Miss Claire started a job in 2001 on a salary of Kshs. 40,000 per year. Her salary increases by 3% annually. Calculate her salary in 2022 and 2030. Also find her total earnings from 2001 to 2030 inclu

Here's a step-by-step solution:

Part (a): Miss Claire's Salary

• Initial salary (Year 2001): $S_1 = \text{Kshs. } 40,000$

• Annual pay rise: 3%

• Common ratio: $r = 1 + 0.03 = 1.03$

• The salary forms a geometric progression with the $n$-th term given by $S_n = S_1 \cdot r^{n-1}$.

• a) (i) Her salary in the year 2022 and 2030.

  • Salary in 2022: The year 2022 is the $(2022 - 2001 + 1) = 22^{\text{nd}}$ year of employment. We need to find $S_{22}$.

    $$S_{22} = S_1 \cdot r^{22-1} = 40,000 \cdot (1.03)^{21}$$ $$S_{22} = 40,000 \cdot 1.86029906 \approx 74411.9624$$

    Rounding to the nearest shilling: Her salary in 2022 is $\boxed{\text{Kshs. } 74,412}$.

  • Salary in 2030: The year 2030 is the $(2030 - 2001 + 1) = 30^{\text{th}}$ year of employment. We need to find $S_{30}$.

    $$S_{30} = S_1 \cdot r^{30-1} = 40,000 \cdot (1.03)^{29}$$ $$S_{30} = 40,000 \cdot 2.35639906 \approx 94255.9624$$

    Rounding to the nearest shilling: Her salary in 2030 is $\boxed{\text{Kshs. } 94,256}$.

• a) (ii) Her total earnings in employment for the year 2001 to 2030, inclusive. This is the sum of a geometric series for $n=30$ terms. The formula for the sum of the first $n$ terms is $S_n = \frac{S_1(r^n - 1)}{r - 1}$.

  $$S_{30} = \frac{40,000((1.03)^{30} - 1)}{1.03 - 1}$$ $$S_{30} = \frac{40,000((1.03)^{30} - 1)}{0.03}$$

  First, calculate $(1.03)^{30}$:

  $$(1.03)^{30} \approx 2.42726247$$

  Now substitute this value:

  $$S_{30} = \frac{40,000(2.42726247 - 1)}{0.03}$$ $$S_{30} = \frac{40,000(1.42726247)}{0.03}$$ $$S_{30} = \frac{57090.4988}{0.03}$$ $$S_{30} \approx 1903016.626$$

  Rounding to the nearest shilling: Her total earnings are $\boxed{\text{Kshs. } 1,903,017}$.

Part (b): Geometric Sequence

• The first four even numbers less than ten: These are 2, 4, 6, 8.

• The first three consecutive terms of a geometric sequence are:

  • 1st term = 1st even number = 2
  • 2nd term = 2nd even number = 4
  • 3rd term = 4th even number = 8

• So, the geometric sequence is 2, 4, 8, ...

• Determine the $n^{th}$ term of the sequence.

  • The first term is $a = 2$.
  • The common ratio $r$ is found by dividing a term by its preceding term: $$r = \frac{4}{2} = 2$$
  • The formula for the $n^{th}$ term of a geometric sequence is $T_n = a \cdot r^{n-1}$.
  • Substitute $a=2$ and $r=2$: $$T_n = 2 \cdot 2^{n-1}$$
  • Using the rule of exponents $x^a \cdot x^b = x^{a+b}$: $$T_n = 2^1 \cdot 2^{n-1} = 2^{1 + (n-1)} = 2^n$$

  The $n^{th}$ term of the sequence is $\boxed{T_n = 2^n}$.