Sequences And Series

Identify the Sequence for : The terms in are 1, 4, 8, 13, 19, 26, ..., which represents a sequence with second differences that are constant. This indicates a quadratic sequence. Let the general term of this sequence be . Using the terms:

Set up equations:

Solving these, we get:

General Term for : The n-th term of is:

Therefore, .

Expression for : Expand and simplify .

Calculate : , which can be computed directly.

Find : Substitute into the expression:

Solving for , we find:

1. Each row in the pattern contains numbers equal to the row number. The sequence follows the triangular numbers series.
2. The sum of the first n natural numbers is given by the formula: .
3. To find the row where the number 5310 is located, determine the smallest n such that: .
4. Starting from the inequality:
   
5. Multiply by 2 to clear the fraction: .
6. Rearrange into a quadratic inequality: .
7. Using the quadratic formula:
   
8. Calculate discriminant: .
9. .
10. Substitute back:
    
11. Take the positive root:
    
12. Round up to the next integer since must be a whole number: .
13. Verify: Calculate , and .
14. Since , the number 5310 is in the 103rd row.
15. Confirm the row number falls within the expected range: 103.

To solve the given problem, we need to find the value of for the series:

The sum of the series is given as 5:

Observe that each term in the series is of the form:

This resembles the formula for a telescoping series:

We can rewrite the series as:

Simplify the expression:

Equating it to the given sum:

Cross-multiply to find :

Therefore, the value of is 5.

The correct option is:

5

Define the Terms of the G.P.:
Let the first term of the G.P. be and the common ratio be . Then the terms of the G.P. are:

Thus: is the th term of the A.P., is the th term of the A.P., is the th term of the A.P.

Set Up the A.P. Terms:
Let the first term of the A.P. be and the common difference of the A.P. be .

Then:
Solving the Equations:
Using the equations, subtract the first equation from the second and the second from the third:


Substitute into :

Rearranging gives:
Solving this quadratic equation for :
Since , we discard , so .

Calculate :
The th term of the G.P. is .
We are given that the th term of the A.P. is :
Substitute and (from previous calculations):

Simplifying, we get:

Given:

The general term:

Given:

Simplifying:

Also, given:

Simplifying:

Subtracting equation (1) from equation (2):

Finally:

To solve , we recognize a geometric series involving the common ratio . Let the series be , where . We split it into two separate series: The first series is geometric with first term and common ratio . The sum of this infinite geometric series is: The second series is an arithmetico-geometric series. Its sum can be found using the formula: Plugging in, we get: Substituting back into the equation, the series sum becomes: Simplifying gives us: Thus, the value of is 9. This value falls within the given range , confirming our solution is correct.

To solve this problem, we need to analyze the given infinite series and the relationship between them:

The given information is:

• The series is equal to .
• The odd term series, .
• The even term series, .

We need to find the ratio .

Step-by-step Solution:

1. The complete series can be split into the sum of the odd series and the even series:
   • Given , hence .
2. Now, we observe the even series :
   • It can be expressed as:
   • Factoring out , we have:
     
   • The series in the parenthesis is the same format as the original series , but starting with .
   • Therefore, .
3. Using the above found values, we compute the value of :
   • We first bring them to a common denominator:
     
4. Finally, we find the ratio :

Therefore, the ratio is 15.

To solve for , we need to analyze the recurrence relation and find the form of . The sequence is defined by , , and the recurrence relation: . Assume a solution of the form in the recurrence relation: Dividing throughout by (assuming ): Rearranging gives the characteristic equation: Solving this quadratic equation using the quadratic formula with , , , yields: So, and . Thus, the general solution is: With initial conditions and : Solving these: , substitute : Thus, . Now calculate the sum: Sum of first 100 terms of a geometric series with , : This matches the correct answer choice: .

We are given the points that lie inside or on the boundary of the triangle formed by the lines , , and .
Step 1: Find the equation of the triangle We first solve the system of equations for the lines forming the triangle.
- The line is the first boundary.
- The second line intersects the first line at a point we need to find.
- The third line intersects the first two lines at another set of points.
Step 2: Solve for the points of intersection We solve these systems of linear equations to find the boundaries of the triangle and determine the limits for , the y-coordinate of the point . The values of that satisfy the condition for the points to lie inside or on the triangle will give the smallest and largest values of .
Step 3: Find the product of the smallest and largest values of Once the smallest and largest values of are identified, we compute their product. After solving, we find that the product of the smallest and largest values of is .
Final Answer: .

Solution to the Geometric Progression Problem

We are given a geometric progression (G.P.) where:

Step 1: Express the Terms of the G.P. in Terms of and

In a geometric progression, the terms are given by: - - - - - - Now, let's use the given conditions:

Step 2: Use the Given Conditions

Condition 1:

Condition 2:

Step 3: Solve the System of Equations

From equation (1): From equation (2): Dividing equation (2) by equation (1): Therefore:

Step 4: Solve for

Now substitute into equation (1):

Step 5: Calculate

Now, we calculate : Substituting and :

Conclusion

Therefore, the value of is:

To find the sum of the series up to 20 terms, we start by analyzing the pattern in the given sequence.

Let's observe the differences between successive terms:

• 3 - 1 = 2
• 11 - 3 = 8
• 25 - 11 = 14
• 45 - 25 = 20
• 71 - 45 = 26

We can observe that the differences themselves form an arithmetic sequence: 2, 8, 14, 20, 26, etc., with a common difference of 6.

The general term of this sequence can be expressed as:

Simplify as follows:

Using the formula for the sum of the first natural numbers, , we have:

So,

This is the nth term formula for the given sequence. Now, we need to find the sum of the first 20 terms:

Breaking it down:

Using the formulae:

Substituting these expressions for :

Thus, the sum is:

Therefore, the sum of the series up to 20 terms is 7240.

Step 1: To find the area , analyze the given inequalities: - describes a parabolic region. - and describe linear constraints on the values of .
Step 2: Using the above inequalities, integrate over the appropriate region to calculate the area. By calculating the bounds for and , and performing the necessary integration, you will find the area .
Step 3: After calculating the area, multiplying by 6 gives the result . Thus, the correct answer is (4).

To solve the series up to infinity, let’s break it down step by step.

The pattern in the series appears to be an arithmetic sequence numerator over factorial denominators. Specifically, each term can be expressed as:

• First term:
• Second term:
• Third term:
• Fourth term:

Observe that the sequence of the numerators follows where is the term index: .

Hence, the general term is:

Thus, the series becomes:

Now, let's evaluate the sum:

We know an essential series expansion for , which is:

Considering , it becomes:

To compute , we can derive from properties of series that involve similar factorial terms.

In fact, let's consider the function:

Differentiate with respect to :

Multiplying by and differentiating again will provide the relationship to resolve:

Simplifying our specific requirement :

Thus, the sum of the given infinite series is indeed:

We are given an Arithmetic Progression (A.P.) defined by . Two conditions are provided: a sum involving specific terms of the A.P. and the sum of the first terms being zero. We need to find the value of .

Concept Used:

The solution relies on the fundamental formulas for an Arithmetic Progression:

1. The k-th term of an A.P.: The general term is given by , where is the first term and is the common difference.

2. Sum of n terms of an A.P.: The sum of the first terms, , is given by the formula:

Alternatively, the sum of an A.P. with terms can be expressed as .

Step-by-Step Solution:

We start with the first given condition:

The terms in this summation are . This is a sequence of 12 terms which also forms an A.P. The first term of this new sequence is , and the last term is . The common difference of this sequence is .

We can find the sum of these 12 terms using the formula :

We express in terms of and : .

Now, we equate this expression with the given condition:

We solve this equation to find a relationship between and .

Dividing both sides by 132, we get:

Next, we use the second condition:

Using the sum formula for the first terms of the A.P.:

Since represents the number of terms, cannot be 0. Thus, the term in the brackets must be zero:

Final Computation & Result:

Now we substitute the relationship into the equation derived from the second condition:

We are given that , so we can divide the entire equation by :

Now, we solve for :

Hence, the value of is 11.

To solve this problem, we first need to understand and apply the properties of a geometric progression (G.P.). In a G.P., each term is obtained by multiplying the previous term by a fixed, non-zero number known as the common ratio, .

Given that is a G.P. of positive numbers, the terms are as follows:

• (first term)

From the problem, we have two key pieces of information:

We'll use these pieces of information to find and finally compute .

Step 1: Use the first piece of information:

Therefore, we can write:

Step 2: Use the second piece of information:

Step 3: Find the relation between using the values obtained:

From Step 1, . Let's use it in the equation from Step 2:

Let , hence:

With :

Solving this equation:

Hence, the common ratio , and from the equation , we have:

Step 4: Now, calculate :

Sum, .

Finally, compute :

Thus, the value of is 129.