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An arithmetic series has first term a and common difference d - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 1
Question 2
An arithmetic series has first term a and common difference d. (a) Prove that the sum of the first n terms of the series is \( \frac{n}{2}[2a + (n - 1)d] \) (4) ... show full transcript
Worked Solution & Example Answer:An arithmetic series has first term a and common difference d - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 1
Step 1
Prove that the sum of the first n terms of the series is \( \frac{n}{2}[2a + (n - 1)d] \)
Answer
To find the sum of the first n terms of an arithmetic series, consider the following: Let ( S_n ) represent the sum of the first n terms, which can be expressed as:
[ S_n = a + (a + d) + (a + 2d) + ... + (a + (n - 1)d) ]
When we write the series in reverse:
[ S_n = (a + (n - 1)d) + (a + (n - 2)d) + ... + a ]
Adding these two equations gives:
[ 2S_n = [2a + (n - 1)d]n ]
This means:
[ S_n = \frac{n}{2}[2a + (n - 1)d] ]
Thus, the proof shows that the sum of the first n terms is indeed ( \frac{n}{2}[2a + (n - 1)d] ).
Step 2
Find the amount Sean repays in the 21st month.
Answer
Sean's repayment schedule decreases by £2 each month, starting from £149. Thus, to find out how much he repays in the 21st month, use:
[ a_{21} = 149 - 2(21 - 1) ]
Calculating this gives:
[ a_{21} = 149 - 2 \times 20 = 149 - 40 = £109 ]
So, the repayment in the 21st month is £109.
Step 3
Form an equation in n, and show that your equation may be written as \( n^2 - 150n + 5000 = 0. \)
Answer
The total repayment over n months is given as £5000. Knowing the last term as ( a_n = 149 - 2(n - 1) ), we can form an equation:
[ S_n = \frac{n}{2}[a_1 + a_n] = \frac{n}{2}[149 + (149 - 2(n - 1))] = \frac{n}{2}[298 - 2n + 2] = \frac{n}{2}[300 - 2n] ]
Setting this equal to £5000:
[ \frac{n(300 - 2n)}{2} = 5000 ]
Multiplying both sides by 2 leads to:
[ n(300 - 2n) = 10000 ]
Expanding gives:
[ 300n - 2n^2 = 10000 ]
Rearranging results in:
[ 2n^2 - 300n + 10000 = 0 ]
Dividing through by 2 gives:
[ n^2 - 150n + 5000 = 0 ].
Step 4
Solve the equation in part (c).
Answer
To solve the quadratic equation ( n^2 - 150n + 5000 = 0 ), we will utilize the quadratic formula:
[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
Substituting a = 1, b = -150, and c = 5000:
[ n = \frac{150 \pm \sqrt{(-150)^2 - 4 \times 1 \times 5000}}{2 \times 1} ]
Calculating the discriminant:
[ (-150)^2 - 4 \times 1 \times 5000 = 22500 - 20000 = 2500 ]
Thus,
[ n = \frac{150 \pm \sqrt{2500}}{2} = \frac{150 \pm 50}{2} ]
This results in two potential solutions:
[ n = \frac{200}{2} = 100 \quad \text{and} \quad n = \frac{100}{2} = 25 ]
Step 5
State, with a reason, which of the solutions to the equation in part (c) is not a sensible solution to the repayment problem.
Answer
The solution ( n = 25 ) is sensible as it indicates that Sean would complete repayment in 25 months. However, the solution ( n = 100 ) is not sensible since the problem states he must make his final repayment after more than 21 months. Therefore, a repayment period of 100 months exceeds this condition.