A sequence $a_1, a_2, a_3, \ldots$ is defined by a_n = k, a_{n+1} = 3a_n + 5, \quad n > 1, where $k$ is a positive integer. (a) Write down an expression for $a_2$ in terms of $k$. (b) Show that $a_3 = 9k + 20$. (c) (i) Find $\sum_{r=1}^{4} a_r$ in terms of $k$. (ii) Show that $\sum_{r=1}^{4} a_r$ is divisible by 10.

A-Level Edexcel Maths Pure Quadratics: A sequence $a_1, a_2, a

A sequence $a_1, a_2, a_3, \ldots$ is defined by a_n = k, a_{n+1} = 3a_n + 5, \quad n > 1, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 1

Question 10

$A-sequence-$a_1,-a_2,-a_3,-\ldots$-is-defined-by----a_n-=-k,----a_{n+1}-=-3a_n-+-5,-\quad-n->-1,---where-$k$-is-a-positive-integer-Edexcel-A-Level Maths Pure-Question 10-2007-Paper 1.png$

A sequence $a_1, a_2, a_3, \ldots$ is defined by a_n = k, a_{n+1} = 3a_n + 5, \quad n > 1, where $k$ is a positive integer. (a) Write down an expression for... show full transcript

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3, \ldots$ is defined by a_n = k, a_{n+1} = 3a_n + 5, \quad n > 1, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 1

Step 1

Write down an expression for $a_2$ in terms of $k$

96%

114 rated

Answer

From the definition, we have: $a_2 = 3a_1 + 5 = 3k + 5$ Thus, an expression for $a_2$ in terms of $k$ is: $a_2 = 3k + 5$

Step 2

Show that $a_3 = 9k + 20$

99%

104 rated

Answer

To find $a_3$ , we use: $a_3 = 3a_2 + 5$ Substituting $a_2 = 3k + 5$ , we get:

a_3 & = 3(3k + 5) + 5 \\ & = 9k + 15 + 5 \\ & = 9k + 20. \end{align*}$$ Thus, it is shown that $a_3 = 9k + 20$.

Step 3

Find $\sum_{r=1}^{4} a_r$ in terms of $k$

96%

101 rated

Answer

We find each term first:

$a_1 = k$
$a_2 = 3k + 5$
$a_3 = 9k + 20$ (shown above)
To find $a_4$ , we have: $a_4 = 3a_3 + 5 = 3(9k + 20) + 5 = 27k + 60 + 5 = 27k + 65$ Now, we can calculate: $\sum_{r=1}^{4} a_r = a_1 + a_2 + a_3 + a_4$ Substituting the values: $= k + (3k + 5) + (9k + 20) + (27k + 65) = (40k + 90)$

Step 4

Show that $\sum_{r=1}^{4} a_r$ is divisible by 10

98%

120 rated

Answer

We have: $\sum_{r=1}^{4} a_r = 40k + 90$ We can express this as: $= 10(4k + 9)$ Since $4k + 9$ is an integer, it follows that $10(4k + 9)$ is divisible by 10.

A sequence $a_1, a_2, a_3, \ldots$ is defined by a_n = k, a_{n+1} = 3a_n + 5, \quad n > 1, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 1

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3, \ldots$ is defined by a_n = k, a_{n+1} = 3a_n + 5, \quad n > 1, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 1

Write down an expression for $a_2$ in terms of $k$

Show that $a_3 = 9k + 20$

Find $\sum_{r=1}^{4} a_r$ in terms of $k$

Show that $\sum_{r=1}^{4} a_r$ is divisible by 10

Join the A-Level students using SimpleStudy...