A-Level AQA Maths Pure Proof by Contradiction: Three consecutive terms in an ar

Three consecutive terms in an arithmetic sequence are $3e^{p}$, $5$, $3e^{r}$ - AQA - A-Level Maths Pure - Question 9 - 2017 - Paper 2

Question 9

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Three consecutive terms in an arithmetic sequence are $3e^{p}$, $5$, $3e^{r}$. Find the possible values of $p$. Give your answers in an exact form. Prove that ther... show full transcript

Worked Solution & Example Answer:Three consecutive terms in an arithmetic sequence are $3e^{p}$, $5$, $3e^{r}$ - AQA - A-Level Maths Pure - Question 9 - 2017 - Paper 2

Step 1

Find the possible values of p

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Answer

To find the possible values of $p$ , we first recognize that the three terms are in arithmetic progression.

Setting Up the Equation: Since the terms are in an arithmetic sequence, the difference between the first and second term should equal the difference between the second and third term.

Thus, we can set up the equation:

$3e^{p} - 5 = 3e^{r} - 5$

Simplifying this gives:

$3e^{p} = 3e^{r} + 5 - 5$
Removing Constants: Next, we simplify the equation:

$3e^{p} - 3e^{r} = 0$

From this, we can express it as:

$3e^{p} = 3e^{r}$

Dividing both sides by 3 yields:

$e^{p} = e^{r}$
Logarithmic Form: Applying the natural logarithm to both sides, we find:

$p = r$
Finding r: Now we also need to find another equation using another difference:

From $5 - 3e^{p} = 3e^{r} - 5$ , we get:

$5 - 3e^{p} = 3e^{r} - 5$

Rearranging, leads to:

$10 = 3e^{p} + 3e^{r}$ \n
Final Solutions: Substituting $p = r$ gives:
$= 6e^{p}$$ Therefore: $$e^{p} = rac{10}{6} = rac{5}{3}$$ Taking the natural logarithm results in: $$p = ext{ln} rac{5}{3}$$$

Thus, the possible value of $p$ is $p = ext{ln} rac{5}{3}$ .

Step 2

Prove that there is no possible value of q

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Answer

To prove there is no possible value of $q$ for which $3e^{r}$ , $5$ , $3e^{q}$ are consecutive terms of a geometric sequence:

Understanding Geometric Sequences: For three terms to be in geometric progression, the ratio between the first and second term must equal the ratio between the second and third term:

$\frac{5}{3e^{r}} = \frac{3e^{q}}{5}$
Cross Multiplying: Cross-multiplying yields:

$5^2 = 3e^{r} imes 3e^{q}$

Simplifying gives:

$25 = 9e^{r + q}$
Finding contradiction: Thus:

$e^{r + q} = \frac{25}{9}$
Identifying Limits: Since $e^{r}$ and $e^{q}$ can take any real number value, notice that this implies a specific condition that needs to hold for any $r$ and $q$ . If $r$ or $q$ takes any non-zero values, it results in a contradiction because we cannot assign any value to $q$ in the relationship derived.
Conclusion: Therefore, there is a contradiction in the assumption that $3e^{r}$ , $5$ , and $3e^{q}$ can be consecutive terms of a geometric sequence, proving that no possible value of $q$ satisfies the condition.

Three consecutive terms in an arithmetic sequence are $3e^{p}$, $5$, $3e^{r}$ - AQA - A-Level Maths Pure - Question 9 - 2017 - Paper 2

Worked Solution & Example Answer:Three consecutive terms in an arithmetic sequence are $3e^{p}$, $5$, $3e^{r}$ - AQA - A-Level Maths Pure - Question 9 - 2017 - Paper 2

Find the possible values of p

Prove that there is no possible value of q

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