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A sequence $a_1, a_2, a_3, ...$ is defined by $a_1 = 2$ $a_{n+1} = 3a_n - c$ where $c$ is a constant - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 2

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A-sequence-$a_1,-a_2,-a_3,-...$-is-defined-by---$a_1-=-2$---$a_{n+1}-=-3a_n---c$---where-$c$-is-a-constant-Edexcel-A-Level Maths Pure-Question 6-2011-Paper 2.png

A sequence $a_1, a_2, a_3, ...$ is defined by $a_1 = 2$ $a_{n+1} = 3a_n - c$ where $c$ is a constant. (a) Find an expression for $a_2$ in terms of $c$. G... show full transcript

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3, ...$ is defined by $a_1 = 2$ $a_{n+1} = 3a_n - c$ where $c$ is a constant - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 2

Step 1

Find an expression for $a_2$ in terms of $c$

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Answer

To find a2a_2, we can use the recursive formula provided:
an+1=3anca_{n+1} = 3a_n - c.
For n=1n = 1:
[ a_2 = 3a_1 - c
= 3(2) - c
= 6 - c.
]

Step 2

find the value of $c$

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104 rated

Answer

Given that i=13ai=0\sum_{i=1}^3 a_i = 0, we need to find a3a_3 first.
We already have a1=2a_1 = 2 and we can use the expression for a2a_2:
[
a_2 = 6 - c.
]
Now, to find a3a_3:
[ a_3 = 3a_2 - c
= 3(6 - c) - c
= 18 - 3c - c
= 18 - 4c.
]
Now we can compute the sum:
[
a_1 + a_2 + a_3 = 2 + (6 - c) + (18 - 4c)
= 2 + 6 - c + 18 - 4c
= 26 - 5c.
]
Setting this equal to zero gives:
[ 26 - 5c = 0
\Rightarrow 5c = 26 \Rightarrow c = \frac{26}{5} = 5.2.
]

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