Given: An Arithmetic Progression (AP) series: $2, 7, 12, \dots$ and the number of terms to be summed, $n = 10$.

To find: The sum of the first $10$ terms of the given AP ($S_{10}$).

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 $a = 2$.
- The common difference $d$ is calculated by subtracting any term from the succeeding term: $d = a_2 - a_1$.
- $d = 7 - 2 = 5$.
- The number of terms $n = 10$.

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.
Substitute $n = 10$, $a = 2$, and $d = 5$ into the formula:
$S_{10} = \frac{10}{2} [2(2) + (10 - 1)5]$

Step 4: Perform the arithmetic calculations.
- First, simplify the fraction outside the brackets:
$\frac{10}{2} = 5$
- Next, simplify the terms inside the brackets:
$2(2) = 4$
$(10 - 1) = 9$
$9 \times 5 = 45$
- Now, combine the values inside the brackets:
$S_{10} = 5 [4 + 45]$
$S_{10} = 5 [49]$

Step 5: Final multiplication.
$S_{10} = 5 \times 49$
$S_{10} = 245$

Final Answer: The sum of the first 10 terms of the AP is 245.

Given:

The second term of the Arithmetic Progression ($a_2$) = $14$.

The third term of the Arithmetic Progression ($a_3$) = $18$.

The number of terms to be summed ($n$) = $51$.

To Find:

The sum of the first $51$ terms ($S_{51}$) of the Arithmetic Progression.

Step 1: Defining the variables and formulas

Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.

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

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

Step 2: Formulating equations based on given terms

Using the $n^{th}$ term formula for the second term ($n=2$):

$a_2 = a + (2 - 1)d$

$14 = a + d$ --- (Equation 1)

Using the $n^{th}$ term formula for the third term ($n=3$):

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

$18 = a + 2d$ --- (Equation 2)

Step 3: Solving for $a$ and $d$

Subtract Equation 1 from Equation 2 to eliminate $a$:

$(a + 2d) - (a + d) = 18 - 14$

$a - a + 2d - d = 4$

$d = 4$

Substitute the value of $d = 4$ into Equation 1 to find $a$:

$a + 4 = 14$

$a = 14 - 4$

$a = 10$

Step 4: Calculating the sum of the first 51 terms

We use the sum formula $S_n = \frac{n}{2} [2a + (n - 1)d]$ with $n = 51$, $a = 10$, and $d = 4$:

$S_{51} = \frac{51}{2} [2(10) + (51 - 1)4]$

$S_{51} = \frac{51}{2} [20 + (50)4]$

$S_{51} = \frac{51}{2} [20 + 200]$

$S_{51} = \frac{51}{2} [220]$

$S_{51} = 51 \times 110$

$S_{51} = 5610$

Final Answer: The sum of the first 51 terms of the Arithmetic Progression is 5610.

Given:

In an Arithmetic Progression (AP):

• First term, $a = 7$
• The $13^{th}$ term, $a_{13} = 35$

To Find:

• Common difference, $d$
• Sum of the first 13 terms, $S_{13}$

Step 1: Finding the common difference ($d$)

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

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

Substituting the given values for the $13^{th}$ term ($n = 13$):

$a_{13} = a + (13 - 1)d$

$35 = 7 + 12d$ [Substituting $a_{13} = 35$ and $a = 7$]

Subtracting 7 from both sides:

$35 - 7 = 12d$

$28 = 12d$

Dividing both sides by 12:

$d = \frac{28}{12}$

Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, 4:

$d = \frac{7}{3}$

Step 2: Finding the sum of the first 13 terms ($S_{13}$)

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

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

Here, $n = 13$, $a = 7$, and $a_{13} = 35$. Substituting these values into the formula:

$S_{13} = \frac{13}{2}(7 + 35)$

$S_{13} = \frac{13}{2}(42)$

Performing the division inside the expression:

$S_{13} = 13 \times \left(\frac{42}{2}\right)$

$S_{13} = 13 \times 21$

Calculating the product:

$S_{13} = 273$

Final Answer: The common difference $d = \frac{7}{3}$ and the sum of the first 13 terms $S_{13} = 273$.

Given:

The $n^{th}$ term of the Arithmetic Progression (AP), $a_n = 4$.

The common difference of the AP, $d = 2$.

The sum of the first $n$ terms of the AP, $S_n = -14$.

To find:

The number of terms, $n$, and the first term, $a$.

Step 1: Formulating the equations based on AP formulas.

We use the standard formulas for an Arithmetic Progression:

(i) The $n^{th}$ term formula: $a_n = a + (n - 1)d$

(ii) The sum of $n$ terms formula: $S_n = \frac{n}{2} [a + a_n]$

Step 2: Expressing $a$ in terms of $n$ using the $n^{th}$ term formula.

Substitute the given values $a_n = 4$ and $d = 2$ into the formula $a_n = a + (n - 1)d$:

$4 = a + (n - 1)(2)$

$4 = a + 2n - 2$

$a = 4 - 2n + 2$

$a = 6 - 2n$ --- (Equation 1)

Step 3: Substituting $a$ into the sum formula.

Substitute $S_n = -14$ and $a_n = 4$ into the formula $S_n = \frac{n}{2} [a + a_n]$:

$-14 = \frac{n}{2} [a + 4]$

Multiply both sides by 2:

$-28 = n(a + 4)$ --- (Equation 2)

Step 4: Solving for $n$.

Substitute Equation 1 into Equation 2:

$-28 = n[(6 - 2n) + 4]$

$-28 = n[10 - 2n]$

$-28 = 10n - 2n^2$

Rearrange the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:

$2n^2 - 10n - 28 = 0$

Divide the entire equation by 2 to simplify:

$n^2 - 5n - 14 = 0$

Factorize the quadratic equation by splitting the middle term:

$n^2 - 7n + 2n - 14 = 0$

$n(n - 7) + 2(n - 7) = 0$

$(n - 7)(n + 2) = 0$

This gives two possible values for $n$: $n = 7$ or $n = -2$.

[Since the number of terms $n$ must be a positive integer, we discard $n = -2$.]

Therefore, $n = 7$.

Step 5: Finding the first term $a$.

Substitute $n = 7$ back into Equation 1:

$a = 6 - 2(7)$

$a = 6 - 14$

$a = -8$

Final Answer:

The number of terms $n = 7$ and the first term $a = -8$.

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.

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:

The last term of the Arithmetic Progression (AP), denoted by $l$ (or $a_n$), is $28$.

The sum of the terms of the AP, denoted by $S_n$, is $144$.

The total number of terms in the AP, denoted by $n$, is $9$.

To find:

The first term of the AP, denoted by $a$.

Step 1: Identifying the relevant formula

For an Arithmetic Progression where the first term ($a$), the last term ($l$), and the total number of terms ($n$) are known, the sum of the terms ($S_n$) is given by the formula:

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

[This formula is derived from the general sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ by substituting $l = a + (n-1)d$].

Step 2: Substituting the given values into the formula

Given $S_n = 144$, $n = 9$, and $l = 28$, we substitute these into the equation:

$144 = \frac{9}{2}(a + 28)$

Step 3: Solving for $a$

To isolate $a$, we first multiply both sides of the equation by $2$ to eliminate the denominator:

$144 \times 2 = 9(a + 28)$

$288 = 9(a + 28)$

Next, divide both sides by $9$:

$\frac{288}{9} = a + 28$

[Performing the division: $288 \div 9 = 32$]

$32 = a + 28$

Finally, subtract $28$ from both sides to solve for $a$:

$a = 32 - 28$

$a = 4$

Step 4: Verification (Optional)

If $a = 4$ and $l = 28$ with $n = 9$, we can find the common difference $d$ using $l = a + (n-1)d$:

$28 = 4 + (9-1)d$

$24 = 8d$

$d = 3$

Checking the sum: $S_9 = \frac{9}{2}[2(4) + (9-1)3] = \frac{9}{2}[8 + 24] = \frac{9}{2}[32] = 9 \times 16 = 144$. The result is consistent.

Final Answer: The first term $a$ is 4.

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 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:

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:

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:

• Total sum of money ($S_n$) = ₹ $700$
• Number of prizes ($n$) = $7$
• The difference between consecutive prizes ($d$) = $-₹ 20$ (since each prize is ₹ 20 less than the preceding one).

To find:

The value of each of the seven prizes.

Step 1: Defining the variables and the Arithmetic Progression (AP)

Let the value of the first prize be $a$. Since each subsequent prize is ₹ 20 less than the previous one, the prizes form an Arithmetic Progression where:

• First term = $a$
• Common difference ($d$) = $-20$
• Number of terms ($n$) = $7$
• Sum of $n$ terms ($S_n$) = $700$

Step 2: Applying the Sum formula for an Arithmetic Progression

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

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

[Substituting the known values into the formula]:

$700 = \frac{7}{2} [2a + (7 - 1)(-20)]$

Step 3: Solving for $a$

Multiply both sides by $\frac{2}{7}$ to isolate the bracketed term:

$700 \times \frac{2}{7} = 2a + (6)(-20)$

$100 \times 2 = 2a - 120$

$200 = 2a - 120$

[Adding 120 to both sides]:

$200 + 120 = 2a$

$320 = 2a$

$a = \frac{320}{2}$

$a = 160$

The first prize is ₹ $160$.

Step 4: Calculating the value of each prize

Using the common difference $d = -20$, we calculate the seven prizes as follows:

• 1st Prize: $a = 160$
• 2nd Prize: $a + d = 160 - 20 = 140$
• 3rd Prize: $a + 2d = 160 - 40 = 120$
• 4th Prize: $a + 3d = 160 - 60 = 100$
• 5th Prize: $a + 4d = 160 - 80 = 80$
• 6th Prize: $a + 5d = 160 - 100 = 60$
• 7th Prize: $a + 6d = 160 - 120 = 40$

Verification:

Sum = $160 + 140 + 120 + 100 + 80 + 60 + 40 = 700$. The sum matches the given total.

Final Answer: The values of the seven prizes are ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, and ₹ 40.

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 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:

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:

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 sequence consists of the first 40 positive integers that are divisible by 6.

To Find: The sum of these first 40 positive integers.

Step 1: Identifying the Arithmetic Progression (AP)

The positive integers divisible by 6 are: $6, 12, 18, 24, \dots$

Let the sequence be represented as an Arithmetic Progression where:

• The first term ($a$) = $6$
• The common difference ($d$) = $12 - 6 = 6$
• The number of terms ($n$) = $40$

Step 2: Stating the Formula for the Sum of $n$ terms

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 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 into the formula

Substitute $n = 40$, $a = 6$, and $d = 6$ into the formula:

$S_{40} = \frac{40}{2} [2(6) + (40 - 1)6]$

Step 4: Performing the arithmetic calculations

First, simplify the fraction outside the brackets:

$S_{40} = 20 [2(6) + (39)6]$

Next, calculate the products inside the brackets:

$S_{40} = 20 [12 + 234]$

Add the terms inside the brackets:

$S_{40} = 20 [246]$

Finally, multiply the result by 20:

$S_{40} = 4920$

Step 5: Verification (Alternative Method)

The $n^{th}$ term ($a_n$) is given by $a_n = a + (n-1)d$.

$a_{40} = 6 + (40-1)6 = 6 + 234 = 240$.

Using the alternative sum formula $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term:

$S_{40} = \frac{40}{2}(6 + 240) = 20(246) = 4920$.

[Since both methods yield the same result, the calculation is verified.]

Final Answer: The sum of the first 40 positive integers divisible by 6 is 4920.

Given:

The sequence consists of the first 15 multiples of 8. This forms an Arithmetic Progression (AP) where:

• The first term ($a$) = $8 \times 1 = 8$
• The second term = $8 \times 2 = 16$
• The common difference ($d$) = $16 - 8 = 8$
• The number of terms ($n$) = $15$

To Find:

The sum of the first 15 multiples of 8 ($S_{15}$).

Step 1: Identifying the Arithmetic Progression

The sequence of the first 15 multiples of 8 is: $8, 16, 24, 32, \dots$

Here, the first term $a = 8$.

The common difference $d = a_2 - a_1 = 16 - 8 = 8$.

The number of terms $n = 15$.

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

Step 3: Substituting the Values

Substitute $n = 15$, $a = 8$, and $d = 8$ into the formula:

$S_{15} = \frac{15}{2} [2(8) + (15 - 1)8]$

Step 4: Performing Arithmetic Calculations

First, simplify the expression inside the parentheses:

$S_{15} = \frac{15}{2} [16 + (14)8]$

[Since $15 - 1 = 14$]

Next, calculate the product inside the bracket:

$S_{15} = \frac{15}{2} [16 + 112]$

[Since $14 \times 8 = 112$]

Add the terms inside the bracket:

$S_{15} = \frac{15}{2} [128]$

[Since $16 + 112 = 128$]

Step 5: Final Simplification

$S_{15} = 15 \times \frac{128}{2}$

$S_{15} = 15 \times 64$

[Since $128 \div 2 = 64$]

$S_{15} = 960$

[Since $15 \times 60 = 900$ and $15 \times 4 = 60$, $900 + 60 = 960$]

Final Answer: The sum of the first 15 multiples of 8 is 960.

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.