Given:

The common difference of the Arithmetic Progression (AP), $d = 5$.

The sum of the first 9 terms of the AP, $S_9 = 75$.

To Find:

The first term ($a$) and the ninth term ($a_9$).

Step 1: Identifying the relevant formulas

To solve for $a$, we use the formula for the sum of the first $n$ terms of an AP:

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

To solve for $a_9$, we use the formula for the $n^{th}$ term of an AP:

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

Step 2: Substituting the given values into the sum formula

Given $n = 9$, $d = 5$, and $S_9 = 75$, we substitute these into the sum formula:

$75 = \frac{9}{2} [2a + (9 - 1)5]$

[Substituting $n=9, d=5, S_9=75$]

Step 3: Solving for $a$

$75 = \frac{9}{2} [2a + (8)5]$

[Simplifying the term inside the parenthesis: $9-1 = 8$]

$75 = \frac{9}{2} [2a + 40]$

[Multiplying $8 \times 5 = 40$]

$75 = 9(a + 20)$

[Factoring out 2 from the bracket: $\frac{9}{2} \times 2(a + 20) = 9(a + 20)$]

$75 = 9a + 180$

[Distributing 9 into the parenthesis]

$9a = 75 - 180$

[Transposing 180 to the left side]

$9a = -105$

$a = \frac{-105}{9}$

$a = \frac{-35}{3}$

[Dividing both numerator and denominator by 3]

Step 4: Calculating the ninth term ($a_9$)

Using the formula $a_n = a + (n - 1)d$ for $n = 9$:

$a_9 = a + (9 - 1)d$

$a_9 = \left(\frac{-35}{3}\right) + 8(5)$

[Substituting $a = \frac{-35}{3}$ and $d = 5$]

$a_9 = \frac{-35}{3} + 40$

$a_9 = \frac{-35 + 120}{3}$

[Finding a common denominator: $40 = \frac{120}{3}$]

$a_9 = \frac{85}{3}$

Final Answer:

The first term $a = -\frac{35}{3}$ and the ninth term $a_9 = \frac{85}{3}$.

Given: An Arithmetic Progression (AP) with the second term $a_2 = 38$ and the sixth term $a_6 = -22$.

To find: The missing terms $a_1$, $a_3$, $a_4$, and $a_5$.

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 terms.
For the second term ($n=2$):
$a_2 = a + (2 - 1)d = 38$
$a + d = 38$ --- (Equation 1)

For the sixth term ($n=6$):
$a_6 = a + (6 - 1)d = -22$
$a + 5d = -22$ --- (Equation 2)

Step 3: Solve the system of linear equations for $a$ and $d$.
Subtract Equation 1 from Equation 2:
$(a + 5d) - (a + d) = -22 - 38$
$a - a + 5d - d = -60$
$4d = -60$
$d = \frac{-60}{4}$
$d = -15$ [Since the common difference is constant]

Substitute $d = -15$ into Equation 1:
$a + (-15) = 38$
$a = 38 + 15$
$a = 53$

Step 4: Calculate the missing terms.
The first term $a_1 = a = 53$.
The third term $a_3 = a + 2d = 53 + 2(-15) = 53 - 30 = 23$.
The fourth term $a_4 = a + 3d = 53 + 3(-15) = 53 - 45 = 8$.
The fifth term $a_5 = a + 4d = 53 + 4(-15) = 53 - 60 = -7$.

Summary of terms:
$a_1 = 53$
$a_2 = 38$
$a_3 = 23$
$a_4 = 8$
$a_5 = -7$
$a_6 = -22$

Final Answer: The missing terms are 53, 23, 8, and -7. The complete AP is 53, 38, 23, 8, -7, -22.

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: 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:

The first term of the Arithmetic Progression (AP), $a = 5$.

The common difference of the AP, $d = 3$.

The $n^{th}$ term of the AP, $a_n = 50$.

To Find:

The number of terms, $n$, and the sum of the first $n$ terms, $S_n$.

Step 1: Finding the value of $n$

We use the standard formula for the $n^{th}$ term of an Arithmetic Progression:

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

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

Substituting the given values into the formula:

$50 = 5 + (n - 1)3$

Subtract 5 from both sides of the equation:

$50 - 5 = (n - 1)3$

$45 = (n - 1)3$

Divide both sides by 3:

$\frac{45}{3} = n - 1$

$15 = n - 1$

Add 1 to both sides to solve for $n$:

$n = 15 + 1$

$n = 16$

Step 2: Finding the sum of $n$ terms ($S_n$)

We use the formula for the sum of the first $n$ terms of an AP when the first and last terms are known:

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

[Where $S_n$ is the sum of $n$ terms, $n$ is the number of terms, $a$ is the first term, and $a_n$ is the last term.]

Substituting the values $n = 16$, $a = 5$, and $a_n = 50$:

$S_{16} = \frac{16}{2}(5 + 50)$

Simplify the fraction and the expression inside the parentheses:

$S_{16} = 8(55)$

Perform the multiplication:

$S_{16} = 440$

Final Answer: The number of terms $n = 16$ and the sum of the terms $S_n = 440$.

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:

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

Substituting the known values:

$a_n = 121 + (n - 1)(-4)$

$a_n = 121 - 4n + 4$

$a_n = 125 - 4n$

Step 3: Formulate the inequality to find the first negative term.

We are looking for the first negative term, which implies we need to solve for $n$ where $a_n < 0$:

$125 - 4n < 0$

Step 4: Solve the inequality for $n$.

Subtract $125$ from both sides:

$-4n < -125$

Divide both sides by $-4$. [Note: When dividing or multiplying an inequality by a negative number, the inequality sign reverses]:

$n > \frac{-125}{-4}$

$n > 31.25$

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

Since $n$ must be a positive integer representing the position of a term in the sequence, and $n > 31.25$, the smallest integer value for $n$ is $32$.

Step 6: Verification (Optional but recommended).

Calculate the $31^{st}$ term and the $32^{nd}$ term to ensure the result is correct:

For $n = 31$: $a_{31} = 125 - 4(31) = 125 - 124 = 1$ (This is the last positive term)

For $n = 32$: $a_{32} = 125 - 4(32) = 125 - 128 = -3$ (This is the first negative term)

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

Given: An arithmetic series $7 + 10\frac{1}{2} + 14 + \dots + 84$.

To Find: The sum of the given arithmetic series.

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

The given series is $7, 10\frac{1}{2}, 14, \dots, 84$.

Let the first term be $a = 7$.

The second term is $10\frac{1}{2} = \frac{21}{2} = 10.5$.

The common difference $d$ is calculated as $d = a_2 - a_1$.

$d = 10.5 - 7 = 3.5$ or $d = \frac{7}{2}$.

The last term (the $n^{th}$ term) is $a_n = l = 84$.

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:

$84 = 7 + (n - 1)\left(\frac{7}{2}\right)$

Subtract 7 from both sides:

$84 - 7 = (n - 1)\left(\frac{7}{2}\right)$

$77 = (n - 1)\left(\frac{7}{2}\right)$

Multiply both sides by $\frac{2}{7}$ to isolate $(n - 1)$:

$77 \times \frac{2}{7} = n - 1$

$11 \times 2 = n - 1$

$22 = n - 1$

$n = 23$

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

The formula for the sum of an arithmetic progression when the last term is known is:

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

Substituting $n = 23$, $a = 7$, and $l = 84$:

$S_{23} = \frac{23}{2}(7 + 84)$

$S_{23} = \frac{23}{2}(91)$

$S_{23} = \frac{23 \times 91}{2}$

Calculating the product $23 \times 91$:

$23 \times 90 = 2070$

$23 \times 1 = 23$

$2070 + 23 = 2093$

$S_{23} = \frac{2093}{2}$

$S_{23} = 1046.5$

Final Answer: The sum of the given arithmetic series is 1046.5 (or $1046\frac{1}{2}$).

Given:

1. Total number of steps ($n$) = $15$.

2. Length of each step ($l$) = $50$ m.

3. Rise of each step ($h$) = $\frac{1}{4}$ m.

4. Tread of each step ($w$) = $\frac{1}{2}$ m.

To Find:

The total volume of concrete required to build the entire terrace.

Step 1: Calculate the volume of individual steps.

The volume of a rectangular prism (each step) is given by the formula: $V = \text{length} \times \text{width} \times \text{height}$.

Volume of the 1st step ($V_1$) = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{50}{8} = 6.25$ m$^3$.

Volume of the 2nd step ($V_2$) = $50 \times \frac{1}{2} \times (\frac{1}{4} + \frac{1}{4}) = 50 \times \frac{1}{2} \times \frac{2}{4} = 2 \times 6.25 = 12.5$ m$^3$.

Volume of the 3rd step ($V_3$) = $50 \times \frac{1}{2} \times (\frac{1}{4} + \frac{1}{4} + \frac{1}{4}) = 50 \times \frac{1}{2} \times \frac{3}{4} = 3 \times 6.25 = 18.75$ m$^3$.

Step 2: Identify the Arithmetic Progression (AP).

The volumes form an AP where:

First term ($a$) = $6.25$

Common difference ($d$) = $V_2 - V_1 = 12.5 - 6.25 = 6.25$

Number of terms ($n$) = $15$

Step 3: Apply the Sum of an AP formula.

The sum of the first $n$ terms of an AP is given by: $S_n = \frac{n}{2} [2a + (n - 1)d]$.

Substituting the values:

$S_{15} = \frac{15}{2} [2(6.25) + (15 - 1)(6.25)]$

$S_{15} = \frac{15}{2} [12.5 + 14(6.25)]$

$S_{15} = \frac{15}{2} [12.5 + 87.5]$

$S_{15} = \frac{15}{2} [100]$

$S_{15} = 15 \times 50$

$S_{15} = 750$

Final Answer: The total volume of concrete required to build the terrace is 750 m$^3$.

Given: An Arithmetic Progression (AP) where the sum of the third term ($a_3$) and the seventh term ($a_7$) is $6$, and their product is $8$.

To Find: The sum of the first sixteen terms ($S_{16}$) of the AP.

Step 1: Defining the variables and formulas.
Let the first term of the AP be $a$ and the common difference be $d$.
The $n^{th}$ term of an AP is given by the formula: $a_n = a + (n - 1)d$.
The sum of the first $n$ terms of an AP is given by: $S_n = \frac{n}{2} [2a + (n - 1)d]$.

Step 2: Expressing the terms in terms of $a$ and $d$.
$a_3 = a + (3 - 1)d = a + 2d$
$a_7 = a + (7 - 1)d = a + 6d$

Step 3: Formulating the equations based on the given conditions.
Condition 1 (Sum): $a_3 + a_7 = 6$
$(a + 2d) + (a + 6d) = 6$
$2a + 8d = 6$
Dividing by 2: $a + 4d = 3 \implies a = 3 - 4d$ --- (Equation 1)

Condition 2 (Product): $a_3 \cdot a_7 = 8$
$(a + 2d)(a + 6d) = 8$ --- (Equation 2)

Step 4: Solving for $d$ and $a$.
Substitute Equation 1 into Equation 2:
$((3 - 4d) + 2d)((3 - 4d) + 6d) = 8$
$(3 - 2d)(3 + 2d) = 8$
Using the algebraic identity $(x - y)(x + y) = x^2 - y^2$:
$3^2 - (2d)^2 = 8$
$9 - 4d^2 = 8$
$-4d^2 = 8 - 9$
$-4d^2 = -1$
$d^2 = \frac{1}{4} \implies d = \pm \frac{1}{2}$

Case 1: If $d = \frac{1}{2}$
$a = 3 - 4(\frac{1}{2}) = 3 - 2 = 1$

Case 2: If $d = -\frac{1}{2}$
$a = 3 - 4(-\frac{1}{2}) = 3 + 2 = 5$

Step 5: Calculating $S_{16}$ for both cases.
Formula: $S_{16} = \frac{16}{2} [2a + (16 - 1)d] = 8[2a + 15d]$

For Case 1 ($a=1, d=1/2$):
$S_{16} = 8[2(1) + 15(\frac{1}{2})] = 8[2 + 7.5] = 8[9.5] = 76$

For Case 2 ($a=5, d=-1/2$):
$S_{16} = 8[2(5) + 15(-\frac{1}{2})] = 8[10 - 7.5] = 8[2.5] = 20$

Final Answer: The sum of the first sixteen terms is either 76 or 20.

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: An Arithmetic Progression (AP) with the first term $a = -4$ and the sixth term $a_6 = 6$. The sequence is $-4, \square, \square, \square, \square, 6$.

To find: The four missing terms in the sequence, which correspond to $a_2, a_3, a_4,$ and $a_5$.

Step 1: Identify the general formula for the $n^{th}$ term of an AP.
The formula for the $n^{th}$ term of an Arithmetic Progression is given by:
$a_n = a + (n - 1)d$
where $a$ is the first term, $d$ is the common difference, and $n$ is the position of the term.

Step 2: Formulate equations based on the given information.
We are given:
$a_1 = a = -4$
$a_6 = 6$
Using the formula $a_n = a + (n - 1)d$ for $n = 6$:
$a_6 = a + (6 - 1)d$
$6 = -4 + 5d$ [Substituting the known values of $a_6$ and $a$]

Step 3: Solve for the common difference ($d$).
$6 + 4 = 5d$ [Adding 4 to both sides]
$10 = 5d$
$d = \frac{10}{5}$ [Dividing both sides by 5]
$d = 2$

Step 4: Calculate the missing terms.
Now that we have $a = -4$ and $d = 2$, we can find the missing terms using $a_n = a + (n - 1)d$:

For the second term ($a_2$):
$a_2 = a + d = -4 + 2 = -2$

For the third term ($a_3$):
$a_3 = a + 2d = -4 + 2(2) = -4 + 4 = 0$

For the fourth term ($a_4$):
$a_4 = a + 3d = -4 + 3(2) = -4 + 6 = 2$

For the fifth term ($a_5$):
$a_5 = a + 4d = -4 + 4(2) = -4 + 8 = 4$

Step 5: Verification.
The sequence is $-4, -2, 0, 2, 4, 6$.
Checking the common difference: $-2 - (-4) = 2$; $0 - (-2) = 2$; $2 - 0 = 2$; $4 - 2 = 2$; $6 - 4 = 2$.
Since the common difference is constant, the values are correct.

Final Answer: The missing terms are -2, 0, 2, and 4. The complete AP is -4, -2, 0, 2, 4, 6.