Given:

• The number of trees planted by a section of a class is equal to the class number.
• There are 12 classes in total (Class I to Class XII).
• Each class has 3 sections.

To Find:

The total number of trees planted by all the students of all classes.

Step 1: Determining the number of trees planted by each class

Let $n$ be the class number, where $n \in \{1, 2, 3, \dots, 12\}$.

Since there are 3 sections for each class, the number of trees planted by a specific class $n$ is given by:

$T_n = 3 \times n$

Calculating the trees for each class:

• Class I: $3 \times 1 = 3$
• Class II: $3 \times 2 = 6$
• Class III: $3 \times 3 = 9$
• ...
• Class XII: $3 \times 12 = 36$

Step 2: Identifying the Arithmetic Progression (AP)

The sequence of the total trees planted by each class is: $3, 6, 9, \dots, 36$.

This sequence forms an Arithmetic Progression where:

• First term ($a$) = $3$
• Common difference ($d$) = $6 - 3 = 3$
• Number of terms ($n$) = $12$
• Last term ($l$ or $a_{12}$) = $36$

Step 3: Calculating the sum of the Arithmetic Progression

To find the total number of trees, we calculate the sum of the first $n$ terms of the AP using the formula:

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

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

Substituting the known values into the formula:

$S_{12} = \frac{12}{2} (3 + 36)$

Step 4: Performing the arithmetic operations

$S_{12} = 6 \times (39)$

$S_{12} = 234$

[Calculation breakdown: $6 \times 30 = 180$ and $6 \times 9 = 54$. Adding these: $180 + 54 = 234$.]

Final Answer: The total number of trees planted by the students is 234.

Given:

The first term of the Arithmetic Progression (AP), denoted by $a = 17$.

The last term of the AP, denoted by $l$ or $a_n = 350$.

The common difference of the AP, denoted by $d = 9$.

To Find:

1. The number of terms in the AP, denoted by $n$.

2. The sum of all terms in the AP, denoted by $S_n$.

Step 1: Finding the number of terms ($n$)

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

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

Substituting the given values into the formula:

$350 = 17 + (n - 1)9$

Subtract $17$ from both sides of the equation:

$350 - 17 = (n - 1)9$

$333 = (n - 1)9$

Divide both sides by $9$:

$\frac{333}{9} = n - 1$

$37 = n - 1$

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

$n = 37 + 1$

$n = 38$

[Since the number of terms must be a positive integer, $n=38$ is valid.]

Step 2: Finding the sum of the 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:

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

Substituting the values $n = 38$, $a = 17$, and $l = 350$:

$S_{38} = \frac{38}{2}(17 + 350)$

Simplify the fraction and the expression inside the parentheses:

$S_{38} = 19(367)$

Perform the multiplication:

$19 \times 367 = 6973$

Step 3: Verification of calculation

$19 \times 300 = 5700$

$19 \times 60 = 1140$

$19 \times 7 = 133$

$5700 + 1140 + 133 = 6973$

Final Answer: There are 38 terms in the AP, and their sum is 6973.

Given: The sum of the first $n$ terms of an Arithmetic Progression (AP) is given by the formula $S_n = 4n - n^2$.

To Find:
1. The first term ($a_1$ or $S_1$).
2. The sum of the first two terms ($S_2$).
3. The second term ($a_2$).
4. The third term ($a_3$).
5. The tenth term ($a_{10}$).
6. The $n$th term ($a_n$).

Step 1: Finding the first term ($a_1$)
By definition, the sum of the first term is the first term itself.
Given $S_n = 4n - n^2$.
For $n = 1$:
$S_1 = 4(1) - (1)^2$
$S_1 = 4 - 1 = 3$
Thus, the first term $a_1 = 3$.

Step 2: Finding the sum of the first two terms ($S_2$)
For $n = 2$:
$S_2 = 4(2) - (2)^2$
$S_2 = 8 - 4 = 4$
Thus, the sum of the first two terms is $4$.

Step 3: Finding the second term ($a_2$)
We know that the sum of the first two terms is the sum of the first term and the second term: $S_2 = a_1 + a_2$.
Substituting the known values:
$4 = 3 + a_2$
$a_2 = 4 - 3 = 1$
Thus, the second term $a_2 = 1$.

Step 4: Finding the third term ($a_3$)
First, calculate the sum of the first three terms ($S_3$):
$S_3 = 4(3) - (3)^2 = 12 - 9 = 3$
The $n$th term of an AP can be found using the relation $a_n = S_n - S_{n-1}$.
For $n = 3$:
$a_3 = S_3 - S_2$
$a_3 = 3 - 4 = -1$
Thus, the third term $a_3 = -1$.

Step 5: Finding the tenth term ($a_{10}$)
Using the relation $a_n = S_n - S_{n-1}$:
$a_{10} = S_{10} - S_9$
Calculate $S_{10}$: $S_{10} = 4(10) - (10)^2 = 40 - 100 = -60$
Calculate $S_9$: $S_9 = 4(9) - (9)^2 = 36 - 81 = -45$
$a_{10} = -60 - (-45)$
$a_{10} = -60 + 45 = -15$
Thus, the tenth term $a_{10} = -15$.

Step 6: Finding the $n$th term ($a_n$)
Using the relation $a_n = S_n - S_{n-1}$:
$S_n = 4n - n^2$
$S_{n-1} = 4(n-1) - (n-1)^2$
Expand $S_{n-1}$:
$S_{n-1} = 4n - 4 - (n^2 - 2n + 1)$
$S_{n-1} = 4n - 4 - n^2 + 2n - 1$
$S_{n-1} = -n^2 + 6n - 5$
Now, subtract $S_{n-1}$ from $S_n$:
$a_n = (4n - n^2) - (-n^2 + 6n - 5)$
$a_n = 4n - n^2 + n^2 - 6n + 5$
$a_n = 5 - 2n$

Final Answer:
The first term ($a_1$) is 3.
The sum of the first two terms ($S_2$) is 4.
The second term ($a_2$) is 1.
The third term ($a_3$) is -1.
The tenth term ($a_{10}$) is -15.
The $n$th term ($a_n$) is $5 - 2n$.

Given:

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

The last term of the AP, denoted by $l$ or $a_n = 45$.

The sum of the $n$ terms of the AP, denoted by $S_n = 400$.

To Find:

1. The number of terms ($n$).

2. The common difference ($d$).

Step 1: Finding the number of terms ($n$)

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

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

[Substituting the given values into the formula]

$400 = \frac{n}{2}(5 + 45)$

$400 = \frac{n}{2}(50)$

[Simplifying the expression inside the parentheses and dividing 50 by 2]

$400 = n \times 25$

$n = \frac{400}{25}$

$n = 16$

[Since $n$ represents the number of terms, it must be a positive integer]

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

We use the formula for the $n^{th}$ term of an AP:

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

[Substituting the known values: $a_n = 45$, $a = 5$, and $n = 16$]

$45 = 5 + (16 - 1)d$

$45 = 5 + 15d$

[Subtracting 5 from both sides of the equation]

$45 - 5 = 15d$

$40 = 15d$

[Dividing both sides by 15 to solve for $d$]

$d = \frac{40}{15}$

[Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, 5]

$d = \frac{8}{3}$

Final Answer: The number of terms is 16 and the common difference is $\frac{8}{3}$.

Given:

A spiral is formed by successive semicircles with centers alternating between points A and B. The radii of these semicircles form an arithmetic progression: $r_1 = 0.5\text{ cm}$, $r_2 = 1.0\text{ cm}$, $r_3 = 1.5\text{ cm}$, $r_4 = 2.0\text{ cm}$, and so on. The total number of semicircles is $n = 13$. The value of $\pi$ is given as $\frac{22}{7}$.

To Find:

The total length of the spiral, which is the sum of the lengths of the 13 consecutive semicircles.

Step 1: Determine the formula for the length of each semicircle.

The circumference of a full circle is $2\pi r$. Therefore, the length of a semicircle with radius $r$ is given by $l = \pi r$.

Let $l_1, l_2, l_3, \dots, l_{13}$ be the lengths of the 13 semicircles.

$l_1 = \pi r_1 = \pi(0.5)$

$l_2 = \pi r_2 = \pi(1.0)$

$l_3 = \pi r_3 = \pi(1.5)$

... and so on.

Step 2: Identify the Arithmetic Progression (AP).

The sequence of lengths is: $\pi(0.5), \pi(1.0), \pi(1.5), \dots$

This is an AP where:

First term ($a$) = $0.5\pi$

Common difference ($d$) = $l_2 - l_1 = \pi(1.0) - \pi(0.5) = 0.5\pi$

Number of terms ($n$) = $13$

Step 3: Apply the sum formula for an AP.

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

Substituting the known values:

$S_{13} = \frac{13}{2} [2(0.5\pi) + (13 - 1)(0.5\pi)]$

$S_{13} = \frac{13}{2} [1.0\pi + 12(0.5\pi)]$

$S_{13} = \frac{13}{2} [1.0\pi + 6.0\pi]$

$S_{13} = \frac{13}{2} [7\pi]$

Step 4: Calculate the final numerical value.

Substitute $\pi = \frac{22}{7}$ into the expression:

$S_{13} = \frac{13}{2} \times 7 \times \frac{22}{7}$

[Canceling the 7 in the numerator and denominator]:

$S_{13} = \frac{13}{2} \times 22$

$S_{13} = 13 \times 11$

$S_{13} = 143$

Final Answer: The total length of the spiral is 143 cm.

Given: The $n^{th}$ term of a sequence is defined by the expression $a_n = 3 + 4n$.

To Find:
1. Show that the sequence forms an Arithmetic Progression (AP).
2. Calculate the sum of the first 15 terms ($S_{15}$).

Step 1: Generating the terms of the sequence
To determine if the sequence is an AP, we calculate the first few terms by substituting $n = 1, 2, 3, ...$ into the given formula $a_n = 3 + 4n$.

For $n = 1$: $a_1 = 3 + 4(1) = 3 + 4 = 7$
For $n = 2$: $a_2 = 3 + 4(2) = 3 + 8 = 11$
For $n = 3$: $a_3 = 3 + 4(3) = 3 + 12 = 15$
For $n = 4$: $a_4 = 3 + 4(4) = 3 + 16 = 19$

Step 2: Verifying the Arithmetic Progression
A sequence is an AP if the difference between consecutive terms is constant. This constant is called the common difference ($d$).

Calculate the differences:
$a_2 - a_1 = 11 - 7 = 4$
$a_3 - a_2 = 15 - 11 = 4$
$a_4 - a_3 = 19 - 15 = 4$

[Since the difference $a_n - a_{n-1} = 4$ is constant for all $n$, the sequence is an AP with first term $a = 7$ and common difference $d = 4$.]

Step 3: Calculating the sum of the first 15 terms
The formula for the sum of the first $n$ terms of an AP is given by:
$S_n = \frac{n}{2} [2a + (n - 1)d]$

Here, $n = 15$, $a = 7$, and $d = 4$. Substituting these values into the formula:

$S_{15} = \frac{15}{2} [2(7) + (15 - 1)4]$

Step 4: Simplifying the expression
$S_{15} = \frac{15}{2} [14 + (14)4]$
$S_{15} = \frac{15}{2} [14 + 56]$
$S_{15} = \frac{15}{2} [70]$
$S_{15} = 15 \times 35$ [Dividing 70 by 2]

$S_{15} = 525$

Final Answer: The sequence forms an AP with a common difference of 4, and the sum of the first 15 terms is 525.

Given:

Total number of logs ($S_n$) = $200$

Number of logs in the bottom row ($a$) = $20$

Number of logs in the next row = $19$

Number of logs in the row after that = $18$

This forms an Arithmetic Progression (AP) where the common difference ($d$) = $19 - 20 = -1$.

To Find:

1. The number of rows ($n$).

2. The number of logs in the top row ($a_n$).

Step 1: Formulating the Equation

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

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

Substituting the given values into the formula:

$200 = \frac{n}{2} [2(20) + (n - 1)(-1)]$

Step 2: Solving for $n$

$200 = \frac{n}{2} [40 - n + 1]$

$200 = \frac{n}{2} [41 - n]$

Multiply both sides by $2$ to eliminate the fraction:

$400 = n(41 - n)$

$400 = 41n - n^2$

Rearranging into a standard quadratic equation form ($ax^2 + bx + c = 0$):

$n^2 - 41n + 400 = 0$

Step 3: Factoring the Quadratic Equation

We need two numbers that multiply to $400$ and add to $-41$. These numbers are $-16$ and $-25$.

$n^2 - 16n - 25n + 400 = 0$

$n(n - 16) - 25(n - 16) = 0$

$(n - 16)(n - 25) = 0$

Therefore, $n = 16$ or $n = 25$.

Step 4: Determining the Valid Value of $n$

Let us check the number of logs in the $n^{th}$ row using the formula $a_n = a + (n - 1)d$.

If $n = 25$:

$a_{25} = 20 + (25 - 1)(-1) = 20 - 24 = -4$

[Since the number of logs cannot be negative, $n = 25$ is rejected.]

If $n = 16$:

$a_{16} = 20 + (16 - 1)(-1) = 20 - 15 = 5$

[This is a valid number of logs.]

Step 5: Final Calculation

The number of rows is $16$.

The number of logs in the top row ($a_{16}$) is $5$.

Final Answer: The 200 logs are placed in 16 rows, and there are 5 logs in the top row.

Given: The $n^{th}$ term of a sequence is defined by the formula $a_n = 9 - 5n$.

To Find:
1. Show that the sequence forms an Arithmetic Progression (AP).
2. Calculate the sum of the first 15 terms ($S_{15}$).

Step 1: Generating the terms of the sequence
To determine if the sequence is an AP, we calculate the first few terms by substituting $n = 1, 2, 3, \dots$ into the given formula $a_n = 9 - 5n$.

For $n = 1$: $a_1 = 9 - 5(1) = 9 - 5 = 4$
For $n = 2$: $a_2 = 9 - 5(2) = 9 - 10 = -1$
For $n = 3$: $a_3 = 9 - 5(3) = 9 - 15 = -6$
For $n = 4$: $a_4 = 9 - 5(4) = 9 - 20 = -11$

Step 2: Verifying the Arithmetic Progression
A sequence is an AP if the difference between consecutive terms is constant. This constant is known as the common difference ($d$).

Calculate the differences:
$d_1 = a_2 - a_1 = -1 - 4 = -5$
$d_2 = a_3 - a_2 = -6 - (-1) = -6 + 1 = -5$
$d_3 = a_4 - a_3 = -11 - (-6) = -11 + 6 = -5$

[Since $a_{n} - a_{n-1} = -5$ for all $n$, the difference is constant.]

Therefore, the sequence $4, -1, -6, -11, \dots$ forms an AP with the first term $a = 4$ and common difference $d = -5$.

Step 3: Calculating the sum of the first 15 terms
The formula for the sum of the first $n$ terms of an AP is given by:
$S_n = \frac{n}{2} [2a + (n - 1)d]$

Here, $n = 15$, $a = 4$, and $d = -5$. Substituting these values into the formula:

$S_{15} = \frac{15}{2} [2(4) + (15 - 1)(-5)]$
$S_{15} = \frac{15}{2} [8 + (14)(-5)]$
$S_{15} = \frac{15}{2} [8 - 70]$
$S_{15} = \frac{15}{2} [-62]$

Step 4: Final Arithmetic Simplification
$S_{15} = 15 \times \left(\frac{-62}{2}\right)$
$S_{15} = 15 \times (-31)$
$S_{15} = -465$

Final Answer: The sequence forms an AP with a common difference of $-5$, and the sum of the first 15 terms is $-465$.

Given:

An Arithmetic Progression (AP) where:

• The common difference, $d = 7$
• The number of terms, $n = 22$
• The 22nd term, $a_{22} = 149$

To find:

The sum of the first 22 terms of the AP, denoted as $S_{22}$.

Step 1: Determine the first term ($a$) of the AP.

We use the general formula for the $n$-th term of an AP:

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

Substituting the given values ($n = 22$, $a_{22} = 149$, $d = 7$):

$149 = a + (22 - 1) \times 7$

$149 = a + (21) \times 7$

$149 = a + 147$

Subtracting 147 from both sides to isolate $a$:

$a = 149 - 147$

$a = 2$

Step 2: Calculate the sum of the first 22 terms ($S_{22}$).

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

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

Here, $n = 22$, $a = 2$, and $a_n = a_{22} = 149$.

Substituting these values into the formula:

$S_{22} = \frac{22}{2} (2 + 149)$

$S_{22} = 11 \times (151)$

Step 3: Perform the final multiplication.

$S_{22} = 11 \times 151$

$S_{22} = 1661$

Final Answer: The sum of the first 22 terms of the AP is 1661.

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: An Arithmetic Progression (AP) series: $-37, -33, -29, \dots$ and the number of terms to be summed, $n = 12$.

To Find: The sum of the first $12$ terms of the given AP ($S_{12}$).

Step 1: Identify the parameters of the Arithmetic Progression.

An Arithmetic Progression is defined by its first term ($a$) and its common difference ($d$).

The first term is given by the first element of the sequence:

$a = -37$

The common difference ($d$) is calculated by subtracting any term from the term that follows it:

$d = a_2 - a_1$

$d = -33 - (-37)$

$d = -33 + 37$

$d = 4$

Step 2: State the formula for the sum of the first $n$ terms of an AP.

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

Step 3: Substitute the known values into the formula.

Given $n = 12$, $a = -37$, and $d = 4$:

$S_{12} = \frac{12}{2} [2(-37) + (12 - 1)(4)]$

Step 4: Perform the arithmetic calculations.

First, simplify the fraction and the terms inside the brackets:

$S_{12} = 6 [2(-37) + (11)(4)]$

[Since $\frac{12}{2} = 6$ and $12 - 1 = 11$]

Next, calculate the products inside the brackets:

$S_{12} = 6 [-74 + 44]$

[Since $2 \times -37 = -74$ and $11 \times 4 = 44$]

Perform the addition inside the brackets:

$S_{12} = 6 [-30]$

[Since $-74 + 44 = -30$]

Finally, multiply the result by 6:

$S_{12} = -180$

Final Answer: The sum of the first 12 terms of the AP is -180.

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: 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.

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: An Arithmetic Progression (AP) series: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$ and the number of terms $n = 11$.

To Find: The sum of the first 11 terms ($S_{11}$) of the given AP.

Step 1: Identify the parameters of the Arithmetic Progression.

The first term ($a$) is the first element of the sequence:
$a = \frac{1}{15}$

The common difference ($d$) is calculated by subtracting the first term from the second term:
$d = a_2 - a_1 = \frac{1}{12} - \frac{1}{15}$

To subtract these fractions, we find the Least Common Multiple (LCM) of the denominators 12 and 15:
Multiples of 12: 12, 24, 36, 48, 60
Multiples of 15: 15, 30, 45, 60
LCM = 60

Converting the fractions:
$d = \frac{1 \times 5}{12 \times 5} - \frac{1 \times 4}{15 \times 4} = \frac{5}{60} - \frac{4}{60} = \frac{1}{60}$

Step 2: State the formula for the sum of the first $n$ terms of an AP.

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

Step 3: Substitute the known values into the formula.

Given $n = 11$, $a = \frac{1}{15}$, and $d = \frac{1}{60}$:
$S_{11} = \frac{11}{2} \left[ 2\left(\frac{1}{15}\right) + (11 - 1)\left(\frac{1}{60}\right) \right]$

Step 4: Perform the arithmetic calculations.

Simplify the expression inside the brackets:
$S_{11} = \frac{11}{2} \left[ \frac{2}{15} + 10\left(\frac{1}{60}\right) \right]$
$S_{11} = \frac{11}{2} \left[ \frac{2}{15} + \frac{10}{60} \right]$

Simplify the fraction $\frac{10}{60}$ to $\frac{1}{6}$:
$S_{11} = \frac{11}{2} \left[ \frac{2}{15} + \frac{1}{6} \right]$

Find the LCM of 15 and 6 to add the fractions inside the brackets:
Multiples of 15: 15, 30
Multiples of 6: 6, 12, 18, 24, 30
LCM = 30

Convert fractions to have a common denominator of 30:
$\frac{2}{15} = \frac{2 \times 2}{15 \times 2} = \frac{4}{30}$
$\frac{1}{6} = \frac{1 \times 5}{6 \times 5} = \frac{5}{30}$

Add the fractions:
$S_{11} = \frac{11}{2} \left[ \frac{4}{30} + \frac{5}{30} \right] = \frac{11}{2} \left[ \frac{9}{30} \right]$

Step 5: Final multiplication.

$S_{11} = \frac{11}{2} \times \frac{9}{30}$
Simplify $\frac{9}{30}$ by dividing both numerator and denominator by 3:
$\frac{9}{30} = \frac{3}{10}$
$S_{11} = \frac{11}{2} \times \frac{3}{10} = \frac{33}{20}$

Final Answer: The sum of the first 11 terms is $\frac{33}{20}$.

Given: The range of numbers is between $0$ and $50$. We are interested in the odd numbers within this range.

To Find: The sum of all odd numbers between $0$ and $50$.

Step 1: Identifying the Sequence

The odd numbers between $0$ and $50$ form an Arithmetic Progression (AP) starting from $1$ and ending at $49$.
The sequence is: $1, 3, 5, 7, \dots, 49$.

Here:
First term ($a$) = $1$
Common difference ($d$) = $3 - 1 = 2$
Last term ($a_n$ or $l$) = $49$

Step 2: Determining the Number of Terms ($n$)

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

Substituting the known values into the formula:
$49 = 1 + (n - 1)2$

Subtract $1$ from both sides:
$49 - 1 = (n - 1)2$
$48 = (n - 1)2$

Divide both sides by $2$:
$\frac{48}{2} = n - 1$
$24 = n - 1$

Add $1$ to both sides:
$n = 24 + 1$
$n = 25$

[Since there are $25$ terms in the sequence]

Step 3: Calculating the Sum of the AP

The formula for the sum of the first $n$ terms of an AP when the last term ($l$) is known is:
$S_n = \frac{n}{2}(a + l)$

Substituting $n = 25$, $a = 1$, and $l = 49$:
$S_{25} = \frac{25}{2}(1 + 49)$

Perform the addition inside the parentheses:
$S_{25} = \frac{25}{2}(50)$

Simplify the expression:
$S_{25} = 25 \times \left(\frac{50}{2}\right)$
$S_{25} = 25 \times 25$

Calculate the product:
$S_{25} = 625$

Final Answer: The sum of the odd numbers between 0 and 50 is 625.

Given:

In an Arithmetic Progression (AP):

1. The 3rd term, $a_3 = 15$

2. The sum of the first 10 terms, $S_{10} = 125$

To find:

1. The common difference, $d$

2. The 10th term, $a_{10}$

Step 1: Formulating equations based on the general term of an AP

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

For $n = 3$:

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

$15 = a + 2d$ --- (Equation 1)

Step 2: Formulating equations based on the sum of $n$ terms of an AP

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

For $n = 10$ and $S_{10} = 125$:

$125 = \frac{10}{2} [2a + (10 - 1)d]$

$125 = 5 [2a + 9d]$

Divide both sides by 5:

$25 = 2a + 9d$ --- (Equation 2)

Step 3: Solving the system of linear equations

From Equation 1, express $a$ in terms of $d$:

$a = 15 - 2d$

Substitute this expression for $a$ into Equation 2:

$25 = 2(15 - 2d) + 9d$

$25 = 30 - 4d + 9d$

$25 = 30 + 5d$

Subtract 30 from both sides:

$25 - 30 = 5d$

$-5 = 5d$

$d = -1$

Step 4: Finding the first term $a$

Substitute $d = -1$ back into the expression for $a$:

$a = 15 - 2(-1)$

$a = 15 + 2$

$a = 17$

Step 5: Calculating the 10th term $a_{10}$

Using the formula $a_n = a + (n - 1)d$ for $n = 10$:

$a_{10} = a + (10 - 1)d$

$a_{10} = 17 + 9(-1)$

$a_{10} = 17 - 9$

$a_{10} = 8$

Final Answer: The common difference $d = -1$ and the 10th term $a_{10} = 8$.

Given:

First term of the Arithmetic Progression ($a$) = $2$

Common difference ($d$) = $8$

Sum of $n$ terms ($S_n$) = $90$

To Find:

Number of terms ($n$) and the $n^{th}$ term ($a_n$).

Step 1: Formulating the equation for $n$ using the Sum formula

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

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

Substituting the given values into the formula:

$90 = \frac{n}{2} [2(2) + (n - 1)8]$

Step 2: Simplifying the algebraic expression

$90 = \frac{n}{2} [4 + 8n - 8]$

$90 = \frac{n}{2} [8n - 4]$

Factor out $2$ from the bracket:

$90 = \frac{n}{2} \cdot 2(4n - 2)$

$90 = n(4n - 2)$

$90 = 4n^2 - 2n$

Step 3: Solving the quadratic equation

Rearrange the equation into the standard form $an^2 + bn + c = 0$:

$4n^2 - 2n - 90 = 0$

Divide the entire equation by $2$ to simplify:

$2n^2 - n - 45 = 0$

Using the splitting the middle term method, we look for two numbers that multiply to $(2 \times -45) = -90$ and add to $-1$:

The numbers are $-10$ and $9$.

$2n^2 - 10n + 9n - 45 = 0$

$2n(n - 5) + 9(n - 5) = 0$

$(2n + 9)(n - 5) = 0$

This gives two possible values for $n$:

$2n + 9 = 0 \implies n = -\frac{9}{2}$

$n - 5 = 0 \implies n = 5$

[Since the number of terms $n$ must be a positive integer, we reject $n = -\frac{9}{2}$]

Therefore, $n = 5$.

Step 4: Finding the $n^{th}$ term ($a_n$)

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

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

Substitute $a = 2$, $n = 5$, and $d = 8$:

$a_5 = 2 + (5 - 1)8$

$a_5 = 2 + (4 \times 8)$

$a_5 = 2 + 32$

$a_5 = 34$

Final Answer: The number of terms $n = 5$ and the $n^{th}$ term $a_n = 34$.

Given: An Arithmetic Progression (AP) with the sequence $9, 17, 25, \dots$ and the sum of $n$ terms $S_n = 636$.

To find: The number of terms $n$ required such that the sum of the AP equals $636$.

Step 1: Identify the parameters of the Arithmetic Progression.

The first term ($a$) is the first number in the sequence: $a = 9$.

The common difference ($d$) is the difference between any two consecutive terms: $d = a_2 - a_1 = 17 - 9 = 8$.

The sum of $n$ terms is given as $S_n = 636$.

Step 2: State the formula for the sum of $n$ terms of an AP.

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

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

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

Step 3: Substitute the known values into the formula.

$636 = \frac{n}{2} [2(9) + (n - 1)8]$

$636 = \frac{n}{2} [18 + 8n - 8]$

$636 = \frac{n}{2} [8n + 10]$

Step 4: Simplify the equation to form a quadratic equation.

$636 = n(4n + 5)$ [Dividing the terms inside the bracket by 2]

$636 = 4n^2 + 5n$

$4n^2 + 5n - 636 = 0$

Step 5: Solve the quadratic equation using the quadratic formula.

The quadratic formula is $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=4, b=5, c=-636$.

$n = \frac{-5 \pm \sqrt{(5)^2 - 4(4)(-636)}}{2(4)}$

$n = \frac{-5 \pm \sqrt{25 + 10176}}{8}$

$n = \frac{-5 \pm \sqrt{10201}}{8}$

Since $\sqrt{10201} = 101$:

$n = \frac{-5 \pm 101}{8}$

Step 6: Evaluate the two possible values for $n$.

Case 1: $n = \frac{-5 + 101}{8} = \frac{96}{8} = 12$

Case 2: $n = \frac{-5 - 101}{8} = \frac{-106}{8} = -13.25$

Step 7: Interpret the results.

Since the number of terms ($n$) must be a positive integer, we reject the negative fractional value $n = -13.25$.

Therefore, $n = 12$.

Final Answer: 12 terms must be taken to give a sum of 636.

Given:

• Distance of the first potato from the bucket = $5$ m.
• Distance between consecutive potatoes = $3$ m.
• Total number of potatoes ($n$) = $10$.
• The competitor must run from the bucket to the potato and back to the bucket for each potato.

To Find:

The total distance the competitor has to run to collect all $10$ potatoes.

Step 1: Determine the distance for each potato run.

The competitor runs to the potato and returns to the bucket. Thus, the distance for each potato is $2 \times (\text{distance from bucket to potato})$.

• Distance for 1st potato: $d_1 = 2 \times 5 = 10$ m.
• Distance for 2nd potato: $d_2 = 2 \times (5 + 3) = 2 \times 8 = 16$ m.
• Distance for 3rd potato: $d_3 = 2 \times (5 + 3 + 3) = 2 \times 11 = 22$ m.

Step 2: Identify the Arithmetic Progression (AP).

The sequence of distances is $10, 16, 22, \dots$

• First term ($a$) = $10$.
• Common difference ($d$) = $16 - 10 = 6$.
• Number of terms ($n$) = $10$.

Step 3: Apply the sum formula for an AP.

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

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

Step 4: Substitute the values into the formula.

$S_{10} = \frac{10}{2} [2(10) + (10 - 1)6]$

$S_{10} = 5 [20 + (9 \times 6)]$

$S_{10} = 5 [20 + 54]$

$S_{10} = 5 [74]$

$S_{10} = 370$

Justification:

Since the distance increases by a constant amount ($2 \times 3 = 6$ m) for each subsequent potato, the sequence of distances forms an arithmetic progression. The sum of these distances represents the total path covered by the competitor.

Final Answer: The total distance the competitor has to run is 370 metres.