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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 \[ \frac{1}{2} n [2a + (n - 1)d] \] Se... 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 \[ \frac{1}{2} n [2a + (n - 1)d] \]
Answer
The sum ( S_n ) of the first n terms of an arithmetic series can be expressed as:
[ S_n = a + (a + d) + (a + 2d) + ... + (a + (n-1)d) ]
By rearranging the terms: [ S_n = (S_n) + (a + (n-1)d) + (a + (n-2)d) + ... + a ]
Adding these two equations gives: [ 2S_n = na + (n-1)d + na + (n-1)d ]
Dividing by 2 gives: [ S_n = \frac{1}{2} n [2a + (n - 1)d] ]
Step 2
Step 3
Form an equation in n, and show that your equation may be written as \[ n^3 - 150n + 5000 = 0 \]
Answer
The total repayment over n months is:
[ S_n = \frac{n}{2}(2 \cdot 149 + (n - 1)(-2)) ]
Setting this equal to £5000: [ \frac{n}{2} (298 - 2(n - 1)) = 5000 ]
This leads to: [ n(298 - 2n + 2) = 10000 ]
Thus: [ n(300 - 2n) = 10000 ]
Rearranging gives: [ -2n^2 + 300n - 10000 = 0 ]
Dividing by -2 leads to: [ n^2 - 150n + 5000 = 0 ]
Step 4
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
Only positive values of n that are greater than 21 make sense in the context of the repayment schedule. If any solution is less than 21, it is deemed nonsensical. For example, if ( n = 100 ), this is valid but unrealistic in a repayment scenario.