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

To prove this, we assume that $3e^{q}$, $5$, and $3e^{r}$ 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:

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

or:

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

Thus:

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

Now, since $q$ and $r$ can take any real values, let's express $e^{q}$ in terms of $e^{r}$:

If we solve for $e^{q}$, 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 $3e^{q}$ and $3e^{r}$ to be positive, $e^{r}$ must be positive. Consequently, there cannot be a scenario wherein $3e^{q}$, $5$, and $3e^{r}$ 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 $q$ for which $3e^{q}$, $5$, and $3e^{r}$ can be consecutive terms of a geometric sequence.