A sequence is defined using this term-to-term rule. $u_{n+1} = \sqrt{2u_n + 15}$ If $u_1 = 5$, find $u_2$. Another sequence is defined using this term-to-term rule, $u_{n+1} = ku_n + r$ where $k$ and $r$ are constants. Given that $u_2 = 41$, $u_3 = 206$ and $u_4 = 1031$, find the value of $k$ and the value of $r$.

GCSE OCR Maths Sequences: A sequence is defined using this

A sequence is defined using this term-to-term rule - OCR - GCSE Maths - Question 12 - 2017 - Paper 1

Question 12

A sequence is defined using this term-to-term rule. $u_{n+1} = \sqrt{2u_n + 15}$ If $u_1 = 5$, find $u_2$. Another sequence is defined using this term-to-term rul... show full transcript

Worked Solution & Example Answer:A sequence is defined using this term-to-term rule - OCR - GCSE Maths - Question 12 - 2017 - Paper 1

Step 1

If $u_1 = 5$, find $u_2$

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Answer

To find $u_2$ , substitute $u_1 = 5$ into the term-to-term rule:

$u_2 = \sqrt{2u_1 + 15} = \sqrt{2 \cdot 5 + 15} = \sqrt{10 + 15} = \sqrt{25} = 5.$

Step 2

Find the value of $k$ and the value of $r$

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Answer

We are given the terms:

$u_2 = 41$
$u_3 = 206$
$u_4 = 1031$

Using the term-to-term rule:

For $n=2$ : $u_3 = ku_2 + r \Rightarrow 206 = k(41) + r.$
For $n=3$ : $u_4 = ku_3 + r \Rightarrow 1031 = k(206) + r.$

We now have two equations:

$41k + r = 206$
$206k + r = 1031$

To eliminate $r$ , we can subtract the first equation from the second:

$(206k + r) - (41k + r) = 1031 - 206$ $165k = 825$ $k = \frac{825}{165} = 5.$

Substituting $k = 5$ back into the first equation:

$41(5) + r = 206 \Rightarrow 205 + r = 206 \Rightarrow r = 1.$

Thus, $k = 5$ and $r = 1$ .

A sequence is defined using this term-to-term rule - OCR - GCSE Maths - Question 12 - 2017 - Paper 1

Worked Solution & Example Answer:A sequence is defined using this term-to-term rule - OCR - GCSE Maths - Question 12 - 2017 - Paper 1

If $u_1 = 5$, find $u_2$

Find the value of $k$ and the value of $r$

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