The nth term of a geometric progression is given by the formula 18(4/3)ⁿ. Determine the sum of the first four terms of the progression.

Neno la $n^{th}$ limetolewa kama $T_n = 18\left(\frac{4}{3}\right)^n$.
Kwa $n=1$, neno la kwanza ni:

$$T_1 = 18\left(\frac{4}{3}\right)^1 = 18 \times \frac{4}{3} = 6 \times 4 = 24$$

Kwa $n=2$, neno la pili ni:

$$T_2 = 18\left(\frac{4}{3}\right)^2 = 18 \times \frac{16}{9} = 2 \times 16 = 32$$

Uwiano wa kawaida $r$ ni $\frac{T_2}{T_1}$:

$$r = \frac{32}{24} = \frac{4}{3}$$

Hivyo, $a = 24$ na $r = \frac{4}{3}$.

Fomula ni $S_n = \frac{a(r^n - 1)}{r - 1}$ kwa $r > 1$.
Tunataka jumla ya maneno manne ya kwanza, hivyo $n=4$.

$$S_4 = \frac{24\left(\left(\frac{4}{3}\right)^4 - 1\right)}{\frac{4}{3} - 1}$$ $$S_4 = \frac{24\left(\frac{256}{81} - 1\right)}{\frac{1}{3}}$$ $$S_4 = \frac{24\left(\frac{256 - 81}{81}\right)}{\frac{1}{3}}$$ $$S_4 = \frac{24\left(\frac{175}{81}\right)}{\frac{1}{3}}$$ $$S_4 = 24 \times \frac{175}{81} \times 3$$ $$S_4 = 24 \times \frac{175}{27}$$ $$S_4 = \frac{8 \times 175}{9}$$ $$S_4 = \frac{1400}{9}$$ $$\boxed{\frac{1400}{9}}$$

2. Without using mathematical tables or calculator evaluate $\frac{\log_2 256 - \log_5 625}{\log_3 27 + \log_2 128}$

$$\log_2 256 = \log_2 2^8 = 8$$ $$\log_5 625 = \log_5 5^4 = 4$$

$$\log_3 27 = \log_3 3^3 = 3$$ $$\log_2 128 = \log_2 2^7 = 7$$

$$\frac{\log_2 256 - \log_5 625}{\log_3 27 + \log_2 128} = \frac{8 - 4}{3 + 7}$$ $$= \frac{4}{10}$$ $$= \frac{2}{5}$$ $$\boxed{\frac{2}{5}}$$

3. Express $\cos 510^\circ$ in surd form hence simplify $\frac{\sqrt{5}}{1-\cos 510^\circ}$ leaving your answer in the form $a\sqrt{b}+c\sqrt{d}$ where $a, b, c$ and $d$ are integers.

Tunajua kuwa $\cos(\theta + 360^\circ) = \cos\theta$.

$$\cos 510^\circ = \cos (510^\circ - 360^\circ) = \cos 150^\circ$$

Tunajua pia kuwa $\cos(180^\circ - \theta) = -\cos\theta$.

$$\cos 150^\circ = \cos (180^\circ - 30^\circ) = -\cos 30^\circ$$

Thamani ya $\cos 30^\circ$ ni $\frac{\sqrt{3}}{2}$.
Hivyo,

$$\cos 510^\circ = -\frac{\sqrt{3}}{2}$$

$$\frac{\sqrt{5}}{1 - \left(-\frac{\sqrt{3}}{2}\right)} = \frac{\sqrt{5}}{1 + \frac{\sqrt{3}}{2}}$$

Ili kurahisisha denomineta, pata kiashiria cha kawaida:

$$\frac{\sqrt{5}}{\frac{2}{2} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{5}}{\frac{2+\sqrt{3}}{2}}$$

Geuza na zidisha:

$$= \frac{2\sqrt{5}}{2+\sqrt{3}}$$

Zidisha nambari na denomineta kwa kiambishi cha denomineta, ambacho ni $2-\sqrt{3}$.

$$\frac{2\sqrt{5}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2\sqrt{5}(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$$

Panua nambari:

$$2\sqrt{5}(2-\sqrt{3}) = 4\sqrt{5} - 2\sqrt{5}\sqrt{3} = 4\sqrt{5} - 2\sqrt{15}$$

Panua denomineta (tumia tofauti ya miraba, $(x+y)(x-y) = x^2 - y^2$):

$$(2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$$

Hivyo, usemi uliorahisishwa ni:

$$\frac{4\sqrt{5} - 2\sqrt{15}}{1} = 4\sqrt{5} - 2\sqrt{15}$$

Hii iko katika fomu $a\sqrt{b}+c\sqrt{d}$ ambapo $a=4$, $b=5$, $c=-2$, na $d=15$.

$$\boxed{4\sqrt{5} - 2\sqrt{15}}$$