Tag: Geometric Sequence

A geometric sequence has the general formula $\text{U}_n = ar^{n-1}.$ Notice that the formula of a geometric sequence consists of only one term (there is no addition or subtraction).
Option A: $\text{U}_n = 4^n-5$
The sequence formula contains two terms (there is subtraction), so it is clearly not a geometric sequence.
Option B: $\text{U}_n = 2^n \cdot n^{-2}$
This sequence formula is not a geometric sequence because the variable $n$ appears in different positions, namely as an exponent and as a base.
Option C: $\text{U}_n = 2n^3-1$
The sequence formula contains two terms (there is subtraction), so it is clearly not a geometric sequence.
Option D: $\text{U}_n = n^3 \cdot 2^{-n}$
This sequence formula is not a geometric sequence because the variable $n$ appears in different positions, namely as an exponent and as a base.
Option E: $\text{U}_n = 2^{n+1} \cdot 3^{-n}$
Notice that the sequence formula above can be rewritten as
$\text{U}_n = 2^{n} \cdot 2^1 \cdot \dfrac{1}{3^n} = 2\left(\dfrac23\right)^n.$
The final form of the formula shows that this is a geometric sequence with first term $a = 2$ and common ratio $r = \dfrac23.$
(Answer E)

Given $a = 24$ and $\text{U}_3 = \dfrac{8}{3}$. The first step is to determine the common ratio of this geometric sequence.
$\begin{aligned} \text{U}_n & = ar^{n-1} \\ \text{U}_3 & = \dfrac{8}{3} = 24r^{3-1} \\ \dfrac{8}{3} & = 24r^2 \\ r^2 & = \dfrac{8}{3} \cdot \dfrac{1}{24} \\ r^2 & = \dfrac{1}{9} \\ r & = \dfrac{1}{3} \end{aligned}$
Therefore, we obtain
$\begin{aligned} \text{U}_5 & = ar^4 \\ & = 24\left(\dfrac{1}{3}\right)^4 \\ & = 24 \cdot \dfrac{1}{81} = \dfrac{8}{27}. \end{aligned}$
Thus, the fifth term of the geometric sequence is $\boxed{\dfrac{8}{27}}.$
(Answer D)

Given:
$\begin{aligned} \text{U}_1 & = a = \dfrac{5}{2} \\ \text{U}_4 & = 20. \end{aligned}$
The first step is to find the common ratio of this geometric sequence.
Notice that
$\begin{aligned} \text{U}_4 & = 20 \\ ar^3 & = 20 \\ \dfrac{5}{2}r^3 & = 20 \\ r^3 & = 20 \times \dfrac{2}{5} \\ r^3 & = 8 \\ r & = \sqrt[3]{8} = 2. \end{aligned}$
Next, determine the sixth term.
$\text{U}_6 = ar^5 = \dfrac{5}{2} \times 2^5 = 80$
Therefore, the sixth term of the sequence is $\boxed{80}.$
(Answer A)

Given $\text{U}_5 = 162$ and $\text{U}_2 =-6.$ By comparing the terms, we obtain
$\begin{aligned} \dfrac{\text{U}_5}{\text{U}_2} & = \dfrac{162}{-6} \\ \dfrac{\cancel{a}r^4}{\cancel{a}r} & =-27 \\ r^3 & =-27 \\ r & =-3. \end{aligned}$
Therefore, the common ratio of the geometric sequence is $\boxed{-3}.$
(Answer A)

Given $a = 16$ and $\text{U}_4 = 2$. The first step is to determine the common ratio.
$\begin{aligned} \text{U}_4 & = 2 \\ ar^3 & = 2 \\ 16r^3 & = 2 \\ r^3 & = \dfrac{1}{8} \\ r & = \dfrac{1}{2} \end{aligned}$
Using the formula for the sum of the first $n$ terms of a geometric sequence:
$\boxed{S_n = \dfrac{a(1-r^n)}{1-r}},$
we obtain
$\begin{aligned} S_6 & = \dfrac{16\left(1-\left(\dfrac{1}{2}\right)^6 \right)}{1- \dfrac{1}{2}} \\ & = \dfrac{16\left(1-\dfrac{1}{64}\right)}{\dfrac{1}{2}} \\ & = 16 \cdot \dfrac{63}{64} \cdot \dfrac{2}{1} \\ & = \dfrac{63}{4} \cdot 2 = 31,\!5. \end{aligned}$
Therefore, the sum of the first $6$ terms of the geometric sequence is $\boxed{31,\!5}.$
(Answer B)

In a geometric sequence, the following relationship holds:
$\boxed{\text{U}_2^2 = \text{U}_1 \cdot \text{U}_3}.$
Thus, we obtain
$$\begin{aligned} (x- 4)^2 & = (2x-5)(-3x+10) \\ x^2-8x+16 & =-6x^2 + 35x-50 \\ 7x^2-43x + 66 & = 0 \\ (7x-22)(x-3) & = 0. \end{aligned}$$We obtain $x = \dfrac{22}{7}$ or $x = 3.$ Since the required value of $x$ is an integer, the value taken is $\boxed{3}.$
(Answer A)

In a geometric sequence, the formula for the $n$-th term is given by $\text{U}_n = ar^{n-1}$ where $a$ is the first term. Therefore,
$\begin{aligned} \dfrac{x} {y} & = \dfrac{\text{U}_3-\text{U}_6}{\text{U}_2- \text{U}_4} \\ & = \dfrac{ar^2-ar^5}{ar-ar^3} \\ & = \dfrac{\cancel{ar} (r- r^4)}{\cancel{ar} (1-r^2)} \\ & = \dfrac{r-r^4}{1-r^2} \\ & = \dfrac{\cancel{(1- r)} (r+r^2+r^3)} {\cancel{(1-r)}(1+r)} \\ & = \dfrac{r+r^2+r^3}{1+r}. \end{aligned}$
Therefore, the value of $\boxed{\dfrac{x} {y} = \dfrac{r^3+r^2+r}{r+1}}.$
(Answer C)

Given $\dfrac{\text{U}_6}{\text{U}_8} = 3$, we obtain
$\begin{aligned} \dfrac{\text{U}_6}{\text{U}_8} & = 3 \\ \dfrac{\cancel{a} r^5}{\cancel{a}r^7} & = 3 \\ r^{-2} & = 3 \\ r^2 & = \dfrac13 \\ (r^2)^4 & = \left(\dfrac13\right)^4 \\ r^8 & = \dfrac{1}{81}. \end{aligned}$
It is also given that $\text{U}_2 \cdot \text{U}_8 = \dfrac13$, so we obtain
$\begin{aligned} \text{U}_2 \cdot \text{U}_8 & = \dfrac13 \\ (ar) (ar^7) & = \dfrac13 \\ a^2r^8 & = \dfrac13 \\ \text{Substitute}~r^8 & = \dfrac{1}{81} \\ a^2\left(\dfrac{1}{81}\right) & = \dfrac13 \\ a^2 & = \dfrac13 \cdot 81 \\ a^2 & = 27 \\ a & = \sqrt{27} = 3\sqrt{3}. \end{aligned}$
Since $r^2 = \dfrac13$, it follows that $r = \sqrt{\dfrac13} = \dfrac{1}{3}\sqrt{3}$.
Therefore,
$\begin{aligned} U_{10} & = ar^9 = ar^8 \cdot r \\ & = (3\sqrt{3}) \left(\dfrac{1}{81}\right) \left(\dfrac{1}{3}\sqrt{3}\right) = \dfrac{1}{27}. \end{aligned}$
Therefore, the value of $\boxed{\text{U}_{10} = \dfrac{1}{27}}.$
(Answer A)

It is known that $\text{U}_6- \text{U}_4 = 4$, so we obtain
$\begin{aligned} \text{U}_6-\text{U}_4 & = 4 \\ ar^5- ar^3 & = 4 \\ ar^2(r^3-r) & = 4 \\ ar^2 & = \dfrac{4}{r^3-r}. \end{aligned}$
It is known that $\text{U}_4-\text{U}_3 = \dfrac23$, so we obtain
$\begin{aligned} \text{U}_4-\text{U}_3 & = \dfrac23 \\ ar^3-ar^2 & = \dfrac23 \\ ar^2(r-1) & = \dfrac23 \\ \dfrac{4}{r^3-r} \cdot (r-1) & = \dfrac23 \\ \dfrac{4}{\cancel{(r-1)}(r^2+r)} \cdot \cancel{(r-1)} & = \dfrac23 \\ \dfrac{4}{r^2+r} & = \dfrac23 \\ r^2 + r & = 4 \cdot \dfrac32 \\ r^2+r & = 6 \\ r^2+r-6 & = 0 \\ (r + 3)(r-2) & = 0. \end{aligned}$
We obtain $r =-3$ or $r = 2$. Since the common ratio is positive, we take $r = 2$.
Therefore,
$\begin{aligned} \text{U}_6-\text{U}_4 & = 4 \\ ar^5-ar^3 & = 4 \\ a(r^5-r^3) & = 4 \\ a & = \dfrac{4}{r^5-r^3} \\ \text{Substitute}~r & = 2 \\ a & = \dfrac{4}{2^5-2^3} = \dfrac{4}{24} = \dfrac{1}{6}. \end{aligned}$
Thus,
$\boxed{ \text{U}_5 = ar^4 = \dfrac16(2)^4 = \dfrac83}.$
(Answer B)

The common ratio of the geometric sequence can be determined by dividing the $(n+1)$-th term by the $n$-th term. For example, divide the second term by the first term.
$r= \dfrac{\text{U}_{n+1}} {\text{U}_n} = \dfrac{\text{U}_2}{\text{U}_1} = \dfrac{4^2}{4^1} = 4.$
From this, we also obtain $\text{U}_1 = a = 4.$
Using the formula for the sum of the first $n$ terms of a geometric sequence, we obtain
$\begin{aligned} \text{S}_n & = \dfrac{a(r^n-1)}{r-1} \\ & = \dfrac{4(4^n-1)} {4-1} \\ & = \dfrac{4}{3}(4^n-1) \\ & = \dfrac{1}{3}(4^{n+1}-4). \end{aligned}$
Therefore, the sum of the first $n$ terms of the geometric sequence is $\boxed{\text{S}_n = \dfrac{1}{3}(4^{n+1}-4)}.$
(Answer A)

Given
$\begin{aligned} \text{U}_1 & = a = p^{-2} \\ \text{U}_2 & = p^{2x} \\ \text{U}_{10} & = p^{88}. \end{aligned}$
The common ratio of the geometric series is $r = \dfrac{p^{2x}}{p^{-2}} = p^{2x+2}.$
Since the tenth term is $p^{88}$, using the formula for the $n$-th term of a geometric sequence, we obtain
$\begin{aligned} \text{U}_{10} & = p^{88} \\ ar^9 & = p^{88} \\ p^{-2} \cdot p^{18x+18} & = p^{88} \\ p^{18x+16} & = p^{88} \\ 18x+16 & = 88 \\ 18x & = 72 \\ \therefore x & = 4. \end{aligned}$
Therefore, $\boxed{x=4}.$
(Answer E)

Mathematical Method:
Given the geometric series:
$16-8+4-2+\cdots$
Its first term is $a = 16$. The common ratio of the geometric sequence corresponding to the series is $r =-\dfrac12.$
Therefore, the sum of its first $10$ terms is given by
$\begin{aligned} \text{S}_{n} & = \dfrac{a(1-r^n)}{1-r} \\ \text{S}_{10} & = \dfrac{16\left(1-\left(-\dfrac12\right)^{10}\right)}{1-\left(-\dfrac12\right)} \\ & = \dfrac{16\left(1-\dfrac{1}{1024}\right)}{1+\dfrac12} \\ & = \dfrac{\dfrac{1023}{64}}{\dfrac32} \\ & = \dfrac{\cancelto{341}{1023}}{\cancelto{32}{64}} \times \dfrac{\cancel{2}}{\cancel{3}} = \dfrac{341}{32}. \end{aligned}$
Therefore, the sum of the first $10$ terms of the geometric series is $\boxed{\dfrac{341}{32}}.$
Manual Method:
The manual method means calculating the terms one by one, as is commonly done in elementary school. It may actually be more efficient for this problem because only the first $10$ terms are involved.
$$\begin{aligned} & (16-8+4-2+1)-\dfrac12+\dfrac14-\dfrac18+\dfrac{1}{16}-\dfrac{1}{32} \\ & = 11+\dfrac{-16+8-4+2-1}{32} \\ & = \dfrac{352}{32}-\dfrac{11}{32} = \dfrac{341}{32} \end{aligned}$$(Answer D)

The odd-numbered terms in the geometric sequence are $6, y, 54.$
Since the square of the second term equals the product of the first and third terms (a property of geometric sequences), we obtain
$\begin{aligned} y^2 & = 6 \cdot 54 \\ y^2 & = 6^2 \cdot 3^2 \\ y & = 6 \cdot 3 = 18. \end{aligned}$
Using the same principle, consider the geometric sequence: $6, x, 18.$
$\begin{aligned} x^2 & = 6 \cdot 18 \\ x^2 & = 6^2 \cdot 3 \\ x & = 6\sqrt3 \end{aligned}$
Next, we also obtain $z = 18\sqrt3.$
Therefore, the value of
$\begin{aligned} \dfrac{xz}{y} & = \dfrac{6\sqrt3 \cdot \cancel{18}\sqrt3}{\cancel{18}} \\ & = 6(\sqrt3)^2 = 18. \end{aligned}$
(Answer C)

Since $x, y, z$ form a geometric sequence, the property that the square of the middle term equals the product of the first and third terms applies, written as $y^2 = xz.$
Next, using logarithmic properties, we obtain
$\begin{aligned} \dfrac{1}{^x \log y}+\dfrac{1}{^z \log y} & = \! ^y \log x + \! ^y \log z \\ & = \! ^y \log \color{red}{xz} \\ \text{Substitute}~y^2 & = xz \\ \dfrac{1}{^x \log y}+\dfrac{1}{^z \log y} & = \! ^y \log y^2 = 2. \end{aligned}$
Therefore, the value of $\boxed{\dfrac{1}{^x \log y}+\dfrac{1}{^z \log y} = 2}.$
(Answer A)

The geometric series has first term $a = \dfrac34$ and common ratio $r = 2.$
Based on the geometric series formula, we obtain
$\begin{aligned} \text{S}_n & = \dfrac{a(r^n-1)}{r-1} \\ \dfrac{765}{4} & = \dfrac{\dfrac34\left(2^n-1\right)}{2-1} \\ \dfrac{765}{\cancel{4}} \cdot \dfrac{\cancel{4}}{3} & = 2^n-1 \\ 255 & = 2^n-1 \\ 256 & = 2^n \\ n & = 8. \end{aligned}$
Next, we determine $\text{U}_8 = P$ using the geometric sequence formula $\text{U}_n = ar^{n-1}.$
$P = \text{U}_8 = \dfrac34(2)^{8-1} = \dfrac{3}{4}(2)^7 = 96 $
Therefore, the value of $\boxed{P = 96}.$
(Answer D)

Suppose the three numbers form a geometric sequence written as: $\dfrac{a}{r}, a, ar$.
Their product is:
$\begin{aligned} \dfrac{a}{\cancel{r}} \times a \times a\cancel{r} & = 216 \\ a^3 & = 6^3 \\ a & = 6. \end{aligned}$
Now, the geometric sequence can be written as: $\dfrac{6}{r}, 6, 6r$.
Sum:
$\begin{aligned} \dfrac{6}{r}+6+6r & = 26 \\ \dfrac{6}{r}+6r & = 20 \\ 6\left(\dfrac{1}{r}+r\right) & = 20 \\ \dfrac{1}{r}+r & = \dfrac{20}{6} = \dfrac{10}{3} \\ r + \dfrac{1}{r} & = 3 + \dfrac{1}{3}. \end{aligned}$
The last equation shows that $r = 3$ or $r = \dfrac13.$
Determining the geometric sequence:
For $a = 6$ and $r = 3$, the geometric sequence is $2, 6, 18.$
For $a = 6$ and $r = \dfrac13,$ the geometric sequence is: $18, 6, 2.$
The sum of the first and third numbers is the same even if their positions are swapped, namely $\boxed{2 + 18 = 18 + 2 = 20}.$
(Answer D)

A geometric sequence has the following formula for the $n$-th term.
$\boxed{\text{U}_n = ar^{n-1}}$
where $a$ is the first term and $r$ is the common ratio.
Thus,
$\begin{aligned} \dfrac{\text{U}_4 + \text{U}_3}{\text{U}_2+\text{U}_1} & = 9 \\ \dfrac{ar^3 + ar^2}{ar + a} \\ \dfrac{\cancel{a}(r^3+r^2)}{\cancel{a}(r + 1)} & = 9 \\ \dfrac{r^3+r^2}{r+1} & = 9 \\ \dfrac{r^2\cancel{(r+1)}}{\cancel{r+1}} & = 9 \\ r^2 & = 9 \\ r & = \pm 3. \end{aligned}$
Since all terms are positive, we take $r = 3.$
Next,
$$\begin{aligned} \dfrac{\text{U}_2 + \text{U}_3}{\text{U}_1+\text{U}_2+\text{U}_3 + \text{U}_4} & = \dfrac{ar + ar^2}{a + ar + ar^2 + ar^3} \\ & = \dfrac{\cancel{a}(r+r^2)}{\cancel{a}(1+r+r^2+r^3)} \\ & = \dfrac{r+r^2}{1+r+r^2+r^3} \\ & = \dfrac{3 + 3^2}{1+3+3^2+3^3} \\ & = \dfrac{12}{40} = \dfrac{3}{10}. \end{aligned}$$Therefore, the value of $\boxed{\dfrac{\text{U}_2 + \text{U}_3}{\text{U}_1+\text{U}_2+\text{U}_3 + \text{U}_4} = \dfrac{3}{10}}.$
(Answer A)

$2^0+2^1+2^2+\cdots+2^n$ is a geometric series with first term $a = 2^0 = 1$ and common ratio $r = 2$, and it has $n+1$ terms, so its sum is equal to
$\begin{aligned} \text{S}_{n+1} & = \dfrac{a(r^{n+1}-1)}{r-1} \\ & = \dfrac{1(2^{n+1}-1)}{2-1} \\ & = 2^{n+1}-1. \end{aligned}$
Therefore, we write $2^{n+1}-1 \to 2.004$, or $2^{n+1} \to 2^{11} = 2.048$, so the value of $\boxed{n=10}.$

In Figure 1, the area of the square is $L_1 = x^2$ square units.
The side length of the square in Figure 2 can be determined using the Pythagorean theorem, namely
$\begin{aligned} s & = \sqrt{\left(\dfrac12x\right)^2 + \left(\dfrac12x\right)^2} \\ & = \sqrt{\dfrac12x^2} \\ & = x\sqrt{\dfrac12}. \end{aligned}$
The area of the square in Figure 2 is $L_2 = \left(x\sqrt{\dfrac12}\right)^2 = \dfrac12x^2$, which is half the area of the square in Figure 1.
By analogy, we obtain that the areas of the squares form a geometric sequence with $a = x^2$ and $r = \dfrac{1}{2}$, so
$\begin{aligned} & U_{n} = ar^{n-1} \\ & U_{1.000} = x^2\left(\dfrac{1}{2}\right)^{1.000-1} = x^2\left(\dfrac{1}{2}\right)^{999}. \end{aligned}$
Therefore, the shaded area in the $1,000$-th pattern is $x^2\left(\dfrac{1}{2}\right)^{999}$ square units.