Series / Sequences: Determine whether the number is an element in the sequence

This sequence of numbers is an arithmetic sequence, characterized by a constantly decreasing pattern by 100. Let's start the sequence identification process:

The given terms are ..., 1800, 1700, 1600, 1500.

From this, we observe:

• The common difference $d$ is $-100$.

One way to consider sequence patterns is based on the number-ending zeros repeatedly positioned as 00. By checking common divisibility differentials or inspecting values directly, we see matching with that mode.

Now, let's inspect each of the options:

• Option 1: 1550 is not easily fitting with the visible sequence number pattern.

• Option 2: 1890 does not conform precisely as non-integral multiples divide suspect differentially.

• Option 3: 2000 was determined for matching preceding sequence confirmation tightly.

• Option 4: 2150 also does not pair properly, nor sophisticated excessive multiples derive support.

Given the earlier assessment execution matched with insight into sequence direct scale or normal concept utilization, the available choice 2000 thus coincides with the known term, maintaining steady uniformity visible in sequence pattern components. Hence:

The correct choice is $2000$.

We know that the first term of the series is 15.

From here we can easily write the entire series, until we see if we reach 1.  

15, 13, 11, 9, 7, 5, 3, 1

The number 1 is indeed an element of the series!

To solve this problem, we'll examine the pattern of how squares are arranged in each element:

• Step 1: Identify the initial pattern.
  Typically, the series or pattern of squares starts with fewer counts and increases steadily. For example, the first few elements can be manually visualized or described, for instance, as a straightforward progression such as the first element having one square, the second having three squares, the third having five squares, and so forth.
• Step 2: Recognize the pattern.
  If we observe this pattern, perhaps the difference between sequential numbers is a common increment. If we visualize this numerically, starting from the first element, it reflects 1, 3, 5... from which we can assume an odd number pattern. Next, generalize these as 1, 3, 5, 7 (the sequence of odd numbers) specifically for each sequential element.
• Step 3: Solve for the fourth element.
  Based on the established or observed sequence of odd numbers, the fourth element will match the fourth odd number, which is 7.

Therefore, by identifying this odd-number pattern in the sequence of squares, we confirm that the fourth element contains $7$ squares.

To solve this problem, we'll follow these steps:

• Step 1: Identify the given information
• Step 2: Apply the appropriate formula to determine if 22.5 is in the sequence
• Step 3: Check if the computed position is valid

Now, let's work through each step:
Step 1: We know the first term $a_1$ is 10 and the common difference $d$ is 2.5. Our target is 22.5.
Step 2: Using the formula for the $n^{th}$ term of an arithmetic sequence, we have:
$a_n = a_1 + (n-1) \cdot d$ Substituting the known values to check if 22.5 is in the sequence, we set:
$22.5 = 10 + (n-1) \cdot 2.5$ Step 3: Solve for $n$:
$22.5 - 10 = (n-1) \cdot 2.5$ $12.5 = (n-1) \cdot 2.5$ $\frac{12.5}{2.5} = n-1$ $5 = n-1$ $n = 6$ The computation shows $n$ is a positive integer (6), confirming that 22.5 is indeed the 6th term of the series.

Therefore, the solution to the problem is Yes, $6$.

To determine whether the number 29 is an element of the series, we start by recognizing that the problem involves an arithmetic sequence. In such a sequence, each term is generated by adding a constant difference to the previous term. Here, the first term $a_1$ is $1.5$, and the common difference $d$ is $3$.

The formula for the nth term of an arithmetic sequence is given by:

$a_n = a_1 + (n-1) \cdot d$

We need to check if 29 is one of the terms of this series, so we set $a_n = 29$ and solve for $n$:

$29 = 1.5 + (n-1) \cdot 3$

Subtract 1.5 from both sides:

$29 - 1.5 = (n-1) \cdot 3$

$27.5 = (n-1) \cdot 3$

Divide both sides by 3 to solve for $n - 1$:

$n-1 = \frac{27.5}{3}$

$n-1 = 9.1666\ldots$

Add 1 to find $n$:

$n = 10.1666\ldots$

Since $n$ is not an integer, the number 29 does not appear as an element in this sequence. Arithmetic sequences only have integer positions for their terms, so $n$ must be a whole number for 29 to be a term. As a result, the correct answer is:

No

To determine whether a structure can have exactly 46 squares and identify which element of the series it corresponds to, we must first recognize the growth pattern of the series of squares in the diagram.

From examining patterns in geometric series, commonly encountered shapes include arrangements forming triangles or squares. Typically, triangular numbers are related to sums of consecutive integers:

The $n$-th triangular number is given by:

Let's calculate the first few triangular numbers to understand the sequence:

• $T_1 = \frac{1 \times (1+1)}{2} = 1$
• $T_2 = \frac{2 \times (2+1)}{2} = 3$
• $T_3 = \frac{3 \times (3+1)}{2} = 6$
• $T_4 = \frac{4 \times (4+1)}{2} = 10$
• $T_5 = \frac{5 \times (5+1)}{2} = 15$
• $T_6 = \frac{6 \times (6+1)}{2} = 21$
• $T_7 = \frac{7 \times (7+1)}{2} = 28$
• $T_8 = \frac{8 \times (8+1)}{2} = 36$
• $T_9 = \frac{9 \times (9+1)}{2} = 45$
• $T_{10} = \frac{10 \times (10+1)}{2} = 55$

Now, check if 46 is among these numbers, as this would indicate a structure with exactly that many squares.

Checking the sequence above, $46$ does not appear. $45$ is the highest number before $46$, and $55$ is the nearest higher one.

Since $46$ is not found in the series of triangular numbers, it is not possible for a structure in this series to have exactly 46 squares.

Therefore, the final answer is No.

To determine if it is possible to have a structure in the series with 49 squares, consider the sequence of numbers representing perfect squares: 1, 4, 9, 16, 25, 36, 49, etc. Here, each number represents the total number of 1 cm x 1 cm squares contained in an $n \times n$ square.

The number 49 is indeed a perfect square, since $49 = 7^2$. This means that a structure made up of 49 squares can indeed be represented as a $7 \times 7$ square.

Therefore, it is possible to have a structure in the series that has 49 squares, and it corresponds to the 7th square in the sequence (since $n = 7$), and this matches answer choice Yes, $7$.

To solve this problem, we take the following steps:

• Step 1: Identify the series pattern.
  The provided structures correspond to perfect squares: 1, 4, 9, 16, etc.
• Step 2: Determine if 81 is in this series.
  Since the number of squares follows $n^2$, we calculate whether 81 is a perfect square:
  $n^2 = 81$ implies $n = \sqrt{81}$.
• Step 3: Calculate the square root.
  Calculating $\sqrt{81}$ yields $n = 9$.
• Step 4: Verify the solution.
  Since 9 is a natural number and $9^2 = 81$, the ninth structure in the series has 81 squares.

Therefore, it is possible to have a structure in the series with 81 squares, and it is the ninth element of the series, explicitly identified by computing the sequence.

Based on the choices provided, the correct answer is: Yes, $9$.

To determine whether a structure with 55 squares exists in the sequence, we'll follow these steps:

• Step 1: Determine the pattern or sequence rule from the given sequence of structures.
• Step 2: Develop a formula for the nth term of the sequence.
• Step 3: Check if there's an $n$ such that $a_n = 55$.

Let's work through each step:

Step 1: Analyze the given structures. The observed structure from the problem graphic suggests a pattern in the number of squares. For instance, visualize a small series increasing as follows:
- First structure: $1$ square (1x1)
- Second structure: $1 + 3 = 4$ squares (2x2)
- Third structure: $4 + 5 = 9$ squares (3x3)
- Fourth structure: $9 + 7 = 16$ squares (4x4), and so forth.

Step 2: Recognize the sequence is quadratic in nature (perfect square numbers). The general term structure is the nth square, meaning $a_n = n^2$. Let's verify the sequence progresses by square numbers.

Step 3: Set $n^2 = 55$ and solve for $n$. Solving $n^2 = 55$ to check if $55$ is a perfect square:

$n = \sqrt{55} \approx 7.416$.

Since $n$ should be an integer and 55 isn't a perfect square, $55$ is not a term in this sequence.

Thus, there's no element in the series possessing exactly 55 squares. The final conclusion follows accordingly based on the sequence rule verification.

Therefore, the solution to the problem is No, as there’s no such element that has exactly 55 squares in the series.

To determine the sequence that forms these structures, observe the pattern that these are square grids building up: 1-by-1, 2-by-2, 3-by-3. This creates the series $1^2, 2^2, 3^2, \ldots, n^2$. We are asked to verify if 64 can be represented in this series.

Let's solve $n^2 = 64$ to find out which structure provides 64 squares:

• Start by taking the square root of both sides: $\sqrt{n^2} = \sqrt{64}$.
• This simplifies to $n = 8$, since $\sqrt{64} = 8$.

This confirms that the structure with 64 squares is indeed possible and corresponds to the 8th element in our series.

Therefore, the element of the series with 64 squares is $8$.

To determine if the number 20 is part of the sequence given by the formula $2(2n-2)$, we proceed as follows:

• Step 1: Set up the equation, $2(2n-2) = 20$.
• Step 2: Simplify the equation:
  $2(2n-2) = 4n - 4$, thus we have $4n - 4 = 20$.
• Step 3: Solve for $n$:
  Add 4 to both sides: $4n - 4 + 4 = 20 + 4$ gives $4n = 24$.
  Divide both sides by 4: $n = \frac{24}{4} = 6$.
• Step 4: Check if $n$ is a positive integer.
  $n = 6$ is indeed a positive integer.

Since $n = 6$ is a positive integer, 20 is indeed part of the sequence, and it is the 6th term.

Therefore, the solution to the problem is Yes, 6.

To determine if the number 18 is part of the series described by the formula $a_n = 4n - 2$, follow these steps:

• Step 1: Set up the equation $4n - 2 = 18$.
• Step 2: Solve for $n$ by first adding 2 to both sides: $4n = 20$.
• Step 3: Divide both sides by 4 to isolate $n$: $n = 5$.

We have found $n = 5$, which is an integer, indicating that 18 is indeed part of the series $4n - 2$, where $n$ is a positive integer. Thus, the element 18 is in the 5th position of the series.

Therefore, the number 18 is part of the series, and it is the 5th element.

The correct answer, based on the choices provided, is: Yes, $5$.

To determine if 9 is part of the sequence given by the formula $2n + 2$, we need to solve the equation for $n$:

$2n + 2 = 9$

Subtract 2 from both sides:

$2n = 7$

Divide both sides by 2:

$n = \frac{7}{2}$

The solution $n = \frac{7}{2}$ is not an integer, meaning there is no integer $n$ such that $2n + 2 = 9$. Therefore, the number 9 is not part of the sequence.

In conclusion, the number 9 does not belong in the sequence defined by $2n + 2$, so the answer is No.

To solve this problem, we'll follow these steps:

• Step 1: Simplify the expression $n - 0.5n$.
• Step 2: Set up the equation for 5 using the simplified formula.
• Step 3: Solve for $n$.
• Step 4: Verify the solution.

Now, let's work through each step:
Step 1: Given the expression $n - 0.5n$, it simplifies to $0.5n$.
Step 2: Set up the equation $0.5n = 5$.
Step 3: Solve for $n$:
To find $n$, we multiply both sides of the equation by 2:
$n = 10$.
Step 4: Verification:
Substitute $n = 10$ back into the simplified formula: $0.5 \cdot 10 = 5$, which confirms that 5 is part of the sequence.

Therefore, the number 5 is indeed part of the sequence, and it corresponds to $n = 10$.

Yes, $10$

Determine whether the number 15 is a member of the sequence defined by the following expression:

$$ $a_n=n+5$

This can be achieved in the following way:

Our first requirement is that the value 15 does in fact exist within the sequence regardless of its position.

Hence the following expression:

$a_n=15$

We will proceed to solve the equation obtained from this requirement. Remember that n is the position of the member in the sequence (also called - the index of the member in the sequence), and therefore must be a natural number ( a positive whole number).

Let's check whether these two requirements can be met:

First, let's solve:

$\begin{cases} a_n=n+5\\ a_n=15 \end{cases}\\ \downarrow\\ 15=n+5$

We inserted $a_n$into the first equation with its value from the second equation.

We obtained an equation with one unknown for n. Let's proceed to solve it by moving terms and isolating the unknown as shown below:

$15=n+5 \\ -n=5-15\\ -n=-10 \hspace{8pt} \text{/:}(-1)\\ n=10$

In the last step we divided both sides of the equation by the coefficient of the unknown on the left side,

Thus we met the requirement that:

$a_n=15$

Leading to:

$n=10$

This is indeed a natural number, - positive and whole. Therefore we can conclude that the number 15 is indeed present in the sequence defined in the problem, and its position is 10, meaning - in mathematical notation:

$a_{10}=15$

Therefore the correct answer is answer A.

Determine whether the number 30 is a term in the sequence defined by the given general term:

$$ $a_n= 15n$,

This can be achieved in the following way:

To begin with we require that such a term exists in the sequence, regardless of its position. Hence the expression below.

$a_n=30$

Next we will proceed to solve the equation obtained from this requirement. Remember that n is the position of the term in the sequence (also called - the index of the term in the sequence) N must therefore be a natural number,( a positive whole number).

Let's check if these two requirements can both be met:

First, let's solve:

$\begin{cases} a_n= 15n \\ a_n=30 \end{cases}\\ \downarrow\\ 30=15n$

When we substituted $a_n$in the first equation with its value from the second equation,

we obtained an equation with one unknown for n. Let's solve it by moving terms and isolating the unknown as shown below:

$30=15n \\ -15n=-30 \hspace{8pt} \text{/:}(-15)\\ n=2$

In the last step we divided both sides of the equation by the coefficient of the unknown on the left side,

We thus met the requirement that:

$a_n=30$

Which is turn equals:

$n=2$

This is indeed a natural number - positive as well as whole. Therefore we can conclude that in the sequence defined in the problem by the given general term, the number 30 is indeed a term and its position is 2, meaning - in mathematical notation:

$a_{2}=30$

Therefore the correct answer is answer B.

To determine if the number 0 is part of the sequence given by $3n - 3$, we will follow these steps:

• Step 1: Set the sequence expression equal to 0, as follows: $3n - 3 = 0$.
• Step 2: Solve for $n$ by adding 3 to both sides: $3n = 3$.
• Step 3: Divide both sides by 3 to isolate $n$: $n = 1$.

Since $n = 1$ is a valid integer, it indicates that the term 0 is indeed part of the sequence. Specifically, 0 is the value of the sequence when $n = 1$.

Therefore, the number 0 is the first term in the sequence when $n = 1$. The correct answer is:

Yes, it's the first term.

The correct choice based on the options given is choice 1.

To determine if 10 is a term in the given sequence $a_n = n-4$, we need to solve the equation for $n$.

First, write down the equation based on the rule of the sequence:

• $a_n = n-4$

We want to see if $a_n = 10$, so set the equation equal to 10:

• $n-4 = 10$

Next, solve for $n$ by isolating it on one side of the equation. Add 4 to both sides:

• $n = 10 + 4$
• $n = 14$

This calculation shows that the 14th term in the sequence equals 10.

Therefore, the number 10 is indeed a term in the sequence, specifically the 14th term.

Yes, it is the 14th term.

To determine if Daniel can save exactly $29, we model his savings as an arithmetic sequence.</p> <ul> <li><strong>Step 1: Identify the given information</strong></li> </ul> <p>Initially, Daniel puts in$15, and he adds $2 daily. We are looking for the day when the total savings equals$29.

• Step 2: Set up the sequence formula

The $n$-th term of an arithmetic sequence is given by:

$a_n = a_1 + (n-1) \cdot d$

Where:

• $a_1 = 15$, the amount on the first day
• $d = 2$, the daily addition
• $a_n = 29$, the amount we want to achieve

• Step 3: Solve for $n$

Set up the equation:

$29 = 15 + (n-1) \cdot 2$

Simplifying:

$29 = 15 + 2n - 2$

$29 = 13 + 2n$

$16 = 2n$

$n = 8$

Therefore, Daniel will have exactly $29 on the eighth day.</p> <p><strong>Conclusion:</strong></p> <p>Yes, it is possible for Daniel to save exactly$29, and it will occur on the eighth day.

To solve this problem, we must visually inspect the sequence of circles depicted in the image and identify where the shape "◯" appears.

We start by examining each element in the sequence. Each element is a circular pattern with varying overlaps and radii. Our goal is to locate the shape "◯" within one of these elements.

Upon careful inspection, it's observed that the fifth element contains a circle that matches the description of ◯ in terms of its standalone characteristic, without complex overlapping or pattern. Framed next simply as a single circle, it captures the intended characteristic of "◯" as initially described.

This analysis leads us to the conclusion that the shape "◯" is indeed present within the fifth element of the sequence.

Therefore, the solution to the problem is Yes, in the fifth element.