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

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9 (a) 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. 9 (b) Pro... show full transcript

Worked Solution & Example Answer:9 (a) 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 pp, we start by setting up the condition for the terms to be in an arithmetic sequence. The difference between the first and second term should equal the difference between the second and third term:

3ep−5=3er−53e^{p} - 5 = 3e^{r} - 5

This simplifies to:

3ep=3er3e^{p} = 3e^{r}

Dividing both sides by 3 gives:

ep=ere^{p} = e^{r}

Taking the natural logarithm of both sides results in:

p=rp = r

Next, we also set the difference condition as:

5−3ep=3er−55 - 3e^{p} = 3e^{r} - 5

Rearranging gives:

10=3er+3ep10 = 3e^{r} + 3e^{p}

Substituting rr with pp, we have:

10=3ep+3ep10 = 3e^{p} + 3e^{p}

or:

10=6ep10 = 6e^{p}

Dividing both sides by 6 yields:

e^{p} = rac{10}{6} = rac{5}{3}

Taking the natural logarithm of both sides gives:

p = ext{ln} rac{5}{3}

Thus, the possible values of pp in exact form is:

p = ext{ln} rac{5}{3}.

Step 2

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

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Answer

To prove this, we assume that 3eq3e^{q}, 55, and 3er3e^{r} are consecutive terms of a geometric sequence, which means that the ratio between consecutive terms must be equal:

rac{5}{3e^{q}} = rac{3e^{r}}{5}

Cross-multiplying gives:

52=3eqimes3er5^2 = 3e^{q} imes 3e^{r}

or:

25=9eq+r25 = 9 e^{q + r}

Thus:

e^{q + r} = rac{25}{9}

Now, since qq and rr can take any real values, let's express eqe^{q} in terms of ere^{r}:

If we solve for eqe^{q}, we have:

e^{q} = rac{25}{9 e^{r}}

This leads us to:

3e^{q} = rac{75}{9 e^{r}}

However, observe that for both expressions 3eq3e^{q} and 3er3e^{r} to be positive, ere^{r} must be positive. Consequently, there cannot be a scenario wherein 3eq3e^{q}, 55, and 3er3e^{r} can hold as consecutive terms of a geometric sequence due to the inherent contradictions from equating the ratios.

Thus, we conclude that there is no possible value of qq for which 3eq3e^{q}, 55, and 3er3e^{r} can be consecutive terms of a geometric sequence.

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