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 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:

1. The distance between consecutive rungs is $d_{gap} = 25$ cm.

2. The length of the bottom rung ($a_1$) = $45$ cm.

3. The length of the top rung ($a_n$) = $25$ cm.

4. The total distance between the top and bottom rungs = $2\frac{1}{2}$ m = $250$ cm.

To Find:

The total length of the wood required for all the rungs, which is the sum of the lengths of all rungs ($S_n$).

Step 1: Determining the total number of rungs ($n$)

The total number of rungs is calculated by dividing the total distance by the gap between rungs and adding 1 (to account for the starting rung).

$n = \frac{\text{Total Distance}}{\text{Gap between rungs}} + 1$

$n = \frac{250}{25} + 1$

$n = 10 + 1 = 11$

[Since there are 11 intervals of 25 cm to cover 250 cm, there must be 11 rungs.]

Step 2: Identifying the Arithmetic Progression (AP) parameters

Let the lengths of the rungs form an Arithmetic Progression where:

First term ($a$) = $45$ cm

Last term ($l$ or $a_n$) = $25$ cm

Number of terms ($n$) = $11$

Step 3: Calculating the total length of wood required

The total length of wood required is the sum of the lengths of all $n$ rungs. We use the formula for the sum of an AP when the first and last terms are known:

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

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

Substituting the known values:

$S_{11} = \frac{11}{2} (45 + 25)$

$S_{11} = \frac{11}{2} (70)$

$S_{11} = 11 \times 35$

$S_{11} = 385$

Step 4: Verification of units

Since the lengths were provided in centimeters (cm), the sum is also in centimeters.

Total length = $385$ cm.

Final Answer: The total length of the wood required for the rungs is 385 cm.

Given: A row of houses numbered consecutively from $1, 2, 3, \dots, 49$. Let $x$ be the number of a specific house in this row.

To Find: The value of $x$ such that the sum of the house numbers preceding $x$ is equal to the sum of the house numbers following $x$.

Step 1: Defining the Sums

The sequence of house numbers is an Arithmetic Progression (AP) where the first term $a = 1$ and the common difference $d = 1$.

The sum of the first $n$ terms of an AP is given by the formula: $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term.

The sum of all house numbers from $1$ to $49$ is: $S_{49} = \frac{49}{2}(1 + 49) = \frac{49}{2}(50) = 49 \times 25 = 1225$.

Step 2: Formulating the Equation

Let the sum of the houses preceding house $x$ be $S_{x-1}$. This is the sum of numbers from $1$ to $x-1$. $S_{x-1} = \frac{(x-1)}{2}(1 + x - 1) = \frac{(x-1)x}{2}$.

The sum of the houses following house $x$ is the total sum $S_{49}$ minus the sum of houses up to $x$. Sum following $x = S_{49} - S_x$. $S_x = \frac{x(x+1)}{2}$.

According to the problem, the sum of houses preceding $x$ equals the sum of houses following $x$: $S_{x-1} = S_{49} - S_x$.

Step 3: Algebraic Deduction

Substitute the expressions into the equation: $\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}$.

Multiply the entire equation by $2$ to clear the denominators: $x(x-1) = 2450 - x(x+1)$.

Expand the terms: $x^2 - x = 2450 - (x^2 + x)$. $x^2 - x = 2450 - x^2 - x$.

Add $x$ to both sides: $x^2 = 2450 - x^2$.

Add $x^2$ to both sides: $2x^2 = 2450$.

Step 4: Solving for $x$

Divide by $2$: $x^2 = 1225$.

Take the square root of both sides: $x = \sqrt{1225}$. Since $x$ must be a positive integer representing a house number: $x = 35$.

Justification: Sum of houses preceding $35$ (i.e., $1$ to $34$): $S_{34} = \frac{34(35)}{2} = 17 \times 35 = 595$. Sum of houses following $35$ (i.e., $36$ to $49$): $S_{49} - S_{35} = 1225 - \frac{35(36)}{2} = 1225 - (35 \times 18) = 1225 - 630 = 595$. Since $595 = 595$, the condition is satisfied.

Final Answer: The value of $x$ is 35.

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$^3$.