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The rth term of an arithmetic series is $(2r - 5)$ - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 2
Question 7
The rth term of an arithmetic series is $(2r - 5)$. (a) Write down the first three terms of this series. (b) State the value of the common difference. (c) Show th... show full transcript
Worked Solution & Example Answer:The rth term of an arithmetic series is $(2r - 5)$ - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 2
Step 1
Write down the first three terms of this series.
Answer
To find the first three terms of the arithmetic series defined by the rth term ( (2r - 5) ), we can compute:
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For ( r = 1 ): [ 2(1) - 5 = 2 - 5 = -3 ]
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For ( r = 2 ): [ 2(2) - 5 = 4 - 5 = -1 ]
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For ( r = 3 ): [ 2(3) - 5 = 6 - 5 = 1 ]
Thus, the first three terms are: -3, -1, 1.
Step 2
State the value of the common difference.
Answer
The common difference in an arithmetic series is the difference between any two successive terms. From the previously calculated terms:
- The difference between the second term (-1) and the first term (-3) is: [ -1 - (-3) = -1 + 3 = 2 ]
Thus, the common difference is 2.
Step 3
Show that \( \sum_{r=1}^n (2r - 5) = n(n - 4) \)
Answer
To show this, we start by evaluating the sum:
[ S = \sum_{r=1}^n (2r - 5) ]
This can be split into two separate sums:
[ S = \sum_{r=1}^n 2r - \sum_{r=1}^n 5 ]
The first part, using the formula for the sum of the first n natural numbers: [ \sum_{r=1}^n r = \frac{n(n + 1)}{2} ] Thus, [ \sum_{r=1}^n 2r = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) ]
The second part is simply: [ \sum_{r=1}^n 5 = 5n ]
Now, substituting back into the equation for S: [ S = n(n + 1) - 5n ] [ S = n^2 + n - 5n = n^2 - 4n ] [ S = n(n - 4) ]
Hence, we have shown that: [ \sum_{r=1}^n (2r - 5) = n(n - 4) ]