Find the fourteenth term of the geometric sequences 1/8,1/4,1/2

asked by guest
on Jan 28, 2025 at 4:49 pm



You asked:

Determine the 14th14^{\text{th}} term in the sequence 18,14,12,\frac{1}{8}, \frac{1}{4}, \frac{1}{2}, \ldots.

MathBot Answer:

18,14,12,...\frac{1}{8},\frac{1}{4},\frac{1}{2},...The 14th term in this geometric sequence is 10241024.


The nth term in this sequence is given by the formula:

Explicit Formula: an=2n4a_n=2^{n - 4}

Recursive Formula: an=2an1,where a1=18a_n=2 a_{n-1}, \text{where } a_{1}=\frac{1}{8}


181412×2×2\begin{matrix}\frac{1}{8}&&\frac{1}{4}&&\frac{1}{2}\\[6pt]&\times2&&\times2\\[6pt]\end{matrix}

Explicit Formula

The formula for a geometric sequence where a1a_1 is the 1st term, rr is the common ratio, and nn is the term number is an=a1rn1a_n=a_1 r^{n-1}

Find a1a_1 and rr: a1=18r=2\begin{aligned} a_1&=\frac{1}{8} \\ r&=2 \end{aligned}

The nth term rule is:an=a1rn1=18(2)n1=2n4\begin{aligned} a_n&=a_1 r^{n-1} \\ &=\frac{1}{8} \left(2\right)^{n-1} \\ &=2^{n - 4} \end{aligned}

Recursive Formula

The formula for a geometric sequence where an1a_{n-1} is the (n-1)th term, rr is the common ratio, and n>1n>1 is an=ran1a_n=r a_{n-1}

Find rr: r=2\begin{aligned} r=2 \end{aligned}

The nth term rule is:an=ran1=(2)an1=2an1\begin{aligned} a_n&=r a_{n-1} \\ &=\left(2\right) a_{n-1} \\ &=2 a_{n-1} \end{aligned}