Evaluate 15.47 x 0.085² / (254)¹/3 using log table.

1. Evaluate $\frac{15.47 \times 0.085^2}{(254)^{1/3}}$ using log table.

Let $P = \frac{15.47 \times 0.085^2}{(254)^{1/3}}$.
Taking the logarithm of both sides:

$$\log P = \log(15.47) + \log(0.085^2) - \log((254)^{1/3})$$ $$\log P = \log(15.47) + 2 \log(0.085) - \frac{1}{3} \log(254)$$

Step 1: Find the logarithms of the individual numbers using a log table.

• $\log(15.47) \approx 1.1895$
• $\log(0.085) \approx \bar{2}.9294$ (which is $-2 + 0.9294 = -1.0706$)
• $\log(254) \approx 2.4048$

Step 2: Substitute these values into the equation for $\log P$.

$$\log P = 1.1895 + 2(\bar{2}.9294) - \frac{1}{3}(2.4048)$$ $$\log P = 1.1895 + (\bar{4}.8588) - 0.8016$$ $$\log P = 1.1895 + (-4 + 0.8588) - 0.8016$$ $$\log P = 1.1895 - 3.1412 - 0.8016$$ $$\log P = 1.1895 - 3.9428$$ $$\log P = -2.7533$$

Step 3: Convert the negative logarithm to characteristic-mantissa form and find the antilog.

$$\log P = -2.7533 = -3 + 0.2467 = \bar{3}.2467$$

Using an antilog table, the number corresponding to the mantissa 0.2467 is approximately 1.765.
The characteristic $\bar{3}$ indicates that the decimal point is after three zeros.

$$P \approx 1.765 \times 10^{-3}$$

The final answer is $\boxed{0.001765}$.

2. An arithmetic progression has 11 as its first term and the fourth term is 32. Find the sum of the first nine terms.

Step 1: Use the formula for the $n$-th term of an arithmetic progression, $a_n = a_1 + (n-1)d$, to find the common difference $d$.
Given $a_1 = 11$ and $a_4 = 32$.

$$a_4 = a_1 + (4-1)d$$ $$32 = 11 + 3d$$ $$3d = 32 - 11$$ $$3d = 21$$ $$d = 7$$

Step 2: Use the formula for the sum of the first $n$ terms of an arithmetic progression, $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, to find $S_9$.

$$S_9 = \frac{9}{2}(2(11) + (9-1)7)$$ $$S_9 = \frac{9}{2}(22 + 8 \times 7)$$ $$S_9 = \frac{9}{2}(22 + 56)$$ $$S_9 = \frac{9}{2}(78)$$ $$S_9 = 9 \times 39$$ $$S_9 = 351$$

The final answer is $\boxed{351}$.

3. The third term of a GP is 24 and its seventh term is $\frac{20}{27}$. Find: a) Its first term b) The sum of its first five terms.

Assumption: Based on common notation in such problems, the seventh term is interpreted as a mixed fraction $4 \frac{20}{27}$, which is $\frac{4 \times 27 + 20}{27} = \frac{108 + 20}{27} = \frac{128}{27}$. If it was meant as a simple fraction $\frac{20}{27}$, the common ratio would be irrational.

Step 1: Use the formula for the $n$-th term of a geometric progression, $a_n = a_1 r^{n-1}$, to set up equations for $a_1$ and $r$.
Given $a_3 = 24$ and $a_7 = \frac{128}{27}$.

$$a_3 = a_1 r^{3-1} \Rightarrow a_1 r^2 = 24 \quad (1)$$ $$a_7 = a_1 r^{7-1} \Rightarrow a_1 r^6 = \frac{128}{27} \quad (2)$$

Step 2: Divide equation (2) by equation (1) to find the common ratio $r$.

$$\frac{a_1 r^6}{a_1 r^2} = \frac{128/27}{24}$$ $$r^4 = \frac{128}{27 \times 24}$$ $$r^4 = \frac{128}{648}$$

Divide both numerator and denominator by 8:

$$r^4 = \frac{16}{81}$$ $$r = \left(\frac{16}{81}\right)^{1/4}$$ $$r = \frac{2}{3}$$