Question

1. for the geometric sequence where $a_2 = 1$ and $a_5 = 64$, find the following.

a. the common ratio, $r$.
b. $a_0$

1. the first two terms of a sequence are 8 and 12.

a. find the third and fourth term if the sequence is arithmetic.
b. find the third and fourth term if the sequence is geometric.

Explanation:

Problem 2a

Step1: Recall geometric sequence formula

The formula for the \(n\)-th term of a geometric sequence is \(a_n = a_1r^{n - 1}\), where \(a_1\) is the first term and \(r\) is the common ratio. We know \(a_2=1\) and \(a_5 = 64\). So, \(a_2=a_1r^{2 - 1}=a_1r = 1\) and \(a_5=a_1r^{5 - 1}=a_1r^{4}=64\).

Step2: Divide the two equations

Divide the equation for \(a_5\) by the equation for \(a_2\): \(\frac{a_1r^{4}}{a_1r}=\frac{64}{1}\). Simplify the left - hand side: \(r^{3}=64\).

Step3: Solve for \(r\)

Take the cube root of both sides: \(r=\sqrt[3]{64}=4\).

Step1: Find \(a_1\) using \(a_2\)

We know from \(a_2=a_1r\) and \(a_2 = 1\), \(r = 4\). So, \(1=a_1\times4\), then \(a_1=\frac{1}{4}\).

Step2: Use geometric sequence formula for \(a_0\)

The formula for the \(n\)-th term is \(a_n=a_1r^{n - 1}\). For \(n = 0\), \(a_0=a_1r^{0 - 1}=a_1\times\frac{1}{r}\). Substitute \(a_1=\frac{1}{4}\) and \(r = 4\): \(a_0=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\).

Step1: Find the common difference \(d\)

For an arithmetic sequence, the common difference \(d=a_2 - a_1\). Given \(a_1 = 8\) and \(a_2=12\), then \(d=12 - 8 = 4\).

Step2: Find the third term \(a_3\)

The formula for the \(n\)-th term of an arithmetic sequence is \(a_n=a_1+(n - 1)d\). For \(n = 3\), \(a_3=a_1+(3 - 1)d=8 + 2\times4=8 + 8 = 16\).

Step3: Find the fourth term \(a_4\)

For \(n = 4\), \(a_4=a_1+(4 - 1)d=8+3\times4=8 + 12 = 20\).

Answer:

The common ratio \(r = 4\)

Problem 2b