
$8000 = 8 \times 1000 = 2^3 \times 10^3$.
$\sqrt{10} = 10^{1/2}$.

The expression is $\frac{\log (2^3 \times 10^3)}{\log (10^{1/2})}$.
This is $\frac{\log (2^3) + \log (10^3)}{\log (10^{1/2})} = \frac{3 \log 2 + 3 \log 10}{\frac{1}{2} \log 10}$.
This is the same result as before.

Is there a common factor that I'm missing?
Let's try to express 8000 in terms of powers of 10.
$8000 = 10^{x}$. This is not an integer power.
$8000 = (10^{1/2})^y$?
$8000 = \sqrt{10}^y$?
$8000^2 = 10^y$?
$64,000,000 = 10^y$? This is $6.4 \times 10^7$. Not a simple power of 10.

Let's try to simplify the numerator and denominator in a different way.
Numerator: $\log 125 + \log 64 = \log (5^3) + \log (2^6)$.
Denominator: $\log \sqrt{5} + \log \sqrt{2} = \log (5^{1/2}) + \log (2^{1/2})$.
Let $L_5 = \log 5$ and $L_2 = \log 2$.
The expression is $\frac{3 L_5 + 6 L_2}{\frac{1}{2} L_5 + \frac{1}{2} L_2}$.
Factor out 3 from the numerator: $3(L_5 + 2 L_2)$.
Factor out $\frac{1}{2}$ from the denominator: $\frac{1}{2}(L_5 + L_2)$.
So we have $\frac{3(L_5 + 2 L_2)}{\frac{1}{2}(L_5 + L_2)}$.
This is $\frac{3(L_5 + L_2^2)}{\frac{1}{2}(L_5 + L_2)} = \frac{3(\log 5 + \log 4)}{\frac{1}{2}(\log 5 + \log 2)} = \frac{3 \log (5 \times 4)}{\frac{1}{2} \log (5 \times 2)} = \frac{3 \log 20}{\frac{1}{2} \log 10}$.
This is $6 \log 20 / \log 10$.
If the base is 10, then $\log 10 = 1$. So it's $6 \log 20$. This is still not a simple number.

Let's consider the possibility that the base of the logarithm is not 10. However, usually, if no base is specified, it's assumed to be 10 or $e$. For these types of problems, it's usually 10.

Is there a mistake in my simplification or understanding of the problem?
$\log 125 + \log 64 = \log (125 \times 64) = \log (8000)$.
$\log \sqrt{5} + \log \sqrt{2} = \log (\sqrt{5} \times \sqrt{2}) = \log (\sqrt{10})$.
So the expression is $\frac{\log 8000}{\log \sqrt{10}}$.
Using the change of base formula, $\frac{\log_b A}{\log_b C} = \log_C A$.
So, $\frac{\log 8000}{\log \sqrt{10}} = \log_{\sqrt{10}} 8000$.
Let $x = \log_{\sqrt{10}} 8000$.
Then $(\sqrt{10})^x = 8000$.
$(10^{1/2})^x = 8000$.
$10^{x/2} = 8000$.
We know $10^3 = 1000$. $10^4 = 10000$. So $x/2$ is between 3 and 4.
This means $x$ is between 6 and 8.
This still doesn't give an integer value.

Let's re-examine the numbers $125, 64, \sqrt{5}, \sqrt{2}$.
$125 = 5^3$.
$64 = 2^6$.
$\sqrt{5} = 5^{1/2}$.
$\sqrt{2} = 2^{1/2}$.

Numerator: $\log(5^3) + \log(2^6) = 3 \log 5 + 6 \log 2$.
Denominator: $\log(5^{1/2}) + \log(2^{1/2}) = \frac{1}{2} \log 5 + \frac{1}{2} \log 2$.

The expression is $\frac{3 \log 5 + 6 \log 2}{\frac{1}{2} \log 5 + \frac{1}{2} \log 2}$.
Let's try to factor out a common term from the numerator that matches the denominator.
The denominator is $\frac{1}{2}(\log 5 + \log 2)$.
The numerator is $3 \log 5 + 6 \log 2$.
Notice that $6 \log 2 = 3 \times (2 \log 2)$.
So, $3 \log 5 + 6 \log 2 = 3 \log 5 + 3 (2 \log 2) = 3 (\log 5 + 2 \log 2