Find the 18th term in the sequence:

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Discussion Overview

The discussion revolves around finding the 18th term in a given sequence, with participants exploring different formulations and interpretations of the sequence's general term. The scope includes mathematical reasoning and exploration of sequence definitions.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes that the 18th term can be calculated using the formula \(a_n = a_1 \cdot r^{n-1}\) and arrives at \(131072\).
  • Another participant suggests an alternative formulation \(a_n = 2^{n-2}\) and calculates \(a_{18} = 65536\).
  • Several participants reference the equation from the book, questioning the ratio used in the sequence.
  • One participant simplifies the expression to confirm that \(a_n = 2^{n-2}\) is indeed correct based on their interpretation.
  • There is a clarification on the relationship between the initial term and the common ratio, with some participants agreeing on the simplification process.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of the sequence's general term, with no clear consensus on the value of the 18th term. Multiple competing interpretations remain unresolved.

Contextual Notes

Participants reference the initial term and common ratio, but there is uncertainty regarding the definitions and assumptions underlying the sequence. The discussion does not resolve these aspects.

karush
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Find the $18th$ term in the sequence:

$$\frac{1}{2},1,2 $$
$$a_1= \frac{1}{2}\ \ \ \ n=18\ \ \ \ r=2 $$
$$a_n=a_1\cdot r^{n-1}=131072$$
 
Last edited:
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I'm thinking:

$$a_n=2^{n-2}$$

And so:

$$a_{18}=2^{16}=65536$$
 
$$a_n=a_1 \cdot r^{n-1}$$

was the eq in the book unless the ratio is wrong
 
$$a_n=\frac{1}{2}\cdot 2^{n-1}=\frac{2^{n-1}}{2}=2^{(n-1)-1}=2^{n-2}$$
 
karush said:
$$a_n=a_1 \cdot r^{n-1}$$

was the eq in the book unless the ratio is wrong

No, that is correct...I just simplified:

$$a_n=a_1r^{n-1}=2^{-1}\cdot2^{n-1}=2^{n-2}$$
 
ok got it..
 
Last edited:

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