A sequence $a_n$, $a_{n+1}$, $a_{n+2}$, ... is defined by a_n = 4a_{n-1} - 3, \, n > 1\ a_1 = k, \, \text{where } k \text{ is a positive integer.} (a) Write down an expression for $a_2$ in terms of $k$. Given that $\sum_{n=1}^3 a_n = 66$ (b) find the value of $k$.

Photo AI

A sequence $a_n$, $a_{n+1}$, $a_{n+2}$, .. - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 2

Question 5

$A-sequence-$a_n$,-$a_{n+1}$,-$a_{n+2}$,-..-Edexcel-A-Level Maths Pure-Question 5-2014-Paper 2.png$

A sequence $a_n$, $a_{n+1}$, $a_{n+2}$, ... is defined by a_n = 4a_{n-1} - 3, \, n > 1\ a_1 = k, \, \text{where } k \text{ is a positive integer.} (a) Write down a... show full transcript

Worked Solution & Example Answer:A sequence $a_n$, $a_{n+1}$, $a_{n+2}$, .. - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 2

Step 1

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

96%

114 rated

Answer

To find $a_2$ , we use the given recursive formula:

$a_2 = 4a_{1} - 3 = 4k - 3.$

Step 2

find the value of $k$

99%

104 rated

Answer

We know that:

$\sum_{n=1}^3 a_n = 66.$

Calculating each term:

$a_1 = k$
$a_2 = 4k - 3$
To find $a_3$ :

$a_3 = 4a_{2} - 3 = 4(4k - 3) - 3 = 16k - 12 - 3 = 16k - 15.$

Now, summing up:

$a_1 + a_2 + a_3 = k + (4k - 3) + (16k - 15) = 21k - 18.$

Setting this equal to $66$ gives:

21k = 66 + 18 \ 21k = 84 \ k = 4.$$

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;