Geometric Sequences
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Question 1 of 3
1. Question
Given the sequence `3,6,12,24…`, find:`(i)` The `8th` term`(ii)` If `1536` is part of the sequenceFor part `ii`, write `Y` for yes and `N` for no-
`(i)` `U_8=` (384)`(ii)` (Y, y)
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General Rule of a Geometric Sequence
$$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$$Common Ratio Formula
$$\color{#00880A}{r}=\frac{U_2}{U_1}=\frac{U_3}{U_2}$$`(i)` Finding the `8th` termFirst, solve for the value of `r`.$$\color{#00880A}{r}$$ `=` $$\frac{U_2}{U_1}$$ `=` $$\frac{6}{3}$$ Substitute the first and second term `=` `2` Next, substitute the known values to the general rule`\text(Number of terms)``[n]` `=` `8` `\text(First term)``[a]` `=` `3` `\text(Common Ratio)``[r]` `=` `2` $$U_{\color{#9a00c7}{n}}$$ `=` $$\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$$ $$U_{\color{#9a00c7}{8}}$$ `=` $$\color{#e65021}{3}\cdot\color{#00880A}{2}^{\color{#9a00c7}{8}-1}$$ Substitute known values `=` $$3\cdot(2^7)$$ Evaluate `=` $$3\cdot128$$ `=` `384` `(ii)` Finding if `1536` is part of the sequenceNext, substitute the known values to the general rule and solve for `n``\text(Nth term)``[U_n]` `=` `1536` `\text(First term)``[a]` `=` `3` `\text(Common Ratio)``[r]` `=` `2` $$U_{\color{#9a00c7}{n}}$$ `=` $$\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$$ $$1536$$ `=` $$\color{#e65021}{3}\cdot\color{#00880A}{2}^{\color{#9a00c7}{n}-1}$$ Substitute known values `1536``divide3` `=` `3(2^(n-1))``divide3` Divide both sides by `3` `512` `=` `2^(n-1)` `2^9` `=` `2^(n-1)` `512=2^9` `9` `=` `n-1` Equate the exponents `9` `+1` `=` `n-1` `+1` Add `1` to both sides `10` `=` `n` `n` `=` `10` `1536` is the value of the `10th` term. Therefore, it is a part of the sequence`(i) U_8=384``(ii) \text(Yes)` -
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Question 2 of 3
2. Question
Find the value of `n``6,3,1 1/2…3/(128)`- `n=` (9)
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General Rule of a Geometric Sequence
$$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$$Common Ratio Formula
$$\color{#00880A}{r}=\frac{U_2}{U_1}=\frac{U_3}{U_2}$$First, solve for the value of `r`.$$\color{#00880A}{r}$$ `=` $$\frac{U_2}{U_1}$$ `=` $$\frac{3}{6}$$ Substitute the first and second term `=` `1/2` Next, substitute the known values to the general rule and solve for `n``\text(Nth term)``[U_n]` `=` `3/128` `\text(First term)``[a]` `=` `6` `\text(Common Ratio)``[r]` `=` `1/2` $$U_{\color{#9a00c7}{n}}$$ `=` $$\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$$ $$\frac{3}{128}$$ `=` $$\color{#e65021}{6}\cdot\color{#00880A}{\frac{1}{2}}^{\color{#9a00c7}{n}-1}$$ Substitute known values `3/128``times1/6` `=` `[6(1/2)^(n-1)]``times1/6` Multiply both sides by `1/6` `1/256` `=` `(1/2)^(n-1)` `6/6=1` `1/(2^8)` `=` `1/(2^(n-1))` `256=2^8` `8` `=` `n-1` Equate the exponents `8` `+1` `=` `n-1` `+1` Add `1` to both sides `9` `=` `n` `n` `=` `9` `n=9` -
Question 3 of 3
3. Question
Find the `8th` term`2/7+3/7+9/14…`Hint
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General Rule of a Geometric Sequence
$$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$$Common Ratio Formula
$$\color{#00880A}{r}=\frac{U_2}{U_1}=\frac{U_3}{U_2}$$First, solve for the value of `r`.$$\color{#00880A}{r}$$ `=` $$\frac{U_2}{U_1}$$ `=` $$\frac{\frac{3}{7}}{\frac{2}{7}}$$ Substitute the first and second term `=` `3/2` Next, substitute the known values to the general rule`\text(Number of terms)``[n]` `=` `8` `\text(First term)``[a]` `=` `2/7` `\text(Common Ratio)``[r]` `=` `3/2` $$U_{\color{#9a00c7}{n}}$$ `=` $$\color{#e65021}{a}\color{#00880A}{r}^{\color{#9a00c7}{n}-1}$$ $$U_{\color{#9a00c7}{8}}$$ `=` $$\color{#e65021}{\frac{2}{7}}\cdot\color{#00880A}{\frac{3}{2}}^{\color{#9a00c7}{8}-1}$$ Substitute known values `=` $$\frac{2}{7}\cdot\left(\frac{3}{2}\right)^2$$ Evaluate `=` $$\frac{2}{7}\cdot\frac{2187}{128}$$ `=` `2187/448` `=` `4 395/448` Simplify `U_8=4 395/448`