{"id":1675,"date":"2025-07-25T19:12:28","date_gmt":"2025-07-25T19:12:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1675"},"modified":"2026-03-25T22:02:23","modified_gmt":"2026-03-25T22:02:23","slug":"geometric-sequences-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/geometric-sequences-learn-it-3\/","title":{"raw":"Geometric Sequences: Learn It 3","rendered":"Geometric Sequences: Learn It 3"},"content":{"raw":"

Using Explicit Formulas for Geometric Sequences<\/h2>\r\nBecause a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.\r\n

[latex]{a}_{n}={a}_{1}{r}^{n - 1}[\/latex]<\/p>\r\nLet\u2019s write the first few terms of the sequence where the first term is [latex]a_1[\/latex] and the common ratio is<\/span>[latex]r[\/latex] to see this pattern.<\/span>\r\n\r\n\"\"\r\n\r\n

\r\n

explicit formula for a geometric sequence<\/h3>\r\nThe explicit formula of a geometric sequence with first term [latex]a_1[\/latex] and the common ratio [latex]r[\/latex] is\r\n

[latex]{a}_{n}={a}_{1}{r}^{n - 1}[\/latex]<\/p>\r\n\r\n<\/section>

Let\u2019s take a look at the sequence [latex]\\left\\{18\\text{, }36\\text{, }72\\text{, }144\\text{, }288\\text{, }...\\right\\}[\/latex].\r\n[latex]\\\\[\/latex]\r\nThis is a geometric sequence with a common ratio of [latex]2[\/latex] and an exponential function with a base of [latex]2[\/latex]. An explicit formula for this sequence is\r\n

[latex]{a}_{n}=18\\cdot {2}^{n - 1}[\/latex]<\/p>\r\n\"Graph\r\n\r\n<\/section>

Given a geometric sequence with [latex]{a}_{1}=3[\/latex] and [latex]{a}_{4}=24[\/latex], find [latex]{a}_{2}[\/latex].[reveal-answer q=\"602906\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"602906\"]The sequence can be written in terms of the initial term and the common ratio [latex]r[\/latex].\r\n

[latex]3,3r,3{r}^{2},3{r}^{3},\\dots[\/latex]<\/p>\r\nFind the common ratio using the given fourth term.\r\n

[latex]\\begin{align}&{a}_{n}={a}_{1}{r}^{n - 1} \\\\ &{a}_{4}=3{r}^{3} && \\text{Write the fourth term of sequence in terms of }{a}_{1}\\text{ and }r \\\\ &24=3{r}^{3} && \\text{Substitute }24\\text{ for }{a}_{4} \\\\ &8={r}^{3} && \\text{Divide} \\\\ &r=2 && \\text{Solve for the common ratio} \\end{align}[\/latex]<\/p>\r\nFind the second term by multiplying the first term by the common ratio.\r\n

[latex]\\begin{align}{a}_{2} & =2{a}_{1} \\\\ & =2\\left(3\\right) \\\\ & =6 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Write an explicit formula for the [latex]n\\text{th}[\/latex] term of the following geometric sequence.\r\n

[latex]\\left\\{2,10,50,250,\\dots\\right\\}[\/latex]<\/p>\r\n[reveal-answer q=\"202402\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"202402\"]\r\n\r\nThe first term is [latex]2[\/latex]. The common ratio can be found by dividing the second term by the first term.\r\n

[latex]\\dfrac{10}{2}=5[\/latex]<\/p>\r\nThe common ratio is [latex]5[\/latex]. Substitute the common ratio and the first term of the sequence into the formula.\r\n

[latex]\\begin{align}&{a}_{n}={a}_{1}{r}^{\\left(n - 1\\right)} \\\\ &{a}_{n}=2\\cdot {5}^{n - 1} \\end{align}[\/latex]<\/p>\r\nThe graph of this sequence shows an exponential pattern.\r\n\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Find the ninth term of the sequence [latex]6, 18, 54, 162, 486, 1458, \\dots[\/latex] Then find the general term for the sequence.<\/span>[reveal-answer q=\"334733\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"334733\"][latex]\\begin{align*} \\text{Let's first determine } a_1 \\text{ and the common ratio } r: & \\\\ \\text{The first term is } 6, \\text{ so } a_1 &= 6 & \\\\ \\text{The ratio is: } \\frac{18}{6} &= \\frac{54}{18} = \\frac{162}{54} = \\frac{486}{162} = \\frac{1458}{486} = 3 & r &= 3 \\end{align*}[\/latex][latex]\\begin{align*} \\text{To find the 9th term, use the formula with } a_1 = 6, \\, r = 3, \\text{ and } n=9: & \\\\ \\text{Substitute these values and simplify:} & & \\\\ a_n &= a_1 r^{n-1} & \\\\ a_9 &= 6(3)^{9-1} & \\\\ a_9 &= 6(3)^8 & \\\\ a_9 &= 39366 & \\end{align*}[\/latex][latex]\\begin{align*} \\text{To find the general term, substitute } a_1 = 6 \\text{ and } r = 3 \\text{ into the formula:} & \\\\ a_n &= a_1 r^{n-1} & \\\\ a_n &= 6(3)^{n-1} & \\end{align*}[\/latex][\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]321969[\/ohm_question]<\/section>
[ohm_question hide_question_numbers=1]321970[\/ohm_question]<\/section>","rendered":"

Using Explicit Formulas for Geometric Sequences<\/h2>\n

Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.<\/p>\n

[latex]{a}_{n}={a}_{1}{r}^{n - 1}[\/latex]<\/p>\n

Let\u2019s write the first few terms of the sequence where the first term is [latex]a_1[\/latex] and the common ratio is<\/span>[latex]r[\/latex] to see this pattern.<\/span><\/p>\n

\"\"<\/p>\n

\n

explicit formula for a geometric sequence<\/h3>\n

The explicit formula of a geometric sequence with first term [latex]a_1[\/latex] and the common ratio [latex]r[\/latex] is<\/p>\n

[latex]{a}_{n}={a}_{1}{r}^{n - 1}[\/latex]<\/p>\n<\/section>\n

Let\u2019s take a look at the sequence [latex]\\left\\{18\\text{, }36\\text{, }72\\text{, }144\\text{, }288\\text{, }...\\right\\}[\/latex].
\n[latex]\\\\[\/latex]
\nThis is a geometric sequence with a common ratio of [latex]2[\/latex] and an exponential function with a base of [latex]2[\/latex]. An explicit formula for this sequence is<\/p>\n

[latex]{a}_{n}=18\\cdot {2}^{n - 1}[\/latex]<\/p>\n

\"Graph<\/p>\n<\/section>\n

Given a geometric sequence with [latex]{a}_{1}=3[\/latex] and [latex]{a}_{4}=24[\/latex], find [latex]{a}_{2}[\/latex].<\/p>\n