A sequence $u_1, u_2, u_3, \ldots$ is defined as follows, for $n \in \mathbb{N}$: $u_1 = 2, \quad u_2 = 64, \quad u_{n+1} = \frac{u_n}{\sqrt{u_{n-1}}}$ Write $u_3$ in the form $2^p$, where $p \in \mathbb{R}$ - Leaving Cert Mathematics - Question 4 - 2022

In this arithmetic sequence, we start with:

• The first term is $5e^{-k}$, the second term is $13$, and the third term is $5e^{k}$.

• Recognizing that these terms must maintain a common difference, we set:

$T_2 - T_1 = T_3 - T_2$

• Let’s substitute the terms:

$13 - 5e^{-k} = 5e^{k} - 13$

• Rearranging gives:

$5e^{-k} + 5e^{k} = 26$

• Multiplying through by $e^k$ (to eliminate $e^{-k}$):

$5 + 5e^{2k} = 26e^{k}$

• Rearranging the equation:

$5e^{2k} - 26e^{k} + 5 = 0$

• Substituting $y = e^k$ leads to:

$5y^2 - 26y + 5 = 0$

To solve the quadratic equation $5y^2 - 26y + 5 = 0$, we can use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 5$, $b = -26$, and $c = 5$. Plugging in the values:

• The discriminant is:

$b^2 - 4ac = (-26)^2 - 4(5)(5) = 676 - 100 = 576$

• Therefore, we have:

$y = \frac{26 \pm \sqrt{576}}{2 \cdot 5}$

• Simplifying further:

$y = \frac{26 \pm 24}{10}$

• This gives two solutions:

1. $y_1 = \frac{50}{10} = 5$
2. $y_2 = \frac{2}{10} = \frac{1}{5}$

• Recalling $y = e^k$, we get:

1. For $y_1 = 5$:
   • $e^k = 5 \Rightarrow k = \ln(5)$
2. For $y_2 = \frac{1}{5}$:
   • $e^k = \frac{1}{5} \Rightarrow k = \ln\left(\frac{1}{5}\right) = -\ln(5)$

• Thus, the two possible values of $k$ in the required form are:

• $k = \ln(5)$ and $k = -\ln(5)$.