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 given conditions algebraically
Using the formula for the $n^{th}$ term:
$a_3 = a + (3 - 1)d = a + 2d$
$a_7 = a + (7 - 1)d = a + 6d$

Condition 1: Sum of the terms is $6$
$(a + 2d) + (a + 6d) = 6$
$2a + 8d = 6$
Dividing by $2$: $a + 4d = 3 \implies a = 3 - 4d$ --- (Equation 1)

Condition 2: Product of the terms is $8$
$(a + 2d)(a + 6d) = 8$ --- (Equation 2)

Step 3: Solving for $a$ and $d$
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 4: Calculating $S_{16}$ for both cases
The formula for $S_{16}$ is $S_{16} = \frac{16}{2} [2a + (16 - 1)d] = 8[2a + 15d]$.

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

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

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:

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.

Visual Representation:

Step 1: Determine the volume of individual steps.

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

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: $6.25, 12.5, 18.75, \dots$

Here, the first term ($a$) = $6.25$.

The common difference ($d$) = $12.5 - 6.25 = 6.25$.

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

Step 3: Apply the sum formula for an AP.

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

Substituting the known 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:

1. The distance between consecutive rungs = $25\text{ cm}$.

2. The length of the bottom rung ($a_1$) = $45\text{ cm}$.

3. The length of the top rung ($a_n$) = $25\text{ cm}$.

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

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: Determine the number of rungs ($n$)

The total distance between the top and bottom rungs is $250\text{ cm}$. Since the rungs are placed $25\text{ cm}$ apart, the number of intervals is $\frac{250}{25} = 10$.

According to the hint provided, the number of rungs ($n$) is given by:

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

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

$n = 10 + 1 = 11$

[Since there is one more rung than the number of intervals in a sequence of equally spaced items]

Step 2: Identify the Arithmetic Progression (AP) parameters

The lengths of the rungs form an Arithmetic Progression where:

First term ($a$) = $45\text{ cm}$

Last term ($l$ or $a_n$) = $25\text{ cm}$

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

Step 3: Calculate the sum of the lengths of the rungs

To find the total length of the wood required, we use the formula for the sum of the first $n$ terms of an Arithmetic Progression when the first and last terms are known:

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

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

Substituting the known values into the formula:

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

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

$S_{11} = 11 \times 35$

$S_{11} = 385$

Step 4: Conclusion

The total length of the wood required for the rungs is the sum of the lengths of all 11 rungs, which is $385\text{ cm}$.

Final Answer: 385 cm

Given: A row of houses numbered consecutively from $1$ to $49$. Let the house number in question be $x$.

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

Step 1: Defining the Sums

The house numbers form 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 houses preceding $x$ be $S_{x-1}$.

$S_{x-1} = \frac{(x-1)}{2}(1 + (x-1)) = \frac{(x-1)x}{2}$

The sum of houses following $x$ is the total sum minus the sum up to $x$:

Sum following $x = S_{49} - S_x$

Where $S_x = \frac{x(x+1)}{2}$.

According to the problem, $S_{x-1} = S_{49} - S_x$.

Step 3: Solving for $x$

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:

$x(x-1) = 2450 - x(x+1)$ $x^2 - x = 2450 - (x^2 + x)$ $x^2 - x = 2450 - x^2 - x$

Add $x$ to both sides:

$x^2 = 2450 - x^2$

Add $x^2$ to both sides:

$2x^2 = 2450$

Divide by $2$:

$x^2 = 1225$

Take the square root of both sides (since $x$ must be positive):

$x = \sqrt{1225}$ $x = 35$

Step 4: Verification

Sum of houses preceding $35$ ($1$ to $34$):

$S_{34} = \frac{34}{2}(1 + 34) = 17 \times 35 = 595$

Sum of houses following $35$ ($36$ to $49$):

$S_{49} - S_{35} = 1225 - \frac{35(36)}{2} = 1225 - (35 \times 18) = 1225 - 630 = 595$

Since $595 = 595$, the value is verified.

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