Answer:
Common difference:
[tex]\displaystyle d = \frac{3}{4}\, a_{1}[/tex],
Where [tex]a_{1}[/tex] is the first term of the arithmetic sequence.
Apply the following steps to show that the specified items are consecutive items in the geometric sequence:
- Find an expression for each of the specified item.
- Find the ratio between consecutive items.
- Show that the ratios are the same and are equal to the requested common ratio.
Step-by-step explanation:
A few values are consecutive items of a geometric sequence if and only if the ratios between the items are equal. In this question, for the [tex]\text{$5$th}[/tex], [tex]\text{$9$th}[/tex], and [tex]\text{$16$th}[/tex] items of the arithmetic sequence to be consecutive items of a geometric sequence, the following ratios should be equal:
- ratio between the [tex]\text{$9$th}[/tex] and the [tex]\text{$5$th}[/tex] item of the arithmetic sequence, and
- ratio between the [tex]\text{$16$th}[/tex] and the [tex]\text{$9$th}[/tex] item of the arithmetic sequence.
Let [tex]a_{1}[/tex] denote the first term of the arithmetic sequence, and let [tex]d[/tex] denote the common difference of that sequence. The [tex]\text{$n$th}[/tex] item of this sequence would be:
[tex]a_{1} + (n - 1)\, d[/tex].
The [tex]\text{$5$th}[/tex], [tex]\text{$9$th}[/tex], and [tex]\text{$16$th}[/tex] of this sequence would be [tex](a_{1} + (5 - 1)\, d)[/tex], [tex](a_{1} + (9 - 1)\, d)[/tex], and [tex](a_{1} + (16 - 1)\, d)[/tex], respectively.
Since these values are consecutive items of a geometric sequence, the ratios between the consecutive items should be equal:
[tex]\displaystyle \frac{a_{1} + (9 - 1)\, d}{a_{1} + (5 - 1)\, d} = \frac{a_{1} + (16 - 1)\, d}{a_{1} + (9 - 1)\, d}[/tex].
Rearrange and solve for the common difference [tex]d[/tex] in terms of the first term [tex]a_{1}[/tex]:
[tex]{a_{1}}}^{2} + 19\, a_{1}\, d + 60\, d^{2} = {a_{1}}^{2} + 16\, a_{1}\, d + 64\, d^{2}[/tex].
[tex]3\, a_{1}\, d = 4\, d^{2}[/tex].
[tex]\displaystyle d = \frac{3}{4}\, a_{1}[/tex].
In other words, [tex]d = (3/4)\, a_{1}[/tex] would ensure that the [tex]\text{$5$th}[/tex], [tex]\text{$9$th}[/tex], and [tex]\text{$16$th}[/tex] of this arithmetic sequence are consecutive items in a geometric sequence.
To show that the [tex]\text{$21$st}[/tex], [tex]\text{$37$th}[/tex], and [tex]\text{$65$th}[/tex] items of this arithmetic sequence are consecutive items of a geometric sequence, start by finding an expression for each of these values.
Expression for the [tex]\text{$21$st}[/tex] item:
[tex]\begin{aligned}& a_{1} + (21 - 1)\, d \\ =\; & a_{1} + 20\, d \\ =\; & a_{1} + 20\, \left(\frac{3}{4}\, a_{1}\right) \\ =\; & 16\, a_{1}\end{aligned}[/tex].
Expression for the [tex]\text{$37$th}[/tex] item:
[tex]\begin{aligned}& a_{1} + (37 - 1)\, d \\ =\; & a_{1} + 36\, \left(\frac{3}{4}\, a_{1}\right) \\ =\; & 28\, a_{1}\end{aligned}[/tex].
Expression for the [tex]\text{$65$th}[/tex] item:
[tex]\begin{aligned}& a_{1} + (65 - 1)\, d \\ =\; & a_{1} +64\, \left(\frac{3}{4}\, a_{1}\right) \\ =\; & 49\, a_{1}\end{aligned}[/tex].
Evaluate the ratio between the consecutive items.
Ratio between the [tex]\text{$37$th}[/tex] and the [tex]\text{$21$st}[/tex] items:
[tex]\displaystyle \frac{28\, a_{1}}{16\, a_{1}} = \frac{7}{4}[/tex].
Ratio between the [tex]\text{$65$th}[/tex] and the [tex]\text{$37$th}[/tex] items:
[tex]\displaystyle \frac{49\, a_{1}}{28\, a_{1}} = \frac{7}{4}[/tex].
