Photo AI
An arithmetic series has first term a and common difference d - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 1
Question 10
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 \[ \frac{n}{2} [2a + (n - 1)d] \] Sean... show full transcript
Worked Solution & Example Answer:An arithmetic series has first term a and common difference d - Edexcel - A-Level Maths Pure - Question 10 - 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 ( S_n ) of the first n terms of an arithmetic series, we begin with the formula for an arithmetic series:
[ S_n = a + (a + d) + (a + 2d) + ... + (a + (n - 1)d) ]
This can also be expressed as:
[ S_n = n imes a + d imes (0 + 1 + 2 + ... + (n - 1)) ]
Using the formula for the sum of the first n-1 integers, we can simplify this further:
[ S_n = n imes a + d \left(\frac{(n - 1)n}{2}\right) ]
This ultimately leads to:
[ S_n = \frac{n}{2} [2a + (n - 1)d] ]
which proves the formula.
Step 2
Find the amount Sean repays in the 21st month.
Answer
From the repayment sequence, we see that the amount repaid in the 21st month is given by the formula:
[ a + (21 - 1)d ]
Given that ( a = 149 ) and ( d = -2 ):
[ ext{Amount} = 149 + (21 - 1)(-2) ]
This simplifies to:
[ 149 - 40 = 109 ]
Thus, the amount Sean repays in the 21st month is £109.
Step 3
Form an equation in n, and show that your equation may be written as \[ n^3 - 150n + 5000 = 0 \]
Answer
The total amount repaid over n months sums to £5000. Hence, we write:
[ S_n = 5000 ]
Utilizing the sum formula derived earlier:
[ \frac{n}{2} [2a + (n - 1)d] = 5000 ]
Inserting our known values ( a = 149 ) and ( d = -2 ):
[ \frac{n}{2} [2(149) + (n - 1)(-2)] = 5000 ]
This simplifies to:
[ n[149 - (n - 1)] = 10000 ]
or
[ n(150 - n) = 10000 ]
Rearranging gives us:
[ n^2 - 150n + 5000 = 0 ]
which can be rewritten as:
[ n^3 - 150n + 5000 = 0 ]
showing our desired equation.
Step 4
Solve the equation in part (c).
Answer
To solve the equation ( n^3 - 150n + 5000 = 0 ), we can use methods such as synthetic division or finding rational roots.
Through test values, we try n = 25:
[ 25^3 - 150 \cdot 25 + 5000 = 15625 - 3750 + 5000 = 0 ]
Finding that n = 25 is a solution. We can perform polynomial long division or use synthetic division to factor:
[ (n - 25)(n^2 + 25n - 200) = 0 ]
Next, we solve the quadratic using the quadratic formula:
[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
Substituting ( a = 1, b = 25, c = -200 ):
[ n = \frac{-25 \pm \sqrt{625 + 800}}{2} = \frac{-25 \pm \sqrt{1425}}{2} ]
The remaining two roots can be further evaluated.
Ultimately, n could also yield other values depending on the solutions of the quadratic.
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
From the context, n represents the number of months Sean is repaying the loan. Therefore, n must be a positive integer. Any negative value or non-integer derived from the equation is not sensible for this scenario.
For this reason, any solution that yields n less than or equal to 0, or is not a whole number, would be deemed unacceptable.