Given:

1. The third term of the Arithmetic Progression ($a_3$) = $16$.

2. The seventh term ($a_7$) exceeds the fifth term ($a_5$) by $12$, which can be written as: $a_7 = a_5 + 12$.

To Find:

The Arithmetic Progression (AP), which is defined by its first term ($a$) and common difference ($d$).

Step 1: Establishing the General Formula

The $n^{th}$ term of an Arithmetic Progression is given by the formula:

$a_n = a + (n - 1)d$

where $a$ is the first term and $d$ is the common difference.

Step 2: Formulating Equations based on Given Conditions

For the third term ($n=3$):

$a_3 = a + (3 - 1)d$

$16 = a + 2d$ --- (Equation 1)

For the relationship between the seventh and fifth terms:

$a_7 = a + (7 - 1)d = a + 6d$

$a_5 = a + (5 - 1)d = a + 4d$

Given $a_7 = a_5 + 12$, substitute the expressions:

$(a + 6d) = (a + 4d) + 12$

Step 3: Solving for the Common Difference ($d$)

Subtract $a$ from both sides:

$6d = 4d + 12$

Subtract $4d$ from both sides:

$6d - 4d = 12$

$2d = 12$

$d = \frac{12}{2}$

$d = 6$

Step 4: Solving for the First Term ($a$)

Substitute $d = 6$ into Equation 1:

$16 = a + 2(6)$

$16 = a + 12$

$a = 16 - 12$

$a = 4$

Step 5: Constructing the Arithmetic Progression

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

Term 1 ($a_1$) = $4$

Term 2 ($a_2$) = $a + d = 4 + 6 = 10$

Term 3 ($a_3$) = $a + 2d = 4 + 2(6) = 4 + 12 = 16$

Term 4 ($a_4$) = $a + 3d = 4 + 3(6) = 4 + 18 = 22$

Final Answer: The Arithmetic Progression is 4, 10, 16, 22, ...

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: An Arithmetic Progression (AP) with terms $121, 117, 113, \dots$

To Find: The value of $n$ such that the $n^{th}$ term ($a_n$) is the first negative term of the sequence.

Step 1: Identify the parameters of the Arithmetic Progression.

The general form of an AP is $a, a+d, a+2d, \dots$ where $a$ is the first term and $d$ is the common difference.

From the given sequence:

First term ($a$) = $121$

Common difference ($d$) = $a_2 - a_1 = 117 - 121 = -4$

Step 2: State the formula for the $n^{th}$ term of an AP.

The formula for the $n^{th}$ term of an AP is given by:

$a_n = a + (n - 1)d$

[Where $a_n$ is the $n^{th}$ term, $a$ is the first term, $n$ is the position of the term, and $d$ is the common difference.]

Step 3: Set up the inequality to find the first negative term.

We are looking for the first term that is less than zero. Therefore, we set $a_n < 0$:

$a + (n - 1)d < 0$

Substitute the known values $a = 121$ and $d = -4$ into the inequality:

$121 + (n - 1)(-4) < 0$

Step 4: Solve the inequality for $n$.

$121 - 4n + 4 < 0$ [Distributing $-4$ into the parentheses]

$125 - 4n < 0$ [Combining like terms $121 + 4 = 125$]

$-4n < -125$ [Subtracting $125$ from both sides]

$4n > 125$ [Multiplying by $-1$ reverses the inequality sign]

$n > \frac{125}{4}$ [Dividing both sides by $4$]

$n > 31.25$

Step 5: Determine the integer value for $n$.

Since $n$ must be a positive integer representing the position of a term in the sequence, and we require the smallest integer $n$ such that $n > 31.25$, we conclude that $n = 32$.

Step 6: Verification (Optional but recommended).

Calculate the $31^{st}$ term: $a_{31} = 121 + (31 - 1)(-4) = 121 + 30(-4) = 121 - 120 = 1$. (This is positive)

Calculate the $32^{nd}$ term: $a_{32} = 121 + (32 - 1)(-4) = 121 + 31(-4) = 121 - 124 = -3$. (This is the first negative term)

Final Answer: The $32^{nd}$ term of the AP is its first negative term.

Given:

The penalty for the first day ($a_1$) = ₹ $200$.

The penalty for the second day ($a_2$) = ₹ $250$.

The penalty for the third day ($a_3$) = ₹ $300$.

The common difference ($d$) between consecutive days = $250 - 200 = 50$ and $300 - 250 = 50$.

The total number of days of delay ($n$) = $30$.

To Find:

The total penalty amount to be paid for $30$ days, which is the sum of the first $30$ terms of the arithmetic progression ($S_{30}$).

Step 1: Identifying the Progression

The sequence of penalties forms an Arithmetic Progression (AP) because the difference between consecutive terms is constant.

The sequence is: $200, 250, 300, \dots$

Here, the first term $a = 200$.

The common difference $d = 50$.

The number of terms $n = 30$.

Step 2: Selecting the Formula

To find the sum of the first $n$ terms of an arithmetic progression, we use the formula:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

[Where $S_n$ is the sum of $n$ terms, $a$ is the first term, $n$ is the number of terms, and $d$ is the common difference.]

Step 3: Substituting the Values

Substitute $n = 30$, $a = 200$, and $d = 50$ into the formula:

$S_{30} = \frac{30}{2} [2(200) + (30 - 1)50]$

Step 4: Performing the Calculations

First, simplify the fraction outside the brackets:

$S_{30} = 15 [2(200) + (29)50]$

Next, calculate the values inside the brackets:

$S_{30} = 15 [400 + 1450]$

[Since $2 \times 200 = 400$ and $29 \times 50 = 1450$]

Add the values inside the brackets:

$S_{30} = 15 [1850]$

Finally, multiply the result by $15$:

$S_{30} = 27750$

Conclusion:

The total penalty for a delay of $30$ days is calculated by summing the arithmetic series of the daily penalties.

Final Answer: The contractor has to pay a total penalty of ₹ 27,750.

Given:

The first term of the Arithmetic Progression (AP), $a = 8$.

The $n^{th}$ term of the AP, $a_n = 62$.

The sum of the first $n$ terms of the AP, $S_n = 210$.

To find:

The number of terms, $n$, and the common difference, $d$.

Step 1: Formulating the equation for $n$ using the sum formula.

The formula for the sum of the first $n$ terms of an AP when the first term ($a$) and the last term ($a_n$) are known is given by:

$S_n = \frac{n}{2}(a + a_n)$

[Substituting the given values into the formula]:

$210 = \frac{n}{2}(8 + 62)$

$210 = \frac{n}{2}(70)$

$210 = n \times 35$

$n = \frac{210}{35}$

$n = 6$

[Since $210 \div 35 = 6$].

Step 2: Formulating the equation for $d$ using the $n^{th}$ term formula.

The formula for the $n^{th}$ term of an AP is given by:

$a_n = a + (n - 1)d$

[Substituting the known values $a_n = 62$, $a = 8$, and $n = 6$]:

$62 = 8 + (6 - 1)d$

$62 = 8 + 5d$

[Subtracting 8 from both sides of the equation]:

$62 - 8 = 5d$

$54 = 5d$

[Dividing both sides by 5]:

$d = \frac{54}{5}$

$d = 10.8$

Summary of Results:

We have determined the number of terms $n$ by utilizing the sum formula for an AP, and subsequently determined the common difference $d$ by substituting $n$ into the general term formula.

Final Answer: $n = 6$ and $d = 10.8$

Given: An Arithmetic Progression (AP) with the first term $a_1 = 2$ and the third term $a_3 = 26$.

To find: The missing term in the box, which is the second term of the AP, denoted as $a_2$.

Step 1: Understanding the properties of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between consecutive terms is constant. This constant is called the common difference, denoted by $d$.
The general form of an AP is $a, a+d, a+2d, a+3d, \dots$
The $n^{th}$ term of an AP is given by the formula: $a_n = a + (n - 1)d$, where $a$ is the first term and $d$ is the common difference.

Step 2: Formulating equations based on the given terms
We are given:
$a_1 = a = 2$
$a_3 = a + (3 - 1)d = a + 2d = 26$

Step 3: Solving for the common difference ($d$)
Substitute the value of $a = 2$ into the equation for $a_3$:
$2 + 2d = 26$
Subtract 2 from both sides of the equation:
$2d = 26 - 2$
$2d = 24$
Divide both sides by 2:
$d = \frac{24}{2}$
$d = 12$

Step 4: Calculating the missing term ($a_2$)
The missing term is the second term of the AP, $a_2$.
Using the formula $a_n = a + (n - 1)d$ for $n = 2$:
$a_2 = a + (2 - 1)d$
$a_2 = a + d$
Substitute the known values $a = 2$ and $d = 12$:
$a_2 = 2 + 12$
$a_2 = 14$

Verification:
If the sequence is $2, 14, 26$, the common difference is:
$14 - 2 = 12$
$26 - 14 = 12$
Since the common difference is constant, the value is correct.

Final Answer: The missing term is 14. The AP is 2, 14, 26.

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: $0.2, 0.22, 0.222, 0.2222, \dots$

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:

The first term of the Arithmetic Progression (AP), $a = 3$.

The number of terms in the AP, $n = 8$.

The sum of the first $n$ terms, $S_n = 192$.

To find:

The common difference, $d$.

Step 1: Stating the relevant formula for the sum of an AP.

The sum of the first $n$ terms of an Arithmetic Progression is given by the formula:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

[Where $S_n$ is the sum, $n$ is the number of terms, $a$ is the first term, and $d$ is the common difference.]

Step 2: Substituting the given values into the formula.

Substitute $S_n = 192$, $n = 8$, and $a = 3$ into the equation:

$192 = \frac{8}{2} [2(3) + (8 - 1)d]$

Step 3: Simplifying the equation.

First, simplify the fraction outside the bracket:

$192 = 4 [2(3) + (8 - 1)d]$

Next, perform the arithmetic operations inside the bracket:

$192 = 4 [6 + 7d]$

Step 4: Solving for $d$.

Divide both sides of the equation by 4:

$\frac{192}{4} = 6 + 7d$

$48 = 6 + 7d$

Subtract 6 from both sides to isolate the term containing $d$:

$48 - 6 = 7d$

$42 = 7d$

Divide both sides by 7 to find the value of $d$:

$d = \frac{42}{7}$

$d = 6$

Final Answer: The common difference $d$ is 6.

Given: An arithmetic progression (AP) series: $-5 + (-8) + (-11) + \dots + (-230)$.

To Find: The sum of the given arithmetic series.

Step 1: Identify the parameters of the Arithmetic Progression.

The given series is $-5, -8, -11, \dots, -230$.

Let the first term be $a$. Thus, $a = -5$.

Let the common difference be $d$. The common difference is calculated as the difference between any two consecutive terms:

$d = a_2 - a_1 = (-8) - (-5) = -8 + 5 = -3$.

The last term (or $n^{th}$ term) is $a_n = -230$.

Step 2: Determine the number of terms ($n$).

We use the formula for the $n^{th}$ term of an AP: $a_n = a + (n - 1)d$.

Substituting the known values into the formula:

$-230 = -5 + (n - 1)(-3)$

Add $5$ to both sides of the equation:

$-230 + 5 = (n - 1)(-3)$

$-225 = (n - 1)(-3)$

Divide both sides by $-3$:

$\frac{-225}{-3} = n - 1$

$75 = n - 1$

Add $1$ to both sides:

$n = 76$

[Since there are 76 terms in the series].

Step 3: Calculate the sum of the series ($S_n$).

The formula for the sum of the first $n$ terms of an AP when the last term ($l$) is known is:

$S_n = \frac{n}{2}(a + l)$

Where $n = 76$, $a = -5$, and $l = -230$.

Substitute the values into the formula:

$S_{76} = \frac{76}{2}(-5 + (-230))$

Simplify the fraction and the expression inside the parentheses:

$S_{76} = 38(-5 - 230)$

$S_{76} = 38(-235)$

Step 4: Perform the final multiplication.

$38 \times (-235) = -8930$

[Calculation: $38 \times 200 = 7600$; $38 \times 30 = 1140$; $38 \times 5 = 190$; $7600 + 1140 + 190 = 8930$].

Final Answer: The sum of the series is -8930.

Given:

Two Arithmetic Progressions (APs):

AP 1: $63, 65, 67, \dots$

AP 2: $3, 10, 17, \dots$

To Find:

The value of $n$ for which the $n$th term of AP 1 is equal to the $n$th term of AP 2.

Step 1: Identify the parameters of the first AP.

For the first AP: $63, 65, 67, \dots$

Let the first term be $a_1$ and the common difference be $d_1$.

$a_1 = 63$

$d_1 = 65 - 63 = 2$

The formula for the $n$th term of an AP is $a_n = a + (n - 1)d$.

Therefore, the $n$th term of the first AP ($A_n$) is:

$A_n = 63 + (n - 1)2$

$A_n = 63 + 2n - 2$

$A_n = 61 + 2n$ --- (Equation 1)

Step 2: Identify the parameters of the second AP.

For the second AP: $3, 10, 17, \dots$

Let the first term be $a_2$ and the common difference be $d_2$.

$a_2 = 3$

$d_2 = 10 - 3 = 7$

Therefore, the $n$th term of the second AP ($B_n$) is:

$B_n = 3 + (n - 1)7$

$B_n = 3 + 7n - 7$

$B_n = 7n - 4$ --- (Equation 2)

Step 3: Equate the $n$th terms and solve for $n$.

According to the problem, the $n$th terms are equal, so $A_n = B_n$.

Substituting Equation 1 and Equation 2:

$61 + 2n = 7n - 4$

Rearranging the terms to isolate $n$:

$61 + 4 = 7n - 2n$ [Transposing $-4$ to the left and $2n$ to the right]

$65 = 5n$

$n = \frac{65}{5}$ [Dividing both sides by 5]

$n = 13$

Step 4: Verification (Optional but recommended).

For $n = 13$ in AP 1: $A_{13} = 63 + (13 - 1)2 = 63 + 12(2) = 63 + 24 = 87$.

For $n = 13$ in AP 2: $B_{13} = 3 + (13 - 1)7 = 3 + 12(7) = 3 + 84 = 87$.

Since $87 = 87$, the value is verified.

Final Answer: The value of $n$ for which the $n$th terms of the two APs are equal is 13.

Given:

The sum of the first $7$ terms of an Arithmetic Progression (AP), denoted as $S_7 = 49$.

The sum of the first $17$ terms of the same AP, denoted as $S_{17} = 289$.

To Find:

The sum of the first $n$ terms of the AP, denoted as $S_n$.

Formulae Used:

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

Where $a$ is the first term and $d$ is the common difference.

Step 1: Formulating the equation for the first 7 terms

Using the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$ for $n = 7$:

$49 = \frac{7}{2} [2a + (7 - 1)d]$

$49 = \frac{7}{2} [2a + 6d]$

Divide both sides by $7$:

$7 = \frac{1}{2} [2a + 6d]$

Multiply by $2$:

$14 = 2a + 6d$

Divide the entire equation by $2$ to simplify:

$7 = a + 3d$ --- (Equation 1)

Step 2: Formulating the equation for the first 17 terms

Using the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$ for $n = 17$:

$289 = \frac{17}{2} [2a + (17 - 1)d]$

$289 = \frac{17}{2} [2a + 16d]$

Divide both sides by $17$ (since $17^2 = 289$):

$17 = \frac{1}{2} [2a + 16d]$

Multiply by $2$:

$34 = 2a + 16d$

Divide the entire equation by $2$ to simplify:

$17 = a + 8d$ --- (Equation 2)

Step 3: Solving the system of linear equations

Subtract Equation 1 from Equation 2:

$(a + 8d) - (a + 3d) = 17 - 7$

$a - a + 8d - 3d = 10$

$5d = 10$

$d = 2$

Substitute $d = 2$ into Equation 1:

$7 = a + 3(2)$

$7 = a + 6$

$a = 7 - 6$

$a = 1$

Step 4: Finding the sum of the first $n$ terms ($S_n$)

Substitute $a = 1$ and $d = 2$ into the general formula $S_n = \frac{n}{2} [2a + (n - 1)d]$:

$S_n = \frac{n}{2} [2(1) + (n - 1)(2)]$

$S_n = \frac{n}{2} [2 + 2n - 2]$

$S_n = \frac{n}{2} [2n]$

$S_n = n \times n$

$S_n = n^2$

Final Answer: The sum of the first $n$ terms is $n^2$.

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: The range of numbers is between 10 and 250. We are looking for multiples of 4 within this range.

To find: The total number of multiples of 4 that lie between 10 and 250.

Step 1: Identifying the first and last terms of the Arithmetic Progression (AP)

To find the first multiple of 4 after 10: $10 \div 4 = 2$ with a remainder of $2$. The next multiple is $4 \times 3 = 12$. Thus, the first term ($a$) is $12$.

To find the last multiple of 4 before 250: $250 \div 4 = 62.5$. The largest integer multiple is $4 \times 62 = 248$. Thus, the last term ($a_n$) is $248$.

Step 2: Defining the Arithmetic Progression

The sequence of multiples of 4 between 10 and 250 is: $12, 16, 20, \dots, 248$.

Here, the first term $a = 12$.
The common difference $d = 16 - 12 = 4$.
The $n^{th}$ term $a_n = 248$.

Step 3: Applying the formula for the $n^{th}$ term of an AP

The formula for the $n^{th}$ term of an Arithmetic Progression is given by: $a_n = a + (n - 1)d$

Substituting the known values into the formula: $248 = 12 + (n - 1)4$

Step 4: Solving for $n$

Subtract 12 from both sides: $248 - 12 = (n - 1)4$ $236 = (n - 1)4$

Divide both sides by 4: $\frac{236}{4} = n - 1$ $59 = n - 1$

Add 1 to both sides: $n = 59 + 1$ $n = 60$

Justification: Since the sequence is finite and follows a constant common difference, the number of terms $n$ represents the count of multiples of 4 within the specified interval.

Final Answer: There are 60 multiples of 4 that lie between 10 and 250.

Given:

An Arithmetic Progression (AP) where:

• The sum of the 4th term ($a_4$) and the 8th term ($a_8$) is $24$. That is, $a_4 + a_8 = 24$.
• The sum of the 6th term ($a_6$) and the 10th term ($a_{10}$) is $44$. That is, $a_6 + a_{10} = 44$.

To Find:

The first three terms of the AP, which are $a$, $a+d$, and $a+2d$.

Step 1: Define the general term of an AP

The $n^{th}$ term of an Arithmetic Progression is given by the formula:

$a_n = a + (n - 1)d$

where $a$ is the first term and $d$ is the common difference.

Step 2: Formulate equations based on the given conditions

Using the formula $a_n = a + (n - 1)d$:

For the first condition ($a_4 + a_8 = 24$):

$a_4 = a + 3d$

$a_8 = a + 7d$

$(a + 3d) + (a + 7d) = 24$

$2a + 10d = 24$

Dividing the entire equation by 2 to simplify:

$a + 5d = 12$ --- (Equation 1)

For the second condition ($a_6 + a_{10} = 44$):

$a_6 = a + 5d$

$a_{10} = a + 9d$

$(a + 5d) + (a + 9d) = 44$

$2a + 14d = 44$

Dividing the entire equation by 2 to simplify:

$a + 7d = 22$ --- (Equation 2)

Step 3: Solve the system of linear equations

Subtract Equation 1 from Equation 2:

$(a + 7d) - (a + 5d) = 22 - 12$

$2d = 10$

$d = 5$ [Dividing both sides by 2]

Now, substitute $d = 5$ into Equation 1 to find $a$:

$a + 5(5) = 12$

$a + 25 = 12$

$a = 12 - 25$

$a = -13$

Step 4: Determine the first three terms

The first three terms are defined as $a_1$, $a_2$, and $a_3$:

$a_1 = a = -13$

$a_2 = a + d = -13 + 5 = -8$

$a_3 = a + 2d = -13 + 2(5) = -13 + 10 = -3$

Final Answer: The first three terms of the AP are -13, -8, and -3.

Given: An arithmetic series $34 + 32 + 30 + \dots + 10$.

To find: The sum of the given arithmetic series.

Step 1: Identify the components of the Arithmetic Progression (AP)

The given series is $34, 32, 30, \dots, 10$.

Let the first term be $a = 34$.

Let the common difference be $d$.

$d = a_2 - a_1 = 32 - 34 = -2$.

The last term (or $n^{th}$ term) is $a_n = l = 10$.

Step 2: Determine the number of terms ($n$)

We use the formula for the $n^{th}$ term of an AP: $a_n = a + (n - 1)d$.

Substituting the known values into the formula:

$10 = 34 + (n - 1)(-2)$

Subtract $34$ from both sides:

$10 - 34 = (n - 1)(-2)$

$-24 = (n - 1)(-2)$

Divide both sides by $-2$:

$\frac{-24}{-2} = n - 1$

$12 = n - 1$

$n = 12 + 1 = 13$.

[Since there are 13 terms in the series]

Step 3: Calculate the sum of the AP

The formula for the sum of the first $n$ terms of an AP when the last term is known is:

$S_n = \frac{n}{2}(a + l)$

Substituting $n = 13$, $a = 34$, and $l = 10$:

$S_{13} = \frac{13}{2}(34 + 10)$

$S_{13} = \frac{13}{2}(44)$

Simplify the expression:

$S_{13} = 13 \times \left(\frac{44}{2}\right)$

$S_{13} = 13 \times 22$

Step 4: Final Arithmetic Calculation

$13 \times 22 = 13 \times (20 + 2)$

$= 260 + 26$

$= 286$

Final Answer: The sum of the series 34 + 32 + 30 + . . . + 10 is 286.