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

Given the information about the arithmetic sequence with first three terms:

• Let the three terms be expressed as:
  • $T_1 = 5e^k$,
  • $T_2 = 13$,
  • $T_3 = 5 e^k$

1. The common difference of the arithmetic sequence can be calculated as:

   • $d = T_2 - T_1 = 13 - 5e^k$

2. Expressing the third term:

   • $T_3 = T_2 + d = 13 + (13 - 5e^k)$
   • This can be simplified to:

   $T_3 = 2(13) - 5e^k = 26 - 5e^k$

3. Thus, the sequence can be equated and rearranged as:

   $5e^k - 26 + 5e^k = 0$

   So we have:

   $5y^2 - 26y + 5 = 0$ By substituting $y = e^k$.

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 $c = 5$.

1. First, calculate the discriminant:

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

2. Substitute into the quadratic formula:

   $y = \frac{26 \pm \sqrt{576}}{10} = \frac{26 \pm 24}{10}$

This gives:

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

3. Since $y = e^k$, we have:

   • For $y_1 = 5$:
     • $k_1 = \ln(5)$
   • For $y_2 = \frac{1}{5}$:
     • $k_2 = \ln(\frac{1}{5}) = -\ln(5)$

Therefore, the two possible values of $k$ are:

• $k = \ln(5)$
• $k = -\ln(5)$.