Photo AI
A geometric series has first term $a$ and common ratio $r$ - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 2
Question 10
A geometric series has first term $a$ and common ratio $r$. Prove that the sum of the first $n$ terms of the series is \[ S_n = \frac{a(1 - r^n)}{1 - r} \] (4) Mr ... show full transcript
Worked Solution & Example Answer:A geometric series has first term $a$ and common ratio $r$ - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 2
Step 1
Prove that the sum of the first n terms of the series is $\frac{a(1 - r^n)}{1 - r}$
Answer
To find the sum of the first n terms of a geometric series, we can denote the series as:
[ S_n = a + ar + ar^2 + ... + ar^{n-1} ]
Multiplying the series by the common ratio gives:
[ S_n r = ar + ar^2 + ar^3 + ... + ar^n ]
Subtracting these two equations, we get:
[ S_n (1 - r) = a - ar^n ]
Therefore, by rearranging, we find:
[ S_n = \frac{a(1 - r^n)}{1 - r} ]
This completes the proof.
Step 2
Find, to the nearest £100, Mr King’s salary in the year 2008.
Answer
Mr King’s salary in 2005 is £35,000 and it increases by 4% each year. The salary for any year can be modeled by the formula:
[ S_n = 35000 \times (1.04)^{n-1} ]
For 2008, which is the 4th year, we find:
[ S_4 = 35000 \times (1.04)^3 \approx 35000 \times 1.124864 = 39440.24 ]
Rounding to the nearest £100, Mr King’s salary in 2008 is approximately £39,400.
Step 3
Find, to the nearest £1,000, the total amount of salary he will receive from 2005 until he retires at the end of 2024.
Answer
The total salary from 2005 to 2024 spans 20 years. This can be calculated as:
[ S_{20} = 35000 \times \frac{1 - (1.04)^{20}}{1 - 1.04} ]
Calculating the expression:
[
S_{20} = 35000 \times \frac{1 - (1.04)^{20}}{-0.04} = 35000 \times \frac{1 - 2.208040279}{-0.04}
= 35000 \times \frac{-1.208040279}{-0.04} \approx 35000 \times 30.201
]
The total salary is approximately £1,057,035, which rounds to approximately £1,000,000 when rounded to the nearest £1,000.