Find the 18th term in the sequence:

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SUMMARY

The 18th term in the sequence defined by the first term \( a_1 = \frac{1}{2} \) and a common ratio \( r = 2 \) is calculated using the formula \( a_n = a_1 \cdot r^{n-1} \). For \( n = 18 \), this results in \( a_{18} = \frac{1}{2} \cdot 2^{17} = 2^{16} = 65536 \). The sequence can also be expressed as \( a_n = 2^{n-2} \), confirming the correctness of the calculations. The discussion clarifies the application of geometric sequence formulas in determining specific terms.

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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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