MaxWong

common ratio = (second term)/(first term) = 16/8 = 2.

That means the 53rd term is \(8 \times 2^{53 - 1} = 8\times 2^{52} = 2^{55}\).

We take log_4 of it and get \(\log_4 (x) = \log_4 (2^{55}) = 55 \log_4 (2) = \dfrac{55}2\).

Common ratio = (second term)/(first term) = \(\dfrac{3\sqrt2}3 = \sqrt2\)

Therefore, if the nth term of the sequence is \(768\sqrt 2\), then \(3\left(\sqrt 2\right)^{n - 1} = 768\sqrt 2\)

Then \(\left(\sqrt 2\right)^n = 512\).

You can calculate the value of n, and 768 sqrt(2) will be the nth term, i.e., there are n terms in the sequence, which I will leave it for you to calculate.

Let O = (0, 0, 0), A = (3, 0, 0), B = (0, 5, 1), and C = (2, 0, 5).

Volume of tetrahedron = \(\renewcommand{\vec}{\overrightarrow} \dfrac16\left|\vec{OA} \cdot \left(\vec{OB} \times \vec{OC}\right)\right|\) = \(\dfrac 16 \begin{vmatrix}3&0&0\\0&5&1\\2&0&5\end{vmatrix}\) = 25/2 cubic units.

Supppose the first fraction is \(\dfrac{n + 2}n\). Then the other two fractions are \(\dfrac{n + 3}{n + 1}\) and \(\dfrac{n + 4}{n + 2}\) respectively.

Then 

\(\dfrac{(n + 2)(n + 3)(n + 4)}{n(n + 1)(n + 2)} = \dfrac1{10}\\ n(n + 1)= 10(n + 3)(n + 4)\)

Now solve the quadratic equation for the positive integer value of n. Substituting back into (n + 2)/n gives the value of the first fraction he wrote.

Edit: There are no positive integer solutions for n. I think there are some problems in the question design.

Edit: I guess n = -5 counts because the resulting fractions are indeed positive, but the numerator and the denominator of the other fractions aren't increased by 1, but instead decreased by 1.

Let d be the common difference.

\(\begin{cases}a_{23} = \dfrac23\\a_{53} = \dfrac53\end{cases} \implies \begin{cases}a_1 + 22d = \dfrac23 \\a_1 + 52d = \dfrac53\end{cases}\)

Solving the linear system for a_1 and d gives a_1 = -1/15, d = 1/30.

Therefore, \(a_{35} = a_1 + 34d = -\dfrac1{15} + \dfrac{34}{30} = \dfrac{16}{15}\)

I assume a_1, a_2, a_3, ... is an arithmetic sequence, because there are no a_5 in the problem.

For an arithmetic sequence a_n, \(a_{n - k} + a_{n + k} = 2a_n\). So \(a_1 + a_5 = 2a_3\) by plugging in n = 3, k = 2.

Therefore, 

\(3a_3 = -12\\ a_3 = -4\)

Now you just solve \(\begin{cases}a_1 + a_5 = -8\\a_1a_5 = -30\end{cases}\) to find the values of a_1 and a_5. That way, you will know the possible common differences of the arithmetic sequence. Then you can use \(a_{10} = a_1 + 9(\text{common difference})\) to calculate the possible values of a_{10}. 

Neither 2u nor 2u doesn't make much sense to me. Please check if that is a typo.

Anyways, if u is a number such that 2u is not in the interval (-1, 1), that means \(2u \leq -1\text{ or }2u \geq 1\). Solving each inequality gives \(u \leq -\dfrac12 \text{ or }u \geq \dfrac12\). Then the answer is \(\left(-\infty, -\dfrac12\right] \cup \left[\dfrac12, \infty\right)\)

Moving the 6cm edge outwards to align with the 3cm edge and moving the edge marked with b outwards to align with the 7cm edge gives a rectangle with dimensions 10cm x 9cm, and the sides are unchanged, so the perimeter is unchanged.

Now you can calculate the perimeter of the original figure by calculating the perimeter of the rectangle, which is much easier compared to the original problem. 

I hope this helps.

For the first one, you can look at the terms pairwise:

\((1 - 2) + (3 - 4) + (5 - 6) + \cdots + (2019 - 2020) + 2021\)

Each pair results in -1 and there are 1010 pairs of numbers (that is just 2020/2), so the answer is \((-1)(1010) + 2021 = 1011\).

Some numbers are missing from your second question. Please check again.

For the third question, look at the terms pairwise again:

\(\quad(1 + 4) +(2 + 8) + (3 + 12) + (4 + 16) + \cdots + (24 + 96) + (25 + 100) \\ =5 + 10 + 15 + 20 + \cdots + 120 + 125\)

Now that is just the sum of an arithmetic sequence. You can find the first term, common difference, and the number of terms, then plug it into the formula to calculate it.

If you want to walk from Mytown to Capital City, you go like this (assuming there are no shortcuts):

Mytown -> Little Village -> Capital City

or this:

Mytown -> Yourtown -> Capital City

The first path is (8 + 51) miles long and the second path is (6 + 37) miles long. Whichever one is shorter is the distance required.