Given:

The cost of digging the first metre of the well = ₹ $150$.

The cost of digging each subsequent metre increases by a fixed amount of ₹ $50$.

To Find:

Determine whether the sequence of costs for digging $1, 2, 3, \dots, n$ metres forms an Arithmetic Progression (AP) and provide the justification.

Step 1: Defining the sequence of costs

Let $a_n$ represent the cost of digging the well up to $n$ metres.

According to the problem statement:

Cost for the first metre ($a_1$) = ₹ $150$.

Cost for the first two metres ($a_2$) = Cost of first metre + Cost of second metre = $150 + 50 = ₹ 200$.

Cost for the first three metres ($a_3$) = Cost of first two metres + Cost of third metre = $200 + 50 = ₹ 250$.

Cost for the first four metres ($a_4$) = Cost of first three metres + Cost of fourth metre = $250 + 50 = ₹ 300$.

Step 2: Analyzing the sequence

The list of numbers representing the costs is: $150, 200, 250, 300, \dots$

An Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. This constant is known as the common difference ($d$).

Let us calculate the differences between consecutive terms:

$d_1 = a_2 - a_1 = 200 - 150 = 50$

$d_2 = a_3 - a_2 = 250 - 200 = 50$

$d_3 = a_4 - a_3 = 300 - 250 = 50$

Step 3: Conclusion based on the definition of an AP

[Since the difference between any two consecutive terms is constant, i.e., $a_{n} - a_{n-1} = 50$ for all $n > 1$].

Because the common difference $d = 50$ is constant throughout the sequence, the situation represents an Arithmetic Progression.

Final Answer: Yes, the list of numbers forms an Arithmetic Progression because each term increases by a constant value of ₹ 50, which serves as the common difference.

Given: A sequence of numbers: $0, -4, -8, -12, \dots$

To find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference ($d$).
3. Write the next three terms of the sequence.

Step 1: Definition of an Arithmetic Progression
A sequence $a_1, a_2, a_3, \dots, a_n$ is said to be an Arithmetic Progression if the difference between consecutive terms remains constant. This constant difference is called the common difference ($d$), defined as:
$d = a_{n} - a_{n-1}$ for all $n > 1$.

Step 2: Calculating differences between consecutive terms
Let the given terms be:
$a_1 = 0$
$a_2 = -4$
$a_3 = -8$
$a_4 = -12$

Calculate the differences:
Difference 1 ($d_1$): $a_2 - a_1 = -4 - 0 = -4$
Difference 2 ($d_2$): $a_3 - a_2 = -8 - (-4) = -8 + 4 = -4$
Difference 3 ($d_3$): $a_4 - a_3 = -12 - (-8) = -12 + 8 = -4$

Step 3: Verification of AP
[Since $d_1 = d_2 = d_3 = -4$, the difference between consecutive terms is constant.]
Therefore, the given sequence forms an Arithmetic Progression with a common difference $d = -4$.

Step 4: Finding the next three terms
To find the next terms, we add the common difference ($d = -4$) to the last known term ($a_4 = -12$).
Let the next three terms be $a_5, a_6,$ and $a_7$.

Calculation for $a_5$:
$a_5 = a_4 + d = -12 + (-4) = -16$

Calculation for $a_6$:
$a_6 = a_5 + d = -16 + (-4) = -20$

Calculation for $a_7$:
$a_7 = a_6 + d = -20 + (-4) = -24$

Final Answer:
The sequence forms an AP.
The common difference $d = -4$.
The next three terms are -16, -20, -24.

Given:

The first term of the Arithmetic Progression (AP), denoted by $a = -2$.

The common difference of the AP, denoted by $d = 0$.

To Find:

The first four terms of the Arithmetic Progression ($a_1, a_2, a_3, a_4$).

Step 1: Understanding the definition of an Arithmetic Progression

An Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference ($d$).

The general form of an AP is given by: $a, a+d, a+2d, a+3d, \dots$

Where:

• $a_1 = a$
• $a_2 = a + d$
• $a_3 = a + 2d$
• $a_4 = a + 3d$

Step 2: Calculating the first four terms

We substitute the given values $a = -2$ and $d = 0$ into the general formulas defined in Step 1.

Calculation for the first term ($a_1$):

$a_1 = a = -2$

Calculation for the second term ($a_2$):

$a_2 = a + d$

$a_2 = -2 + 0$

$a_2 = -2$

Calculation for the third term ($a_3$):

$a_3 = a + 2d$

$a_3 = -2 + 2(0)$

$a_3 = -2 + 0$

$a_3 = -2$

Calculation for the fourth term ($a_4$):

$a_4 = a + 3d$

$a_4 = -2 + 3(0)$

$a_4 = -2 + 0$

$a_4 = -2$

Step 3: Conclusion

Since the common difference $d$ is $0$, every term in the sequence remains identical to the first term $a$.

Final Answer: The first four terms of the AP are -2, -2, -2, -2.

Given:

Principal amount ($P$) = ₹ $10000$

Rate of interest ($r$) = $8\%$ per annum

The interest is compounded annually.

To Find:

Whether the sequence of amounts at the end of each year forms an Arithmetic Progression (AP).

Step 1: Understanding the Formula for Compound Interest

The amount ($A$) after $n$ years with compound interest is given by the formula:

$A = P \left(1 + \frac{r}{100}\right)^n$

Where:

• $P = 10000$
• $r = 8$
• $n$ is the number of years ($n = 1, 2, 3, \dots$)

Step 2: Calculating the amount for consecutive years

Let $a_1, a_2, a_3, \dots$ be the amount in the account at the end of the 1st, 2nd, and 3rd year respectively.

For $n = 1$ (Amount at the end of 1st year):

$a_1 = 10000 \left(1 + \frac{8}{100}\right)^1 = 10000(1.08) = 10800$

For $n = 2$ (Amount at the end of 2nd year):

$a_2 = 10000 \left(1 + \frac{8}{100}\right)^2 = 10000(1.08)^2 = 10000(1.1664) = 11664$

For $n = 3$ (Amount at the end of 3rd year):

$a_3 = 10000 \left(1 + \frac{8}{100}\right)^3 = 10000(1.259712) = 12597.12$

Step 3: Checking for Arithmetic Progression

A sequence is an Arithmetic Progression if the difference between consecutive terms is constant (i.e., $a_{n+1} - a_n = d$, where $d$ is the common difference).

Calculate the first difference ($d_1$):

$d_1 = a_2 - a_1 = 11664 - 10800 = 864$

Calculate the second difference ($d_2$):

$d_2 = a_3 - a_2 = 12597.12 - 11664 = 933.12$

Step 4: Conclusion

Since $d_1 \neq d_2$ ($864 \neq 933.12$), the difference between consecutive terms is not constant.

[By definition, a sequence is an AP if and only if the common difference is constant for all terms.]

Final Answer: The list of numbers does not form an Arithmetic Progression because the difference between consecutive terms is not constant.

Given: A sequence of numbers: $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$

To Find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and the next three terms.

Step 1: Simplify the given terms
To analyze the sequence, we first simplify each radical term by extracting perfect square factors:
Term 1 ($a_1$) = $\sqrt{2} = 1\sqrt{2}$
Term 2 ($a_2$) = $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$
Term 3 ($a_3$) = $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$
Term 4 ($a_4$) = $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$
The sequence is: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$

Step 2: Verify if the sequence is an AP
A sequence is an AP if the difference between consecutive terms is constant. This constant is called the common difference ($d$).
Calculate the differences between consecutive terms:
$d_1 = a_2 - a_1 = 2\sqrt{2} - 1\sqrt{2} = (2-1)\sqrt{2} = \sqrt{2}$
$d_2 = a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = (3-2)\sqrt{2} = \sqrt{2}$
$d_3 = a_4 - a_3 = 4\sqrt{2} - 3\sqrt{2} = (4-3)\sqrt{2} = \sqrt{2}$

[Since $d_1 = d_2 = d_3 = \sqrt{2}$, the difference between consecutive terms is constant.]
Therefore, the given sequence is an Arithmetic Progression with common difference $d = \sqrt{2}$.

Step 3: Find the next three terms
The sequence currently has four terms. We need to find the 5th, 6th, and 7th terms using the formula $a_n = a_{n-1} + d$.
Term 5 ($a_5$) = $a_4 + d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50}$
Term 6 ($a_6$) = $a_5 + d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72}$
Term 7 ($a_7$) = $a_6 + d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98}$

Final Answer: The sequence forms an AP with common difference $d = \sqrt{2}$. The next three terms are $\sqrt{50}, \sqrt{72}, \text{ and } \sqrt{98}$.

Given: An Arithmetic Progression (AP) sequence: $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, . . .$

To find: The first term ($a$) and the common difference ($d$) of the given AP.

Step 1: Identifying the First Term

By definition, an Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. The first term is denoted by $a$ or $a_1$.

Looking at the sequence: $a_1 = \frac{1}{3}, a_2 = \frac{5}{3}, a_3 = \frac{9}{3}, a_4 = \frac{13}{3}$.

Therefore, the first term $a = \frac{1}{3}$.

Step 2: Defining the Common Difference

The common difference ($d$) of an AP is defined as the difference between any two consecutive terms. Mathematically, it is expressed as:

$d = a_{n} - a_{n-1}$ for $n > 1$.

Specifically, for this sequence:

$d = a_2 - a_1$

Step 3: Calculating the Common Difference

Substitute the values of $a_2$ and $a_1$ into the formula:

$d = \frac{5}{3} - \frac{1}{3}$

[Since the denominators are the same, we subtract the numerators directly]

$d = \frac{5 - 1}{3}$

$d = \frac{4}{3}$

Step 4: Verification (Optional but Recommended)

To ensure the sequence is indeed an AP, we verify the difference between the third and second terms:

$d = a_3 - a_2 = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}$

[Since the difference is consistent, the value of $d$ is confirmed as $\frac{4}{3}$]

Final Answer: The first term $a = \frac{1}{3}$ and the common difference $d = \frac{4}{3}$.

Given: A sequence of numbers: $2, 4, 8, 16, \dots$

To Find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference $d$ and the next three terms.

Step 1: Definition and Condition for an Arithmetic Progression
A sequence $a_1, a_2, a_3, a_4, \dots$ is said to be an Arithmetic Progression if the difference between consecutive terms remains constant. This constant difference is called the common difference ($d$).
Mathematically, the condition is:
$a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = d$

Step 2: Calculating the differences between consecutive terms
Let the given terms be:
$a_1 = 2$
$a_2 = 4$
$a_3 = 8$
$a_4 = 16$

Now, calculate the differences:
Difference 1 ($d_1$): $a_2 - a_1 = 4 - 2 = 2$
Difference 2 ($d_2$): $a_3 - a_2 = 8 - 4 = 4$
Difference 3 ($d_3$): $a_4 - a_3 = 16 - 8 = 8$

Step 3: Analyzing the results
Comparing the differences obtained:
$d_1 = 2$
$d_2 = 4$
$d_3 = 8$

Since $d_1 \neq d_2 \neq d_3$, the difference between consecutive terms is not constant.
[By the definition of an Arithmetic Progression, a sequence must have a constant common difference to be classified as an AP.]

Step 4: Conclusion
Because the common difference is not constant, the sequence $2, 4, 8, 16, \dots$ does not form an Arithmetic Progression.

Final Answer: The given sequence $2, 4, 8, 16, \dots$ does not form an AP because the difference between consecutive terms is not constant.

Given: An arithmetic progression (AP) series: $0.6, 1.7, 2.8, 3.9, \dots$

To Find: The first term ($a$) and the common difference ($d$) of the given arithmetic progression.

Step 1: Identifying the First Term

In an arithmetic progression, the first term is denoted by the variable $a$ (or $a_1$). By observing the given sequence:

The sequence is: $a_1 = 0.6, a_2 = 1.7, a_3 = 2.8, a_4 = 3.9, \dots$

Therefore, the first term $a = 0.6$.

Step 2: Defining the Common Difference

The common difference ($d$) of an arithmetic progression is the constant value obtained by subtracting any term from the term that immediately follows it. The formula is defined as:

$d = a_{n} - a_{n-1}$

For this sequence, we can calculate $d$ using the first two terms:

$d = a_2 - a_1$

Step 3: Calculating the Common Difference

Substitute the values of $a_2$ and $a_1$ into the formula:

$d = 1.7 - 0.6$

Performing the subtraction:

$d = 1.1$

Step 4: Verification (Optional but Recommended)

To ensure the sequence is indeed an AP, we verify the difference between subsequent terms:

$a_3 - a_2 = 2.8 - 1.7 = 1.1$

$a_4 - a_3 = 3.9 - 2.8 = 1.1$

[Since the difference remains constant at $1.1$, the common difference is confirmed.]

Final Answer: The first term $a = 0.6$ and the common difference $d = 1.1$.

Given: A sequence of numbers: $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$

To find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference ($d$).
3. Write the next three terms of the sequence.

Step 1: Identifying the terms of the sequence

Let the sequence be denoted by $a_1, a_2, a_3, a_4, \dots$
$a_1 = 2$
$a_2 = \frac{5}{2}$
$a_3 = 3$
$a_4 = \frac{7}{2}$

Step 2: Checking for a common difference

A sequence is an AP if the difference between consecutive terms is constant. This constant is called the common difference ($d$).
Calculate the difference between consecutive terms:

Difference $d_1 = a_2 - a_1 = \frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}$

Difference $d_2 = a_3 - a_2 = 3 - \frac{5}{2} = \frac{6}{2} - \frac{5}{2} = \frac{1}{2}$

Difference $d_3 = a_4 - a_3 = \frac{7}{2} - 3 = \frac{7}{2} - \frac{6}{2} = \frac{1}{2}$

Step 3: Conclusion on AP status

[Since $d_1 = d_2 = d_3 = \frac{1}{2}$, the difference between consecutive terms is constant.]
Therefore, the given sequence is an Arithmetic Progression with a common difference $d = \frac{1}{2}$.

Step 4: Finding the next three terms

To find the next terms, we add the common difference $d = \frac{1}{2}$ to the last known term ($a_4 = \frac{7}{2}$).

Fifth term ($a_5$) = $a_4 + d = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$

Sixth term ($a_6$) = $a_5 + d = 4 + \frac{1}{2} = \frac{8}{2} + \frac{1}{2} = \frac{9}{2}$

Seventh term ($a_7$) = $a_6 + d = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5$

Final Answer: The sequence forms an AP with common difference $d = \frac{1}{2}$. The next three terms are $4, \frac{9}{2}, 5$.

Given: A sequence of numbers: $-1.2, -3.2, -5.2, -7.2, \dots$

To find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it is an AP, find the common difference $d$.
3. Write the next three terms of the sequence.

Step 1: Definition of an Arithmetic Progression
A sequence $a_1, a_2, a_3, \dots, a_n$ is an Arithmetic Progression if the difference between consecutive terms is constant. This constant is called the common difference $d$, defined as:
$d = a_{n} - a_{n-1}$ for all $n > 1$.

Step 2: Calculating differences between consecutive terms
Let the given terms be:
$a_1 = -1.2$
$a_2 = -3.2$
$a_3 = -5.2$
$a_4 = -7.2$

Calculate the differences:
$d_1 = a_2 - a_1 = (-3.2) - (-1.2) = -3.2 + 1.2 = -2.0$
$d_2 = a_3 - a_2 = (-5.2) - (-3.2) = -5.2 + 3.2 = -2.0$
$d_3 = a_4 - a_3 = (-7.2) - (-5.2) = -7.2 + 5.2 = -2.0$

Step 3: Verification
Since $d_1 = d_2 = d_3 = -2.0$, the difference between consecutive terms is constant.
[Since the common difference is constant, the sequence is an AP.]

Step 4: Finding the next three terms
To find the next terms, we add the common difference $d = -2.0$ to the last known term ($a_4 = -7.2$).

Fifth term ($a_5$):
$a_5 = a_4 + d = -7.2 + (-2.0) = -9.2$

Sixth term ($a_6$):
$a_6 = a_5 + d = -9.2 + (-2.0) = -11.2$

Seventh term ($a_7$):
$a_7 = a_6 + d = -11.2 + (-2.0) = -13.2$

Final Answer:
The sequence forms an AP.
The common difference $d = -2.0$.
The next three terms are -9.2, -11.2, and -13.2.

Given:

The first term of the Arithmetic Progression (AP), denoted by $a = -1$.

The common difference of the AP, denoted by $d = \frac{1}{2}$.

To Find:

The first four terms of the Arithmetic Progression ($a_1, a_2, a_3, a_4$).

Step 1: Understanding the definition of an Arithmetic Progression

An Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).

The general form of an AP is given by: $a, a+d, a+2d, a+3d, \dots$

Where:

• $a_1 = a$
• $a_2 = a + d$
• $a_3 = a + 2d$
• $a_4 = a + 3d$

Step 2: Calculating the first four terms

Calculation for the first term ($a_1$):

$a_1 = a$

$a_1 = -1$

Calculation for the second term ($a_2$):

$a_2 = a + d$

$a_2 = -1 + \frac{1}{2}$

[Finding a common denominator to add the terms]

$a_2 = \frac{-2}{2} + \frac{1}{2}$

$a_2 = \frac{-2 + 1}{2}$

$a_2 = -\frac{1}{2}$

Calculation for the third term ($a_3$):

$a_3 = a_2 + d$

$a_3 = -\frac{1}{2} + \frac{1}{2}$

$a_3 = 0$

Calculation for the fourth term ($a_4$):

$a_4 = a_3 + d$

$a_4 = 0 + \frac{1}{2}$

$a_4 = \frac{1}{2}$

Step 3: Summary of the sequence

The first four terms of the Arithmetic Progression are $-1, -\frac{1}{2}, 0, \frac{1}{2}$.

Final Answer: The first four terms of the AP are $-1, -\frac{1}{2}, 0, \frac{1}{2}$.

Given: A sequence of numbers: $1^2, 5^2, 7^2, 73, \dots$

To Find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and the next three terms.

Step 1: Simplifying the terms of the sequence

Let the given sequence be denoted by $a_1, a_2, a_3, a_4, \dots$

Calculating the values of each term:

$a_1 = 1^2 = 1$

$a_2 = 5^2 = 25$

$a_3 = 7^2 = 49$

$a_4 = 73$

The sequence is: $1, 25, 49, 73, \dots$

Step 2: Checking for a common difference

A sequence is an AP if the difference between consecutive terms ($a_{n+1} - a_n$) is constant. This constant is called the common difference $d$.

Calculate the difference between the first and second terms ($d_1$):

$d_1 = a_2 - a_1 = 25 - 1 = 24$

Calculate the difference between the second and third terms ($d_2$):

$d_2 = a_3 - a_2 = 49 - 25 = 24$

Calculate the difference between the third and fourth terms ($d_3$):

$d_3 = a_4 - a_3 = 73 - 49 = 24$

[Since $d_1 = d_2 = d_3 = 24$, the difference between consecutive terms is constant.]

Step 3: Conclusion on AP status

Because the difference between consecutive terms is constant ($d = 24$), the given sequence forms an Arithmetic Progression.

Step 4: Finding the next three terms

To find the next three terms ($a_5, a_6, a_7$), we add the common difference $d = 24$ to the preceding term.

Fifth term ($a_5$):

$a_5 = a_4 + d = 73 + 24 = 97$

Sixth term ($a_6$):

$a_6 = a_5 + d = 97 + 24 = 121$

Seventh term ($a_7$):

$a_7 = a_6 + d = 121 + 24 = 145$

Final Answer: The sequence forms an AP with a common difference $d = 24$. The next three terms are 97, 121, and 145.

Given: A cylinder contains an initial amount of air. A vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at each stroke.

To Find: Determine whether the sequence of the amounts of air remaining in the cylinder after each stroke forms an Arithmetic Progression (AP).

Step 1: Defining the Variables
Let the initial amount of air present in the cylinder be $V$ units.
Let $a_1$ be the amount of air after the 0th stroke (initial state).
Let $a_2$ be the amount of air after the 1st stroke.
Let $a_3$ be the amount of air after the 2nd stroke.
Let $a_4$ be the amount of air after the 3rd stroke.

Step 2: Calculating the sequence of air amounts
The pump removes $\frac{1}{4}$ of the air present in the cylinder at each step.

Initial amount: $a_1 = V$

After the 1st stroke ($a_2$):
$a_2 = V - \frac{1}{4}V = \frac{3}{4}V$

After the 2nd stroke ($a_3$):
The pump removes $\frac{1}{4}$ of the air remaining, which is $a_2$.
$a_3 = a_2 - \frac{1}{4}a_2 = \frac{3}{4}a_2$
Substituting $a_2 = \frac{3}{4}V$:
$a_3 = \frac{3}{4} \times (\frac{3}{4}V) = \frac{9}{16}V$

After the 3rd stroke ($a_4$):
$a_4 = a_3 - \frac{1}{4}a_3 = \frac{3}{4}a_3$
Substituting $a_3 = \frac{9}{16}V$:
$a_4 = \frac{3}{4} \times (\frac{9}{16}V) = \frac{27}{64}V$

Step 3: Checking for Arithmetic Progression
A sequence is an Arithmetic Progression if the difference between consecutive terms is constant (i.e., $a_{n+1} - a_n = d$, where $d$ is the common difference).

Calculate the first difference ($d_1$):
$d_1 = a_2 - a_1 = \frac{3}{4}V - V = -\frac{1}{4}V$

Calculate the second difference ($d_2$):
$d_2 = a_3 - a_2 = \frac{9}{16}V - \frac{3}{4}V$
To subtract, find a common denominator (16):
$d_2 = \frac{9}{16}V - \frac{12}{16}V = -\frac{3}{16}V$

Step 4: Comparison and Conclusion
Since $d_1 \neq d_2$ (because $-\frac{1}{4}V \neq -\frac{3}{16}V$), the difference between consecutive terms is not constant.

[Definition of an Arithmetic Progression: A sequence of numbers is an AP if the difference between any two consecutive terms is constant.]

Final Answer: The list of numbers does not form an Arithmetic Progression because the difference between consecutive terms is not constant.

Given: A sequence of numbers: $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$

To find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and write the next three terms.

Step 1: Definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).
Let the terms be $a_1, a_2, a_3, a_4, \dots$
Here, $a_1 = 3$, $a_2 = 3+\sqrt{2}$, $a_3 = 3+2\sqrt{2}$, $a_4 = 3+3\sqrt{2}$.

Step 2: Calculating the differences between consecutive terms
We calculate the difference $d_n = a_{n+1} - a_n$ for consecutive terms:

Difference 1 ($d_1$):
$d_1 = a_2 - a_1 = (3 + \sqrt{2}) - 3$
$d_1 = 3 - 3 + \sqrt{2} = \sqrt{2}$

Difference 2 ($d_2$):
$d_2 = a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2})$
$d_2 = 3 + 2\sqrt{2} - 3 - \sqrt{2}$
$d_2 = (3 - 3) + (2\sqrt{2} - \sqrt{2}) = \sqrt{2}$

Difference 3 ($d_3$):
$d_3 = a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2})$
$d_3 = 3 + 3\sqrt{2} - 3 - 2\sqrt{2}$
$d_3 = (3 - 3) + (3\sqrt{2} - 2\sqrt{2}) = \sqrt{2}$

Step 3: Verification
Since $d_1 = d_2 = d_3 = \sqrt{2}$, the difference between consecutive terms is constant. Therefore, the given sequence is an Arithmetic Progression with common difference $d = \sqrt{2}$.

Step 4: Finding the next three terms
To find the next terms, we add the common difference $d = \sqrt{2}$ to the last known term ($a_4 = 3 + 3\sqrt{2}$):

Fifth term ($a_5$):
$a_5 = a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$

Sixth term ($a_6$):
$a_6 = a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$

Seventh term ($a_7$):
$a_7 = a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$

Final Answer: The sequence forms an AP with common difference $d = \sqrt{2}$. The next three terms are $3+4\sqrt{2}, 3+5\sqrt{2},$ and $3+6\sqrt{2}$.

Given: A sequence of numbers: $0.2, 0.22, 0.222, 0.2222, \dots$

To Find: Determine if the given sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and write the next three terms.

Step 1: Definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).
For a sequence $a_1, a_2, a_3, a_4, \dots$, the sequence is an AP if and only if:
$a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = d$

Step 2: Calculating the differences between consecutive terms
Let the terms be:
$a_1 = 0.2$
$a_2 = 0.22$
$a_3 = 0.222$
$a_4 = 0.2222$

Calculate the difference between the first and second term ($d_1$):
$d_1 = a_2 - a_1 = 0.22 - 0.2 = 0.02$

Calculate the difference between the second and third term ($d_2$):
$d_2 = a_3 - a_2 = 0.222 - 0.22 = 0.002$

Calculate the difference between the third and fourth term ($d_3$):
$d_3 = a_4 - a_3 = 0.2222 - 0.222 = 0.0002$

Step 3: Comparing the differences
We observe that:
$d_1 = 0.02$
$d_2 = 0.002$
$d_3 = 0.0002$
Since $d_1 \neq d_2 \neq d_3$, the difference between consecutive terms is not constant.

Step 4: Conclusion
Because the common difference is not constant, the given sequence $0.2, 0.22, 0.222, 0.2222, \dots$ does not satisfy the condition for an Arithmetic Progression.

Final Answer: The given sequence does not form an AP because the difference between consecutive terms is not constant.

Given: A sequence of numbers $1^2, 3^2, 5^2, 7^2, \dots$

To Find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and the next three terms.

Step 1: Evaluating the terms of the sequence

Let the sequence be denoted by $a_1, a_2, a_3, a_4, \dots$

Calculating the values of the given terms:

$a_1 = 1^2 = 1$

$a_2 = 3^2 = 9$

$a_3 = 5^2 = 25$

$a_4 = 7^2 = 49$

The sequence is $1, 9, 25, 49, \dots$

Step 2: Checking for a common difference

A sequence is an AP if the difference between consecutive terms is constant, i.e., $a_{n+1} - a_n = d$ for all $n$.

Calculate the difference between the first and second terms ($d_1$):

$d_1 = a_2 - a_1 = 9 - 1 = 8$

Calculate the difference between the second and third terms ($d_2$):

$d_2 = a_3 - a_2 = 25 - 9 = 16$

Calculate the difference between the third and fourth terms ($d_3$):

$d_3 = a_4 - a_3 = 49 - 25 = 24$

Step 3: Conclusion on the nature of the sequence

Since $d_1 \neq d_2 \neq d_3$ (specifically, $8 \neq 16 \neq 24$), the difference between consecutive terms is not constant.

[Definition of an Arithmetic Progression: A sequence is an AP if and only if the difference between any two consecutive terms is constant.]

Final Answer: The given sequence $1^2, 3^2, 5^2, 7^2, \dots$ does not form an Arithmetic Progression because the common difference is not constant.

Given: A sequence of numbers: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$

To Find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference ($d$).
3. Write the next three terms of the sequence.

Step 1: Definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).
Let the terms of the sequence be represented as $a_1, a_2, a_3, a_4, \dots$
The sequence is an AP if $(a_2 - a_1) = (a_3 - a_2) = (a_4 - a_3) = d$.

Step 2: Calculating the differences between consecutive terms
Given terms: $a_1 = -\frac{1}{2}$, $a_2 = -\frac{1}{2}$, $a_3 = -\frac{1}{2}$, $a_4 = -\frac{1}{2}$.

Difference $d_1 = a_2 - a_1 = (-\frac{1}{2}) - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$
Difference $d_2 = a_3 - a_2 = (-\frac{1}{2}) - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$
Difference $d_3 = a_4 - a_3 = (-\frac{1}{2}) - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

Step 3: Verification and Conclusion on AP
Since $d_1 = d_2 = d_3 = 0$, the difference between consecutive terms is constant.
[Since the common difference is constant, the sequence forms an Arithmetic Progression.]
The common difference $d = 0$.

Step 4: Finding the next three terms
To find the next terms, we add the common difference $d$ to the last known term.
Let the next three terms be $a_5, a_6,$ and $a_7$.

$a_5 = a_4 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$
$a_6 = a_5 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$
$a_7 = a_6 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$

Final Answer: The sequence forms an AP with common difference $d = 0$. The next three terms are $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$.

Given:

• Fare for the first kilometer ($a_1$) = ₹ $15$.
• Additional fare for every subsequent kilometer ($d$) = ₹ $8$.

To Find:

Determine whether the sequence of taxi fares for $1, 2, 3, 4, \dots$ kilometers forms an Arithmetic Progression (AP) and provide the justification.

Definition:

An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$). If the terms are $a_1, a_2, a_3, \dots, a_n$, then the sequence is an AP if $(a_2 - a_1) = (a_3 - a_2) = (a_4 - a_3) = d$.

Step 1: Calculating the terms of the sequence

Let $a_n$ be the fare for $n$ kilometers.

• Fare for 1st km ($a_1$) = $15$
• Fare for 2nd km ($a_2$) = $a_1 + 8 = 15 + 8 = 23$
• Fare for 3rd km ($a_3$) = $a_2 + 8 = 23 + 8 = 31$
• Fare for 4th km ($a_4$) = $a_3 + 8 = 31 + 8 = 39$

The sequence of fares is: $15, 23, 31, 39, \dots$

Step 2: Verifying the common difference

To check if the sequence is an AP, we calculate the difference between consecutive terms:

• Difference 1 ($d_1$) = $a_2 - a_1 = 23 - 15 = 8$
• Difference 2 ($d_2$) = $a_3 - a_2 = 31 - 23 = 8$
• Difference 3 ($d_3$) = $a_4 - a_3 = 39 - 31 = 8$

Step 3: Conclusion

Since the difference between consecutive terms is constant ($d = 8$), the sequence satisfies the definition of an Arithmetic Progression.

Final Answer: Yes, the situation forms an Arithmetic Progression because the fare increases by a constant amount of ₹ 8 for each additional kilometer, resulting in a common difference of 8.

Given: A sequence of numbers $a, a^2, a^3, a^4, \dots$

To find: Determine if the given sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and the next three terms.

Step 1: Definition of an Arithmetic Progression
A sequence $a_1, a_2, a_3, \dots, a_n$ is said to be an Arithmetic Progression if the difference between consecutive terms is constant. That is, $a_{n+1} - a_n = d$ for all $n \geq 1$, where $d$ is the common difference.

Step 2: Calculating the differences between consecutive terms
Let the terms be:
$a_1 = a$
$a_2 = a^2$
$a_3 = a^3$
$a_4 = a^4$

Calculate the difference between the first and second terms ($d_1$):
$d_1 = a_2 - a_1 = a^2 - a = a(a - 1)$

Calculate the difference between the second and third terms ($d_2$):
$d_2 = a_3 - a_2 = a^3 - a^2 = a^2(a - 1)$

Step 3: Comparing the differences
For the sequence to be an AP, the condition $d_1 = d_2$ must hold true.
Comparing $d_1$ and $d_2$:
$a(a - 1) \neq a^2(a - 1)$ (assuming $a \neq 0$ and $a \neq 1$)

Since $d_1 \neq d_2$, the difference between consecutive terms is not constant.

Step 4: Conclusion on the nature of the sequence
Because the difference between consecutive terms is not constant, the sequence $a, a^2, a^3, a^4, \dots$ does not satisfy the definition of an Arithmetic Progression.

Final Answer: The given sequence $a, a^2, a^3, a^4, \dots$ does not form an Arithmetic Progression because the common difference is not constant.

Given: An Arithmetic Progression (AP) sequence: $-5, -1, 3, 7, \dots$

To find: The first term ($a$) and the common difference ($d$) of the given AP.

Step 1: Identifying the first term ($a$)

In an Arithmetic Progression, the first term is denoted by the variable $a$ (or $a_1$). By observing the given sequence:

Sequence: $-5, -1, 3, 7, \dots$

The first term is the element at the first position in the sequence.

Therefore, $a = -5$.

Step 2: Defining the common difference ($d$)

The common difference ($d$) of an Arithmetic Progression is defined as the difference between any two consecutive terms. Mathematically, it is expressed as:

$d = a_{n} - a_{n-1}$

where $a_n$ is the $n^{th}$ term and $a_{n-1}$ is the preceding term.

Step 3: Calculating the common difference ($d$)

We can calculate $d$ by subtracting the first term from the second term:

$d = a_2 - a_1$

Given $a_1 = -5$ and $a_2 = -1$:

$d = (-1) - (-5)$

[Applying the rule of signs: subtracting a negative number is equivalent to adding its positive counterpart]

$d = -1 + 5$

$d = 4$

Step 4: Verification of the common difference

To ensure the sequence is indeed an AP, we verify the difference between other consecutive terms:

For the third and second terms: $d = a_3 - a_2 = 3 - (-1) = 3 + 1 = 4$

For the fourth and third terms: $d = a_4 - a_3 = 7 - 3 = 4$

[Since the difference is constant throughout the sequence, the value $d = 4$ is confirmed.]

Final Answer: The first term ($a$) is $-5$ and the common difference ($d$) is $4$.