Progressions

Answer: Option B
Solution:
This is an arithmetic sequence, where:

– The first term $a = 20$,
– The common difference $d = 25 – 20 = 5$,
– The last term $l = 140$.

Step 1: Use the formula for the $n$-th term of an arithmetic sequence
The formula for the $n$-th term of an arithmetic sequence is:

$$l = a + (n-1) \times d$$

Where:
– $l$ is the last term,
– $a$ is the first term,
– $n$ is the number of terms,
– $d$ is the common difference.

Step 2: Substitute the known values
Substitute $l = 140$, $a = 20$, and $d = 5$ into the formula:

$$140 = 20 + (n-1) \times 5$$

Step 3: Solve for $n$
Simplify the equation:

$$140 = 20 + 5(n-1)$$ $$140 – 20 = 5(n-1)$$ $$120 = 5(n-1)$$ $$n – 1 = \frac{120}{5} = 24$$ $$n = 24 + 1 = 25$$

Final Answer:
There are 25 terms in the sequence.

Answer: Option C
Solution:
Let’s solve for the first term of the arithmetic progression (AP) using the given information.

Step 1: Define the variables
– Let the first term be $a$,
– The common difference be $d$.

We are given:
– The 8th term ($T_8$) is 39, so:

$$T_8 = a + 7d = 39$$

– The 12th term ($T_{12}$) is 59, so:

$$T_{12} = a + 11d = 59$$

Step 2: Set up the system of equations
We now have the following system of equations:
1. $a + 7d = 39$
2. $a + 11d = 59$

Step 3: Subtract the first equation from the second
By subtracting equation (1) from equation (2), we eliminate $a$:

$$(a + 11d) – (a + 7d) = 59 – 39$$

Simplify:

$$4d = 20$$

Step 4: Solve for $d$
Solve for $d$:

$$d = \frac{20}{4} = 5$$

Step 5: Find $a$
Now that we know $d = 5$, substitute this value into one of the original equations. Let’s use $a + 7d = 39$:

$$a + 7(5) = 39$$ $$a + 35 = 39$$ $$a = 39 – 35 = 4$$

Final Answer:
The first term of the arithmetic progression is 4.

Answer: Option C
Solution:
This is an arithmetic progression (AP), where:

– The first term $a = 20$,
– The common difference $d = 15 – 20 = -5$.

We are asked to find the 15th term of the sequence. The formula for the $n$-th term of an arithmetic progression is:

$$T_n = a + (n – 1) \cdot d$$

Step 1: Substitute the known values
We need to find the 15th term, so substitute $a = 20$, $d = -5$, and $n = 15$ into the formula:

$$T_{15} = 20 + (15 – 1) \cdot (-5)$$

Step 2: Simplify the expression
$$T_{15} = 20 + 14 \cdot (-5)$$ $$T_{15} = 20 – 70$$ $$T_{15} = -50$$

Final Answer:
The 15th term of the sequence is -50.

Answer: Option D
Solution:
We are given an arithmetic progression (AP) with the following information:

– The first term $a = 5$,
– The third term $T_3 = 15$.

We need to find the sum of the first 16 terms of the AP.

Step 1: Find the common difference $d$
The formula for the $n$-th term of an arithmetic progression is:

$$T_n = a + (n – 1) \cdot d$$

We are given the third term, $T_3 = 15$, and we can use the formula for the 3rd term:

$$T_3 = a + 2d$$

Substitute $a = 5$ and $T_3 = 15$:

$$15 = 5 + 2d$$

Solve for $d$:

$$15 – 5 = 2d$$ $$10 = 2d$$ $$d = 5$$

Step 2: Use the formula for the sum of the first $n$ terms
The formula for the sum of the first $n$ terms of an arithmetic progression is:

$$S_n = \frac{n}{2} \cdot \left( 2a + (n – 1) \cdot d \right)$$

We are asked to find the sum of the first 16 terms, so $n = 16$, $a = 5$, and $d = 5$. Substitute these values into the formula:

$$S_{16} = \frac{16}{2} \cdot \left( 2 \cdot 5 + (16 – 1) \cdot 5 \right)$$ $$S_{16} = 8 \cdot \left( 10 + 15 \cdot 5 \right)$$ $$S_{16} = 8 \cdot \left( 10 + 75 \right)$$ $$S_{16} = 8 \cdot 85$$ $$S_{16} = 680$$

Final Answer:
The sum of the first 16 terms of the AP is 680.

Answer: Option E
Solution:
We are given a geometric progression (GP) with the following terms:

– First term $a = 5$,
– Common ratio $r = \frac{20}{5} = 4$,
– The last term is $20480$.

Step 1: Use the formula for the $n$-th term of a geometric progression
The formula for the $n$-th term of a geometric progression is:

$$T_n = a \cdot r^{n-1}$$

We are given that the last term $T_n = 20480$, so:

$$20480 = 5 \cdot 4^{n-1}$$

Step 2: Solve for $n$
Divide both sides of the equation by 5:

$$\frac{20480}{5} = 4^{n-1}$$

$$4096 = 4^{n-1}$$

Now, express 4096 as a power of 4. Since $4^6 = 4096$, we have:

$$4^{n-1} = 4^6$$

This means:

$$n – 1 = 6$$

Solve for $n$:

$$n = 6 + 1 = 7$$

Final Answer:
There are 7 terms in the geometric progression.

Answer: Option B
Solution:
In this scenario, the boy’s wages follow a geometric progression where:

– The first term (on the first day) $a = 1$ rupee,
– The common ratio $r = 2$ (as each day’s wage doubles).

The wage on the $n$-th day is given by the formula for the $n$-th term of a geometric progression:

$$T_n = a \cdot r^{n-1} = 1 \cdot 2^{n-1} = 2^{n-1}$$

Step 1: Calculate the total wages for 20 days
We need to find the sum of the first 20 terms of this geometric progression. The sum $S_n$ of the first $n$ terms of a geometric progression is given by the formula:

$$S_n = \frac{a(r^n – 1)}{r – 1}$$

For this problem:
– $a = 1$,
– $r = 2$,
– $n = 20$.

Substitute these values into the formula:

$$S_{20} = \frac{1 \cdot (2^{20} – 1)}{2 – 1} = 2^{20} – 1$$

Step 2: Calculate $2^{20} – 1$
We know that:

$$2^{20} = 1048576$$

Therefore:

$$S_{20} = 1048576 – 1 = 1048575$$

Final Answer:
The boy will receive Rs. 1,048,575 if he works from February 1st to February 20th.

Answer: Option C
Solution:
In a geometric progression (GP), the general formula for the $n$-th term is:

$$T_n = a \cdot r^{n-1}$$

Where:
– $a$ is the first term,
– $r$ is the common ratio,
– $n$ is the term number.

Step 1: Given values for the fifth term and the first term
– The first term $a = 16$,
– The fifth term $T_5 = 81$.

Using the formula for the fifth term:

$$T_5 = a \cdot r^{5-1} = 16 \cdot r^4 = 81$$

Step 2: Solve for the common ratio $r$
From the equation:

$$16 \cdot r^4 = 81$$

Divide both sides by 16:

$$r^4 = \frac{81}{16}$$

$$r^4 = \left(\frac{9}{4}\right)$$

Now, take the fourth root of both sides:

$$r = \frac{3}{2}$$

So, the common ratio $r = \frac{3}{2}$.

Step 3: Find the fourth term $T_4$
We use the formula for the $n$-th term again to find the fourth term $T_4$:

$$T_4 = a \cdot r^{4-1} = 16 \cdot r^3$$

Substitute $r = \frac{3}{2}$:

$$T_4 = 16 \cdot \left(\frac{3}{2}\right)^3$$

$$T_4 = 16 \cdot \frac{27}{8}$$

$$T_4 = 16 \cdot 3.375 = 54$$

Final Answer:
The fourth term of the geometric progression is 54.

Answer: Option B
Solution:
We are given the following information about the arithmetic progression (AP):

– The 7th term $T_7 = 6$,
– The 21st term $T_{21} = -22$.

We need to find the 26th term $T_{26}$.

Step 1: Use the formula for the $n$-th term of an AP
The formula for the $n$-th term of an arithmetic progression is:

$$T_n = a + (n-1) \cdot d$$

Where:
– $T_n$ is the $n$-th term,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the term number.

Step 2: Set up the equations for $T_7$ and $T_{21}$
For the 7th term:

$$T_7 = a + (7-1) \cdot d = a + 6d = 6$$

For the 21st term:

$$T_{21} = a + (21-1) \cdot d = a + 20d = -22$$

So, we have the system of equations:
1. $a + 6d = 6$
2. $a + 20d = -22$

Step 3: Solve for $a$ and $d$
Subtract equation (1) from equation (2) to eliminate $a$:

$$(a + 20d) – (a + 6d) = -22 – 6$$

Simplify:

$$14d = -28$$

Solve for $d$:

$$d = \frac{-28}{14} = -2$$

Step 4: Substitute $d = -2$ into equation (1)
Substitute $d = -2$ into $a + 6d = 6$:

$$a + 6(-2) = 6$$

$$a – 12 = 6$$

Solve for $a$:

$$a = 6 + 12 = 18$$

Step 5: Find the 26th term $T_{26}$
Now, we can find the 26th term using the formula:

$$T_{26} = a + (26-1) \cdot d = a + 25d$$

Substitute $a = 18$ and $d = -2$:

$$T_{26} = 18 + 25(-2) = 18 – 50 = -32$$

Final Answer:
The 26th term of the arithmetic progression is -32.

Answer: Option C
Solution:
Let’s calculate the total distance the rubber ball travels before coming to rest.

Given:
– The ball is dropped from a height of 120 meters.
– After striking the floor, it rebounds to $\frac{4}{5}$ of the height it has fallen.
– The ball continues to rebound until it comes to rest.

Step 1: Calculate the total distance traveled
The ball follows a repetitive pattern: it falls a certain distance and then rebounds to a height that is $\frac{4}{5}$ of the previous height.

The total distance traveled can be broken into two parts:
1. The distance the ball falls.
2. The distance the ball rebounds after each fall.

Distance Fallen:
– Initially, the ball falls from a height of 120 meters.
– After each rebound, the ball falls again, but each time it falls to a height of $\frac{4}{5}$ of the previous height.

Distance Rebounded:
– After falling 120 meters, the ball rebounds to a height of $\frac{4}{5} \times 120 = 96$ meters.
– Then, it rebounds to $\frac{4}{5} \times 96 = 76.8$ meters, and so on.

Thus, the total distance traveled is the sum of all the falls and rebounds.

Step 2: Write the total distance as a series
The total distance traveled consists of:
– The initial fall of 120 meters,
– Then the ball falls 96 meters, 76.8 meters, and so on after each rebound.

So the total distance is:

$$\text{Total Distance} = 120 + 2 \times (96 + 76.8 + 61.44 + \cdots)$$

The factor of 2 accounts for both the falling and the rebounding distances.

Step 3: Sum the infinite geometric series
The distances after the first fall (96, 76.8, 61.44, …) form a geometric series with the first term $96$ and common ratio $\frac{4}{5}$.

The sum $S$ of an infinite geometric series is given by the formula:

$$S = \frac{a}{1 – r}$$

Where:
– $a = 96$ (the first term),
– $r = \frac{4}{5}$ (the common ratio).

Substitute the values:

$$S = \frac{96}{1 – \frac{4}{5}} = \frac{96}{\frac{1}{5}} = 96 \times 5 = 480$$

Step 4: Calculate the total distance
Now, the total distance is:

$$\text{Total Distance} = 120 + 2 \times 480 = 120 + 960 = 1080 \, \text{meters}$$

Final Answer:
The total distance the ball travels before coming to rest is 1080 meters.

Answer: Option A
Solution:
Let’s break down the problem step by step:

Step 1: Area of the first square
The side of the first square is given as 4 cm. The area of a square is given by the formula:

$$\text{Area of square} = \text{side}^2$$

For the first square:

$$\text{Area of first square} = 4^2 = 16 \, \text{cm}^2$$

Step 2: Side length of the second square
In the second step, a square is drawn by joining the midpoints of the sides of the first square. The diagonal of this new square will be equal to the side length of the first square.

For a square, the relationship between the side length $s$ and the diagonal $d$ is:

$$d = s \sqrt{2}$$

Here, the side length of the first square is 4 cm, so the diagonal of the second square is 4 cm. Now, we use the formula for the diagonal to find the side length $s_2$ of the second square:

$$4 = s_2 \sqrt{2}$$

Solving for $s_2$:

$$s_2 = \frac{4}{\sqrt{2}} = 4 \times \frac{\sqrt{2}}{2} = 2\sqrt{2} \, \text{cm}$$

Step 3: Area of the second square
Now, let’s find the area of the second square. The area of a square is $s^2$, so:

$$\text{Area of second square} = (2\sqrt{2})^2 = 4 \times 2 = 8 \, \text{cm}^2$$

Step 4: Area of the third square
The same process continues indefinitely. For each subsequent square, the side length is reduced by a factor of $\frac{1}{\sqrt{2}}$ because the diagonal of each new square equals the side length of the previous square.

Thus, the area of each successive square will be reduced by a factor of $\frac{1}{2}$ compared to the previous square, since the area is proportional to the square of the side length.

– The area of the third square will be $\frac{8}{2} = 4 \, \text{cm}^2$,
– The area of the fourth square will be $\frac{4}{2} = 2 \, \text{cm}^2$,
– The area of the fifth square will be $\frac{2}{2} = 1 \, \text{cm}^2$, and so on.

Step 5: Sum of all areas
The areas of the squares form a geometric series with the first term $a = 16$ and common ratio $r = \frac{1}{2}$.

The sum of an infinite geometric series is given by the formula:

$$S = \frac{a}{1 – r}$$

Substitute the values $a = 16$ and $r = \frac{1}{2}$:

$$S = \frac{16}{1 – \frac{1}{2}} = \frac{16}{\frac{1}{2}} = 16 \times 2 = 32$$

Final Answer:
The sum of the areas of all the squares is 32 cm².

Answer: Option B
Solution:
We are given the following information about the arithmetic progression (AP):

– The first term $a = 22$,
– The last term $l = -11$,
– The sum of the AP $S = 66$.

We need to find the number of terms $n$ in the AP.

Step 1: Use the formula for the sum of an AP
The sum of the first $n$ terms of an arithmetic progression is given by the formula:

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

Substitute the known values into the formula:

$$66 = \frac{n}{2} \cdot (22 + (-11))$$

Simplify the expression:

$$66 = \frac{n}{2} \cdot 11$$

Multiply both sides by 2 to eliminate the fraction:

$$132 = n \cdot 11$$

Solve for $n$:

$$n = \frac{132}{11} = 12$$

Final Answer:
The number of terms in the arithmetic progression is 12.

Answer: Option C
Solution:
We are given the following information about the arithmetic progression (AP):

– The 2nd term $T_2 = 17$,
– The 8th term $T_8 = -1$.

We need to find the 14th term $T_{14}$.

Step 1: Use the formula for the $n$-th term of an AP
The general formula for the $n$-th term of an arithmetic progression is:

$$T_n = a + (n-1) \cdot d$$

Where:
– $T_n$ is the $n$-th term,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the term number.

Step 2: Set up the equations for $T_2$ and $T_8$
Using the formula, for the 2nd term:

$$T_2 = a + (2-1) \cdot d = a + d = 17$$

This gives us the equation:

$$a + d = 17 \quad \text{(1)}$$

For the 8th term:

$$T_8 = a + (8-1) \cdot d = a + 7d = -1$$

This gives us the equation:

$$a + 7d = -1 \quad \text{(2)}$$ Step 3: Solve the system of equations
Now we have the system of equations:
1. $a + d = 17$,
2. $a + 7d = -1$.

Subtract equation (1) from equation (2):

$$(a + 7d) – (a + d) = -1 – 17$$

Simplify:

$$6d = -18$$

Solve for $d$:

$$d = \frac{-18}{6} = -3$$

Step: Find the first term $a$
Substitute $d = -3$ into equation (1):

$$a + (-3) = 17$$

$$a = 17 + 3 = 20$$

Step 5: Find the 14th term $T_{14}$
Now that we know $a = 20$ and $d = -3$, we can find the 14th term using the formula for the $n$-th term:

$$T_{14} = a + (14-1) \cdot d = a + 13d$$

Substitute the values of $a$ and $d$:

$$T_{14} = 20 + 13(-3) = 20 – 39 = -19$$

Final Answer:
The 14th term of the arithmetic progression is -19.

Answer: Option D
Solution:
We are given the following information about the arithmetic progression (AP):

– The 2nd term $T_2 = 8$,
– The 6th term $T_6 = 20$.

We need to find the 20th term $T_{20}$.

Step 1: Use the formula for the $n$-th term of an AP
The general formula for the $n$-th term of an arithmetic progression is:

$$T_n = a + (n-1) \cdot d$$

Where:
– $T_n$ is the $n$-th term,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the term number.

Step 2: Set up the equations for $T_2$ and $T_6$
For the 2nd term:

$$T_2 = a + (2-1) \cdot d = a + d = 8$$

This gives us the equation:

$$a + d = 8 \quad \text{(1)}$$

For the 6th term:

$$T_6 = a + (6-1) \cdot d = a + 5d = 20$$

This gives us the equation:

$$a + 5d = 20 \quad \text{(2)}$$

Step 3: Solve the system of equations
Now we have the system of equations:
1. $a + d = 8$,
2. $a + 5d = 20$.

Subtract equation (1) from equation (2):

$$(a + 5d) – (a + d) = 20 – 8$$

Simplify:

$$4d = 12$$

Solve for $d$:

$$d = \frac{12}{4} = 3$$

Step 4: Find the first term $a$
Substitute $d = 3$ into equation (1):

$$a + 3 = 8$$

$$a = 8 – 3 = 5$$

Step 5: Find the 20th term $T_{20}$
Now that we know $a = 5$ and $d = 3$, we can find the 20th term using the formula for the $n$-th term:

$$T_{20} = a + (20-1) \cdot d = a + 19d$$

Substitute the values of $a$ and $d$:

$$T_{20} = 5 + 19(3) = 5 + 57 = 62$$

Final Answer:
The 20th term of the arithmetic progression is 62.

Answer: Option A
Solution:
To find the sum of the first 17 terms of an arithmetic progression, we can use the formula for the sum of the first $n$ terms of an arithmetic progression:

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

Where:
– $S_n$ is the sum of the first $n$ terms,
– $n$ is the number of terms,
– $a$ is the first term,
– $l$ is the last term.

Given:
– The first term $a = -20$,
– The last term $l = 28$,
– The number of terms $n = 17$.

Step 1: Substitute the values into the formula

$$S_{17} = \frac{17}{2} \cdot (-20 + 28)$$

$$S_{17} = \frac{17}{2} \cdot 8$$

$$S_{17} = 17 \cdot 4 = 68$$

Final Answer:
The sum of the first 17 terms of the arithmetic progression is 68.

Answer: Option B
Solution:
We are given the following information about the arithmetic progression (AP):

– The 4th term $T_4 = 11$,
– The 7th term $T_7 = -4$.

We need to find the 15th term $T_{15}$.

Step 1: Use the formula for the $n$-th term of an AP
The general formula for the $n$-th term of an arithmetic progression is:

$$T_n = a + (n-1) \cdot d$$

Where:
– $T_n$ is the $n$-th term,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the term number.

Step 2: Set up the equations for $T_4$ and $T_7$
For the 4th term:

$$T_4 = a + (4-1) \cdot d = a + 3d = 11$$

This gives us the equation:

$$a + 3d = 11 \quad \text{(1)}$$

For the 7th term:

$$T_7 = a + (7-1) \cdot d = a + 6d = -4$$

This gives us the equation:

$$a + 6d = -4 \quad \text{(2)}$$

Step 3: Solve the system of equations
Now we have the system of equations:
1. $a + 3d = 11$,
2. $a + 6d = -4$.

Subtract equation (1) from equation (2):

$$(a + 6d) – (a + 3d) = -4 – 11$$

Simplify:

$$3d = -15$$

Solve for $d$:

$$d = \frac{-15}{3} = -5$$

Step 4: Find the first term $a$
Substitute $d = -5$ into equation (1):

$$a + 3(-5) = 11$$

$$a – 15 = 11$$

$$a = 11 + 15 = 26$$

Step 5: Find the 15th term $T_{15}$
Now that we know $a = 26$ and $d = -5$, we can find the 15th term using the formula for the $n$-th term:

$$T_{15} = a + (15-1) \cdot d = a + 14d$$

Substitute the values of $a$ and $d$:

$$T_{15} = 26 + 14(-5) = 26 – 70 = -44$$

Final Answer:
The 15th term of the arithmetic progression is -44.

Answer: Option C
Solution:
We are given the following information about the arithmetic progression (AP):

– The 3rd term $T_3 = -13$,
– The 8th term $T_8 = 2$.

We need to find the 14th term $T_{14}$.

Step 1: Use the formula for the $n$-th term of an AP
The general formula for the $n$-th term of an arithmetic progression is:

$$T_n = a + (n-1) \cdot d$$

Where:
– $T_n$ is the $n$-th term,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the term number.

Step 2: Set up the equations for $T_3$ and $T_8$
For the 3rd term:

$$T_3 = a + (3-1) \cdot d = a + 2d = -13$$

This gives us the equation:

$$a + 2d = -13 \quad \text{(1)}$$

For the 8th term:

$$T_8 = a + (8-1) \cdot d = a + 7d = 2$$

This gives us the equation:

$$a + 7d = 2 \quad \text{(2)}$$

Step 3: Solve the system of equations
Now we have the system of equations:
1. $a + 2d = -13$,
2. $a + 7d = 2$.

Subtract equation (1) from equation (2):

$$(a + 7d) – (a + 2d) = 2 – (-13)$$

Simplify:

$$5d = 15$$

Solve for $d$:

$$d = \frac{15}{5} = 3$$

Step 4: Find the first term $a$
Substitute $d = 3$ into equation (1):

$$a + 2(3) = -13$$

$$a + 6 = -13$$

$$a = -13 – 6 = -19$$

Step 5: Find the 14th term $T_{14}$
Now that we know $a = -19$ and $d = 3$, we can find the 14th term using the formula for the $n$-th term:

$$T_{14} = a + (14-1) \cdot d = a + 13d$$

Substitute the values of $a$ and $d$:

$$T_{14} = -19 + 13(3) = -19 + 39 = 20$$

Final Answer:
The 14th term of the arithmetic progression is 20.

Answer: Option B
Solution:
We are given the following information about the arithmetic progression (AP):

– The 3rd term $T_3 = -1$,
– The 8th term $T_8 = 19$.

We need to find the sum of the first 11 terms, $S_{11}$.

Step 1: Use the formula for the $n$-th term of an AP
The general formula for the $n$-th term of an arithmetic progression is:

$$T_n = a + (n-1) \cdot d$$

Where:
– $T_n$ is the $n$-th term,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the term number.

Step 2: Set up the equations for $T_3$ and $T_8$
For the 3rd term:

$$T_3 = a + (3-1) \cdot d = a + 2d = -1$$

This gives us the equation:

$$a + 2d = -1 \quad \text{(1)}$$

For the 8th term:

$$T_8 = a + (8-1) \cdot d = a + 7d = 19$$

This gives us the equation:

$$a + 7d = 19 \quad \text{(2)}$$

Step 3: Solve the system of equations
Now we have the system of equations:
1. $a + 2d = -1$,
2. $a + 7d = 19$.

Subtract equation (1) from equation (2):

$$(a + 7d) – (a + 2d) = 19 – (-1)$$

Simplify:

$$5d = 20$$

Solve for $d$:

$$d = \frac{20}{5} = 4$$

Step 4: Find the first term $a$
Substitute $d = 4$ into equation (1):

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

$$a + 8 = -1$$

$$a = -1 – 8 = -9$$

Step 5: Find the sum of the first 11 terms $S_{11}$
The sum of the first $n$ terms of an arithmetic progression is given by the formula:

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

Substitute $n = 11$, $a = -9$, and $d = 4$:

$$S_{11} = \frac{11}{2} \cdot \left( 2(-9) + (11-1) \cdot 4 \right)$$

$$S_{11} = \frac{11}{2} \cdot \left( -18 + 40 \right)$$

$$S_{11} = \frac{11}{2} \cdot 22$$

$$S_{11} = 11 \cdot 11 = 121$$

Final Answer:
The sum of the first 11 terms of the arithmetic progression is 121.

Answer: Option A
Solution:
To find the sum of the first 13 terms of an arithmetic progression, we can use the formula for the sum of the first $n$ terms of an arithmetic progression:

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

Where:
– $S_n$ is the sum of the first $n$ terms,
– $a$ is the first term,
– $l$ is the last term,
– $n$ is the number of terms.

Given:
– The first term $a = -10$,
– The last term $l = 26$,
– The number of terms $n = 13$.

Step 1: Substitute the values into the formula

$$S_{13} = \frac{13}{2} \cdot (-10 + 26)$$

$$S_{13} = \frac{13}{2} \cdot 16$$

$$S_{13} = 13 \cdot 8 = 104$$

Final Answer:
The sum of the first 13 terms of the arithmetic progression is 104.

Answer: Option C
Solution:
To find the sum of the first 12 terms of an arithmetic progression (AP), we can use the formula for the sum of an arithmetic series:

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

Where:
– $S_n$ is the sum of the first $n$ terms,
– $n$ is the number of terms,
– $a$ is the first term,
– $l$ is the last term.

Given:
– The first term $a = -19$,
– The last term $l = 36$,
– The number of terms $n = 12$.

Now substitute the values into the formula:

$$S_{12} = \frac{12}{2} \times (-19 + 36)$$

$$S_{12} = 6 \times (17)$$

$$S_{12} = 102$$

So, the sum of the first 12 terms of the arithmetic progression is 102.

Answer: Option C
Solution:
To solve this, we need to find the 15th term of the arithmetic progression (AP). The formula for the $n$-th term of an AP is:

$$a_n = a_1 + (n – 1) \times d$$

Where:
– $a_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

We are given:
– The 3rd term, $a_3 = -9$,
– The 7th term, $a_7 = 11$.

Using the formula for the 3rd term:

$$a_3 = a_1 + (3 – 1) \times d = a_1 + 2d$$ $$-9 = a_1 + 2d \quad \text{(Equation 1)}$$

Using the formula for the 7th term:

$$a_7 = a_1 + (7 – 1) \times d = a_1 + 6d$$ $$11 = a_1 + 6d \quad \text{(Equation 2)}$$

Now, solve these two equations simultaneously.

Step 1: Subtract Equation 1 from Equation 2:

$$(a_1 + 6d) – (a_1 + 2d) = 11 – (-9)$$ $$4d = 20$$ $$d = 5$$

Step 2: Substitute $d = 5$ into Equation 1:

$$-9 = a_1 + 2 \times 5$$ $$-9 = a_1 + 10$$ $$a_1 = -9 – 10 = -19$$

So, the first term $a_1 = -19$ and the common difference $d = 5$.

Step 3: Find the 15th term:

Using the formula for the $n$-th term:

$$a_{15} = a_1 + (15 – 1) \times d$$ $$a_{15} = -19 + 14 \times 5$$ $$a_{15} = -19 + 70$$ $$a_{15} = 51$$

So, the 15th term is 51.

Answer: Option C
Solution:
We are given that the 3rd term of an arithmetic progression (AP) is -13, and the 6th term is -4. We need to find the sum of the first 12 terms.

Step 1: Use the formula for the $n$-th term of an AP
The general formula for the $n$-th term of an arithmetic progression is:

$$a_n = a_1 + (n – 1) \times d$$

Where:
– $a_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

Step 2: Set up equations for the 3rd and 6th terms
For the 3rd term ($a_3 = -13$):

$$a_3 = a_1 + (3 – 1) \times d = a_1 + 2d$$ $$-13 = a_1 + 2d \quad \text{(Equation 1)}$$

For the 6th term ($a_6 = -4$):

$$a_6 = a_1 + (6 – 1) \times d = a_1 + 5d$$ $$-4 = a_1 + 5d \quad \text{(Equation 2)}$$

Step 3: Solve the system of equations
We now have two equations:

1. $-13 = a_1 + 2d$
2. $-4 = a_1 + 5d$

Step 3a: Subtract Equation 1 from Equation 2:**

$$(a_1 + 5d) – (a_1 + 2d) = -4 – (-13)$$ $$3d = 9$$ $$d = 3$$

Step 3b: Substitute $d = 3$ into Equation 1:**

$$-13 = a_1 + 2 \times 3$$ $$-13 = a_1 + 6$$ $$a_1 = -13 – 6 = -19$$

So, the first term $a_1 = -19$ and the common difference $d = 3$.

Step 4: Calculate the sum of the first 12 terms
The formula for the sum of the first $n$ terms of an arithmetic progression is:

$$S_n = \frac{n}{2} \times (a_1 + a_n)$$

Where:
– $S_n$ is the sum of the first $n$ terms,
– $a_1$ is the first term,
– $a_n$ is the $n$-th term.

We are looking for the sum of the first 12 terms, so we need to find $a_{12}$.

Using the formula for the $n$-th term:

$$a_{12} = a_1 + (12 – 1) \times d$$ $$a_{12} = -19 + 11 \times 3$$ $$a_{12} = -19 + 33 = 14$$

Now, substitute into the sum formula:

$$S_{12} = \frac{12}{2} \times (-19 + 14)$$ $$S_{12} = 6 \times (-5)$$ $$S_{12} = -30$$

Final Answer:
The sum of the first 12 terms is -30.

Answer: Option A
Solution:
We are given that the 3rd term of an arithmetic progression (AP) is 13 and the 5th term is 21. We need to find the 13th term.

Step 1: Use the formula for the $n$-th term of an AP
The formula for the $n$-th term of an arithmetic progression is:

$$a_n = a_1 + (n – 1) \times d$$

Where:
– $a_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

Step 2: Set up equations for the 3rd and 5th terms
For the 3rd term ($a_3 = 13$):

$$a_3 = a_1 + (3 – 1) \times d = a_1 + 2d$$ $$13 = a_1 + 2d \quad \text{(Equation 1)}$$

For the 5th term ($a_5 = 21$):

$$a_5 = a_1 + (5 – 1) \times d = a_1 + 4d$$ $$21 = a_1 + 4d \quad \text{(Equation 2)}$$

Step 3: Solve the system of equations
We now have two equations:

1. $13 = a_1 + 2d$
2. $21 = a_1 + 4d$

Step 3a: Subtract Equation 1 from Equation 2:

$$(a_1 + 4d) – (a_1 + 2d) = 21 – 13$$ $$2d = 8$$ $$d = 4$$

Step 3b: Substitute $d = 4$ into Equation 1:

$$13 = a_1 + 2 \times 4$$ $$13 = a_1 + 8$$ $$a_1 = 13 – 8 = 5$$

So, the first term $a_1 = 5$ and the common difference $d = 4$.

Step 4: Find the 13th term
Now, we can use the formula for the $n$-th term to find the 13th term:

$$a_{13} = a_1 + (13 – 1) \times d$$ $$a_{13} = 5 + 12 \times 4$$ $$a_{13} = 5 + 48$$ $$a_{13} = 53$$

Final Answer:
The 13th term is 53.

Answer: Option D
Solution:
We are given that the 3rd term of an arithmetic progression (AP) is 13, and the 6th term is -5. We need to find the 11th term.

Step 1: Use the formula for the $n$-th term of an AP
The formula for the $n$-th term of an arithmetic progression is:

$$a_n = a_1 + (n – 1) \times d$$

Where:
– $a_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

Step 2: Set up equations for the 3rd and 6th terms
For the 3rd term ($a_3 = 13$):

$$a_3 = a_1 + (3 – 1) \times d = a_1 + 2d$$ $$13 = a_1 + 2d \quad \text{(Equation 1)}$$

For the 6th term ($a_6 = -5$):

$$a_6 = a_1 + (6 – 1) \times d = a_1 + 5d$$ $$-5 = a_1 + 5d \quad \text{(Equation 2)}$$

Step 3: Solve the system of equations
We now have two equations:

1. $13 = a_1 + 2d$
2. $-5 = a_1 + 5d$

Step 3a: Subtract Equation 1 from Equation 2:

$$(a_1 + 5d) – (a_1 + 2d) = -5 – 13$$ $$3d = -18$$ $$d = -6$$

Step 3b: Substitute $d = -6$ into Equation 1:

$$13 = a_1 + 2 \times (-6)$$ $$13 = a_1 – 12$$ $$a_1 = 13 + 12 = 25$$

So, the first term $a_1 = 25$ and the common difference $d = -6$.

Step 4: Find the 11th term
Now, we can use the formula for the $n$-th term to find the 11th term:

$$a_{11} = a_1 + (11 – 1) \times d$$ $$a_{11} = 25 + 10 \times (-6)$$ $$a_{11} = 25 – 60$$ $$a_{11} = -35$$

Final Answer:
The 11th term is -35.

Answer: Option D
Solution:
We are given that the 3rd term of an arithmetic progression (AP) is -8 and the 9th term is 10. We need to find the 16th term.

Step 1: Use the formula for the $n$-th term of an AP
The general formula for the $n$-th term of an arithmetic progression is:

$$a_n = a_1 + (n – 1) \times d$$

Where:
– $a_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

Step 2: Set up equations for the 3rd and 9th terms
For the 3rd term ($a_3 = -8$):

$$a_3 = a_1 + (3 – 1) \times d = a_1 + 2d$$ $$-8 = a_1 + 2d \quad \text{(Equation 1)}$$

For the 9th term ($a_9 = 10$):

$$a_9 = a_1 + (9 – 1) \times d = a_1 + 8d$$ $$10 = a_1 + 8d \quad \text{(Equation 2)}$$

Step 3: Solve the system of equations
We now have two equations:

1. $-8 = a_1 + 2d$
2. $10 = a_1 + 8d$

**Step 3a: Subtract Equation 1 from Equation 2:**

$$(a_1 + 8d) – (a_1 + 2d) = 10 – (-8)$$ $$6d = 18$$ $$d = 3$$

**Step 3b: Substitute $d = 3$ into Equation 1:**

$$-8 = a_1 + 2 \times 3$$ $$-8 = a_1 + 6$$ $$a_1 = -8 – 6 = -14$$

So, the first term $a_1 = -14$ and the common difference $d = 3$.

Step 4: Find the 16th term
Now, we can use the formula for the $n$-th term to find the 16th term:

$$a_{16} = a_1 + (16 – 1) \times d$$ $$a_{16} = -14 + 15 \times 3$$ $$a_{16} = -14 + 45$$ $$a_{16} = 31$$

Final Answer:
The 16th term is 31.

Answer: Option D
Solution:
The arithmetic mean of a set of numbers is calculated by summing all the numbers and dividing by the number of terms.

We are given the numbers 7, 5, 13, $x$, and 9, and we are told their arithmetic mean is 10. To find $x$, we use the formula for the arithmetic mean:

$$\text{Arithmetic mean} = \frac{\text{Sum of all terms}}{\text{Number of terms}}$$

The sum of the terms is:

$$7 + 5 + 13 + x + 9$$

So, the arithmetic mean is:

$$\frac{7 + 5 + 13 + x + 9}{5} = 10$$

Simplifying the sum of the numbers:

$$\frac{34 + x}{5} = 10$$

Now, multiply both sides by 5 to eliminate the denominator:

$$34 + x = 50$$

Next, solve for $x$:

$$x = 50 – 34$$ $$x = 16$$

Final Answer:
The value of $x$ is 16.

Answer: Option B
Solution:
To find the sum of the first 9 terms of an arithmetic progression (AP), we can use the formula for the sum of an arithmetic series:

$$S_n = \frac{n}{2} \times (a_1 + a_n)$$

Where:
– $S_n$ is the sum of the first $n$ terms,
– $n$ is the number of terms,
– $a_1$ is the first term,
– $a_n$ is the last term.

We are given:
– The first term $a_1 = 7$,
– The last term $a_9 = 55$,
– The number of terms $n = 9$.

Substitute the values into the formula:

$$S_9 = \frac{9}{2} \times (7 + 55)$$

$$S_9 = \frac{9}{2} \times 62$$

$$S_9 = 9 \times 31$$

$$S_9 = 279$$

Final Answer:
The sum of the first 9 terms is 279.

Answer: Option A
Solution:
We are given that 7 times the 7th term of an arithmetic progression (AP) is equal to 11 times its 11th term. We need to find the 18th term of the AP.

Step 1: Use the formula for the $n$-th term of an AP
The formula for the $n$-th term of an arithmetic progression is:

$$a_n = a_1 + (n – 1) \times d$$

Where:
– $a_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

Step 2: Set up the equation based on the given condition
We are given that:

$$7 \times a_7 = 11 \times a_{11}$$

Using the formula for the 7th term ($a_7$) and the 11th term ($a_{11}$):

$$a_7 = a_1 + (7 – 1) \times d = a_1 + 6d$$ $$a_{11} = a_1 + (11 – 1) \times d = a_1 + 10d$$

Substitute these into the given equation:

$$7 \times (a_1 + 6d) = 11 \times (a_1 + 10d)$$

Step 3: Simplify the equation
Distribute both sides:

$$7a_1 + 42d = 11a_1 + 110d$$

Now, move all terms involving $a_1$ and $d$ to one side:

$$7a_1 – 11a_1 = 110d – 42d$$ $$-4a_1 = 68d$$

Divide both sides by -4:

$$a_1 = -17d$$

Step 4: Find the 18th term
Now that we know $a_1 = -17d$, we can use the formula for the $n$-th term to find the 18th term ($a_{18}$):

$$a_{18} = a_1 + (18 – 1) \times d$$ $$a_{18} = a_1 + 17d$$ Substitute $a_1 = -17d$:

$$a_{18} = -17d + 17d$$ $$a_{18} = 0$$

Final Answer:
The 18th term of the AP is 0.

Answer: Option C
Solution:
We are given that the 7th term of an arithmetic progression (AP) is -15, and the 12th term is 5. We need to find the 16th term.

Step 1: Use the formula for the $n$-th term of an AP
The general formula for the $n$-th term of an arithmetic progression is:

$$a_n = a_1 + (n – 1) \times d$$

Where:
– $a_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

Step 2: Set up equations for the 7th and 12th terms
For the 7th term ($a_7 = -15$):

$$a_7 = a_1 + (7 – 1) \times d = a_1 + 6d$$ $$-15 = a_1 + 6d \quad \text{(Equation 1)}$$

For the 12th term ($a_{12} = 5$):

$$a_{12} = a_1 + (12 – 1) \times d = a_1 + 11d$$ $$5 = a_1 + 11d \quad \text{(Equation 2)}$$

Step 3: Solve the system of equations
We now have two equations:

1. $-15 = a_1 + 6d$
2. $5 = a_1 + 11d$

Step 3a: Subtract Equation 1 from Equation 2:

$$(a_1 + 11d) – (a_1 + 6d) = 5 – (-15)$$ $$5d = 20$$ $$d = 4$$

Step 3b: Substitute $d = 4$ into Equation 1:

$$-15 = a_1 + 6 \times 4$$ $$-15 = a_1 + 24$$ $$a_1 = -15 – 24 = -39$$

So, the first term $a_1 = -39$ and the common difference $d = 4$.

Step 4: Find the 16th term
Now, we can use the formula for the $n$-th term to find the 16th term ($a_{16}$):

$$a_{16} = a_1 + (16 – 1) \times d$$ $$a_{16} = -39 + 15 \times 4$$ $$a_{16} = -39 + 60$$ $$a_{16} = 21$$

Final Answer:
The 16th term is 21.

Answer: Option B
Solution:
To find $T_{18} – T_8$ for an arithmetic progression (AP), we can use the formula for the $n$-th term of an arithmetic progression:

$$T_n = a_1 + (n – 1) \times d$$

Where:
– $T_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

Now, let’s calculate $T_{18}$ and $T_8$.

Step 1: Calculate $T_{18}$
Using the formula for the 18th term:

$$T_{18} = a_1 + (18 – 1) \times d = a_1 + 17d$$

Step 2: Calculate $T_8$
Using the formula for the 8th term:

$$T_8 = a_1 + (8 – 1) \times d = a_1 + 7d$$

Step 3: Find $T_{18} – T_8$
Now subtract $T_8$ from $T_{18}$:

$$T_{18} – T_8 = (a_1 + 17d) – (a_1 + 7d)$$

Simplify:

$$T_{18} – T_8 = a_1 + 17d – a_1 – 7d$$ $$T_{18} – T_8 = 10d$$

Final Answer:
The value of $T_{18} – T_8$ is $10d$, where $d$ is the common difference of the arithmetic progression.

Answer: Option C
Solution:
To find the first negative term of the arithmetic progression (AP) 24, 21, 18, …, we can use the general formula for the $n$-th term of an AP:

$$T_n = a_1 + (n – 1) \times d$$

Where:
– $T_n$ is the $n$-th term,
– $a_1$ is the first term,
– $d$ is the common difference,
– $n$ is the position of the term.

Step 1: Identify the given values
– The first term $a_1 = 24$,
– The common difference $d = 21 – 24 = -3$.

Step 2: Find the first negative term
We need to find the value of $n$ for which $T_n < 0$. Using the formula for the $n$-th term:

$$T_n = 24 + (n – 1) \times (-3)$$

We want to find when $T_n < 0$:

$$24 + (n – 1) \times (-3) < 0$$

Simplify the inequality:

$$24 – 3(n – 1) < 0$$

Distribute the $-3$:

$$24 – 3n + 3 < 0$$

Combine like terms:

$$27 – 3n < 0$$

Subtract 27 from both sides:

$$-3n 9$$

Step 3: Find the smallest integer value for $n$
The smallest integer greater than 9 is $n = 10$.

Step 4: Verify the result
Now let’s check the 10th term:

$$T_{10} = 24 + (10 – 1) \times (-3) = 24 + 9 \times (-3) = 24 – 27 = -3$$

So, the 10th term is $-3$, which is the first negative term.

Final Answer:
The first negative term is the 10th term.

Answer: Option A
Solution:
We are given the arithmetic progression (AP) with terms $x – 7, x – 2, x + 3, \dots$, and we need to find the 15th term.

Step 1: Identify the first term and common difference
The first term $a_1$ is the first term of the sequence:
$$a_1 = x – 7$$

The common difference $d$ is the difference between the second term and the first term:
$$d = (x – 2) – (x – 7) = x – 2 – x + 7 = 5$$

Step 2: Use the formula for the $n$-th term of an AP
The formula for the $n$-th term of an arithmetic progression is:
$$T_n = a_1 + (n – 1) \times d$$

We need to find the 15th term, so substitute $n = 15$, $a_1 = x – 7$, and $d = 5$ into the formula:

$$T_{15} = (x – 7) + (15 – 1) \times 5$$ $$T_{15} = (x – 7) + 14 \times 5$$ $$T_{15} = (x – 7) + 70$$ $$T_{15} = x + 63$$

Final Answer:
The 15th term of the AP is $\boxed{x + 63}$.

Answer: Option C
Solution:
We are given that the first term of the arithmetic progression (AP) is $a = 1$, the $n$-th term $t_n = 20$, and the sum of the first $n$ terms $s_n = 399$. We need to find the value of $n$.

Step 1: Use the formula for the $n$-th term of an AP
The formula for the $n$-th term $t_n$ of an arithmetic progression is:

$$t_n = a + (n – 1) \times d$$

Given:
– $a = 1$,
– $t_n = 20$,
– We need to find $n$ and $d$.

Substitute the given values into the formula:

$$20 = 1 + (n – 1) \times d$$

Simplify:

$$20 = 1 + (n – 1) \times d$$ $$19 = (n – 1) \times d$$ $$d = \frac{19}{n – 1} \quad \text{(Equation 1)}$$

Step 2: Use the formula for the sum of the first $n$ terms of an AP
The formula for the sum of the first $n$ terms $s_n$ of an arithmetic progression is:

$$s_n = \frac{n}{2} \times (2a + (n – 1) \times d)$$

We are given $s_n = 399$ and $a = 1$, so substitute these into the formula:

$$399 = \frac{n}{2} \times (2 \times 1 + (n – 1) \times d)$$ $$399 = \frac{n}{2} \times (2 + (n – 1) \times d)$$

Substitute the expression for $d$ from Equation 1:

$$399 = \frac{n}{2} \times \left(2 + (n – 1) \times \frac{19}{n – 1}\right)$$ $$399 = \frac{n}{2} \times \left(2 + 19\right)$$ $$399 = \frac{n}{2} \times 21$$ $$399 = \frac{21n}{2}$$

Multiply both sides by 2:

$$798 = 21n$$

Solve for $n$:

$$n = \frac{798}{21} = 38$$

Final Answer:
The value of $n$ is $\boxed{38}$.

Answer: Option B
Solution:
To find the value of the equipment after 10 years, we need to apply the depreciation percentages to the original cost of the equipment.

Given:
– Original cost of the equipment = 6,00,000
– Depreciation percentages for each year:
– Year 1: 15%
– Year 2: 13.5%
– Year 3: 12%
– and so on, decreasing by 1.5% per year.

For each year, the value of the equipment after depreciation is calculated by subtracting the depreciation amount from the original cost.

1. Year 1: 15% depreciation
Depreciation for Year 1 = 6,00,000 × 15% = 90,000
Value at the end of Year 1 = 6,00,000 – 90,000 = 5,10,000

2. Year 2: 13.5% depreciation
Depreciation for Year 2 = 6,00,000 × 13.5% = 81,000
Value at the end of Year 2 = 5,10,000 – 81,000 = 4,29,000

3. Year 3: 12% depreciation
Depreciation for Year 3 = 6,00,000 × 12% = 72,000
Value at the end of Year 3 = 4,29,000 – 72,000 = 3,57,000

4. Year 4: 10.5% depreciation
Depreciation for Year 4 = 6,00,000 × 10.5% = 63,000
Value at the end of Year 4 = 3,57,000 – 63,000 = 2,94,000

5. Year 5: 9% depreciation
Depreciation for Year 5 = 6,00,000 × 9% = 54,000
Value at the end of Year 5 = 2,94,000 – 54,000 = 2,40,000

6. Year 6: 7.5% depreciation
Depreciation for Year 6 = 6,00,000 × 7.5% = 45,000
Value at the end of Year 6 = 2,40,000 – 45,000 = 1,95,000

7. Year 7: 6% depreciation
Depreciation for Year 7 = 6,00,000 × 6% = 36,000
Value at the end of Year 7 = 1,95,000 – 36,000 = 1,59,000

8. Year 8: 4.5% depreciation
Depreciation for Year 8 = 6,00,000 × 4.5% = 27,000
Value at the end of Year 8 = 1,59,000 – 27,000 = 1,32,000

9. Year 9: 3% depreciation
Depreciation for Year 9 = 6,00,000 × 3% = 18,000
Value at the end of Year 9 = 1,32,000 – 18,000 = 1,14,000

10. Year 10: 1.5% depreciation
Depreciation for Year 10 = 6,00,000 × 1.5% = 9,000
Value at the end of Year 10 = 1,14,000 – 9,000 = 1,05,000

Final value of the equipment after 10 years: 1,05,000.

Answer: Option B
Solution:
The series you’re asking about is an arithmetic series, where:

– The first term $a = -64$,
– The common difference $d = -66 – (-64) = -2$,
– The last term is $-100$.

Step 1: Find the number of terms in the series.

The $n$-th term of an arithmetic series is given by the formula:

$$a_n = a + (n – 1) \cdot d$$

We know the last term is $a_n = -100$, so we can set up the equation:

$$-100 = -64 + (n – 1) \cdot (-2)$$

Now, solve for $n$:

$$-100 + 64 = (n – 1) \cdot (-2)$$ $$-36 = (n – 1) \cdot (-2)$$ $$n – 1 = \frac{-36}{-2} = 18$$ $$n = 19$$

So, there are 19 terms in the series.

Step 2: Calculate the sum of the series.

The sum $S_n$ of the first $n$ terms of an arithmetic series is given by the formula:

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

Substitute the known values:

$$S_{19} = \frac{19}{2} \cdot (-64 + (-100))$$ $$S_{19} = \frac{19}{2} \cdot (-164)$$ $$S_{19} = 19 \cdot (-82)$$ $$S_{19} = -1558$$

Final Answer:
The sum of the series is -1558.

Answer: Option D
Solution:
We are asked to find the sum of all positive integers up to 1000 that are divisible by 5 but not divisible by 2.

Step 1: Identify numbers divisible by 5 but not by 2.
– A number divisible by 5 is of the form $5k$, where $k$ is a positive integer.
– To ensure the number is not divisible by 2, $k$ must be odd because any multiple of 2 is even.

So, we need to find all numbers of the form $5k$, where $k$ is odd, and $5k \leq 1000$.

Step 2: Determine the possible values for $k$.
We know the numbers divisible by 5 are:
$$5, 10, 15, 20, \ldots, 1000$$

Since we need $k$ to be odd, the corresponding numbers are:
$$5 \times 1, 5 \times 3, 5 \times 5, 5 \times 7, \ldots$$

Thus, the sequence of $k$’s that satisfy this condition is:
$$1, 3, 5, 7, \ldots$$

Step 3: Determine the last value of $k$.
We need the largest $k$ such that $5k \leq 1000$. This means:
$$5k \leq 1000$$ $$k \leq \frac{1000}{5} = 200$$

So, the largest $k$ is 199 (since $k$ must be odd). Thus, the odd numbers from 1 to 199 are the values for $k$.

Step 4: Sum the numbers.
The sum we want is the sum of all numbers of the form $5k$, where $k$ is odd and $k$ ranges from 1 to 199.

First, the sum of the odd numbers from 1 to 199 forms an arithmetic series with the first term 1, the last term 199, and the common difference 2. The number of terms in this series is:
$$\frac{199 – 1}{2} + 1 = 100$$

The sum $S$ of the first $n$ terms of an arithmetic series is given by the formula:
$$S = \frac{n}{2} \cdot ( \text{first term} + \text{last term} )$$ So the sum of the odd numbers from 1 to 199 is:
$$S_{\text{odd}} = \frac{100}{2} \cdot (1 + 199) = 50 \cdot 200 = 10000$$

Now, since we want the sum of the numbers $5k$, we multiply the sum of the odd numbers by 5:
$$S_{\text{total}} = 5 \times 10000 = 50000$$

Final Answer:
The sum of all positive integers up to 1000 that are divisible by 5 and not divisible by 2 is 50,000.

Answer: Option B
Solution:
We are given that the sum of the first $n$ terms of an arithmetic progression (A.P.) is:

$$S_n = 3n^2 + 5n$$

We need to determine which term in the sequence is 164.

Step 1: Formula for the $n$-th term of an A.P.
The $n$-th term of an A.P. is given by:

$$T_n = S_n – S_{n-1}$$

Where $S_n$ is the sum of the first $n$ terms, and $S_{n-1}$ is the sum of the first $n-1$ terms.

Step 2: Express $S_n$ and $S_{n-1}$.

From the given, we know:

$$S_n = 3n^2 + 5n$$

Now, for $S_{n-1}$, we replace $n$ with $n-1$:

$$S_{n-1} = 3(n-1)^2 + 5(n-1)$$ Expanding this:

$$S_{n-1} = 3(n^2 – 2n + 1) + 5(n – 1)$$ $$S_{n-1} = 3n^2 – 6n + 3 + 5n – 5$$ $$S_{n-1} = 3n^2 – n – 2$$

Step 3: Calculate $T_n$.

Now we calculate $T_n$:

$$T_n = S_n – S_{n-1}$$ $$T_n = (3n^2 + 5n) – (3n^2 – n – 2)$$ Simplifying:

$$T_n = 3n^2 + 5n – 3n^2 + n + 2$$ $$T_n = 6n + 2$$

Step 4: Set $T_n = 164$.

We are given that one of the terms is 164, so we set $T_n = 164$:

$$6n + 2 = 164$$

Solving for $n$:

$$6n = 164 – 2$$ $$6n = 162$$ $$n = \frac{162}{6} = 27$$

Final Answer:
The term 164 is the 27th term of the A.P.

Answer: Option D
Solution:
We are given that the numbers $18$, $a$, and $b – 3$ are in an arithmetic progression (A.P.). In an A.P., the difference between consecutive terms is constant. Thus, the difference between the first and second terms is equal to the difference between the second and third terms. This can be expressed as:

$$a – 18 = (b – 3) – a$$

Step 1: Solve for $a + b$

Now, let’s solve for $a$ and $b$ by simplifying the equation:

$$a – 18 = b – 3 – a$$

Add $a$ to both sides to get:

$$2a – 18 = b – 3$$

Now, add 18 to both sides:

$$2a = b + 15$$

Thus, we have the equation:

$$b = 2a – 15$$

Step 2: Find $a + b$

We want to find $a + b$, so substitute $b = 2a – 15$ into $a + b$:

$$a + b = a + (2a – 15)$$ $$a + b = 3a – 15$$

Final Answer:

Thus, $a + b = 3a – 15$.

Answer: Option B
Solution:
We are given that the 18th and 11th terms of an arithmetic progression (A.P.) are in the ratio 3:2. We need to find the ratio of the 21st and 5th terms.

Step 1: General formula for the $n$-th term
The $n$-th term of an arithmetic progression is given by the formula:

$$T_n = a + (n – 1) \cdot d$$

Where:
– $a$ is the first term,
– $d$ is the common difference,
– $T_n$ is the $n$-th term.

Step 2: Express the 18th and 11th terms
For the 18th term:

$$T_{18} = a + (18 – 1) \cdot d = a + 17d$$

For the 11th term:

$$T_{11} = a + (11 – 1) \cdot d = a + 10d$$

We are told that the ratio of the 18th term to the 11th term is $3 : 2$, i.e.,

$$\frac{T_{18}}{T_{11}} = \frac{3}{2}$$

Step 3: Set up the equation
Substitute the expressions for $T_{18}$ and $T_{11}$:

$$\frac{a + 17d}{a + 10d} = \frac{3}{2}$$

Step 4: Solve for $a$ and $d$
Cross-multiply:

$$2(a + 17d) = 3(a + 10d)$$

Expand both sides:

$$2a + 34d = 3a + 30d$$

Simplify:

$$2a – 3a = 30d – 34d$$

$$-a = -4d$$

$$a = 4d$$

Step 5: Find the ratio of the 21st and 5th terms
Now, let’s find the 21st and 5th terms.

For the 21st term:

$$T_{21} = a + (21 – 1) \cdot d = a + 20d$$

Substitute $a = 4d$:

$$T_{21} = 4d + 20d = 24d$$

For the 5th term:

$$T_5 = a + (5 – 1) \cdot d = a + 4d$$

Substitute $a = 4d$:

$$T_5 = 4d + 4d = 8d$$

Step 6: Find the ratio of the 21st term to the 5th term
Now, we calculate the ratio:

$$\frac{T_{21}}{T_5} = \frac{24d}{8d} = 3$$

Final Answer:
The ratio of the 21st term to the 5th term is $\boxed{3}$.

Answer: Option B
Solution:
We are given that $k$, $2k – 1$, and $2k + 1$ are three consecutive terms of an arithmetic progression (A.P.). In an A.P., the difference between consecutive terms is constant.

Step 1: Use the property of an A.P.
In an A.P., the difference between consecutive terms is constant. This means that the difference between the second and first term is equal to the difference between the third and second term. Therefore, we can write the equation:

$$(2k – 1) – k = (2k + 1) – (2k – 1)$$

Step 2: Simplify both sides
Now, let’s simplify both sides of the equation.

On the left side:

$$(2k – 1) – k = 2k – 1 – k = k – 1$$

On the right side:

$$(2k + 1) – (2k – 1) = 2k + 1 – 2k + 1 = 2$$

So, the equation becomes:

$$k – 1 = 2$$

Step 3: Solve for $k$
Now, solve for $k$:

$$k = 2 + 1 = 3$$

Final Answer:
The value of $k$ is $\boxed{3}$.

Answer: Option B
Solution:
We are given an arithmetic progression (A.P.) where:
– The first term $a = 1$,
– The last term $l = 11$,
– The sum of the terms $S_n = 36$.

We need to find the number of terms $n$.

Step 1: Use the formula for the sum of an A.P.
The sum of the first $n$ terms of an A.P. is given by the formula:

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

Where:
– $S_n$ is the sum of the first $n$ terms,
– $a$ is the first term,
– $l$ is the last term,
– $n$ is the number of terms.

Step 2: Substitute the given values into the formula
We are given:
– $S_n = 36$,
– $a = 1$,
– $l = 11$.

Substitute these values into the sum formula:

$$36 = \frac{n}{2} \cdot (1 + 11)$$ $$36 = \frac{n}{2} \cdot 12$$ $$36 = 6n$$

Step 3: Solve for $n$
Now, solve for $n$:

$$n = \frac{36}{6} = 6$$

Final Answer:
The number of terms in the A.P. is $\boxed{6}$.

Answer: Option D
Solution:
We are given the arithmetic progression (A.P.):

$$3, 7, 11, 15, \dots$$

The first term $a = 3$, and the common difference $d = 4$.

We need to find the number of terms $n$ such that the sum of these terms is 406.

Step 1: Use the sum formula of an A.P.
The sum of the first $n$ terms of an A.P. is given by the formula:

$$S_n = \frac{n}{2} \cdot (2a + (n – 1) \cdot d)$$

Where:
– $S_n$ is the sum of the first $n$ terms,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the number of terms.

Step 2: Substitute the known values
We are given:
– $S_n = 406$,
– $a = 3$,
– $d = 4$.

Substitute these values into the sum formula:

$$406 = \frac{n}{2} \cdot \left(2 \cdot 3 + (n – 1) \cdot 4\right)$$ $$406 = \frac{n}{2} \cdot \left(6 + 4(n – 1)\right)$$ $$406 = \frac{n}{2} \cdot \left(6 + 4n – 4\right)$$ $$406 = \frac{n}{2} \cdot (4n + 2)$$ $$406 = \frac{n}{2} \cdot 2(2n + 1)$$ $$406 = n \cdot (2n + 1)$$

Step 3: Solve the quadratic equation
Now, we solve the equation:

$$406 = n(2n + 1)$$ $$406 = 2n^2 + n$$ $$2n^2 + n – 406 = 0$$

This is a quadratic equation in $n$. We can solve it using the quadratic formula:

$$n = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

For the equation $2n^2 + n – 406 = 0$, $a = 2$, $b = 1$, and $c = -406$. Now, substitute these values into the quadratic formula:

$$n = \frac{-1 \pm \sqrt{1^2 – 4(2)(-406)}}{2(2)}$$ $$n = \frac{-1 \pm \sqrt{1 + 3248}}{4}$$ $$n = \frac{-1 \pm \sqrt{3249}}{4}$$ $$n = \frac{-1 \pm 57}{4}$$

So, we have two possible solutions:

$$n = \frac{-1 + 57}{4} = \frac{56}{4} = 14$$ $$n = \frac{-1 – 57}{4} = \frac{-58}{4} = -14.5 \quad \text{(not a valid solution, as \( n \) must be a positive integer)}$$

Final Answer:
The number of terms required is $\boxed{14}$.

Answer: Option D
Solution:
We are given two arithmetic progressions (A.P.s) with the same common difference. The first term of one A.P. is $a_1 = 8$, and the first term of the other is $b_1 = 3$. We need to find the difference between their 30th terms.

Step 1: Formula for the $n$-th term of an A.P.
The $n$-th term of an A.P. is given by the formula:

$$T_n = a + (n – 1) \cdot d$$

Where:
– $T_n$ is the $n$-th term,
– $a$ is the first term,
– $d$ is the common difference.

Step 2: Express the 30th terms of both A.P.s
Let the common difference of both A.P.s be $d$.

– The 30th term of the first A.P. is:

$$T_{30}^{(1)} = a_1 + (30 – 1) \cdot d = 8 + 29d$$

– The 30th term of the second A.P. is:

$$T_{30}^{(2)} = b_1 + (30 – 1) \cdot d = 3 + 29d$$

Step 3: Find the difference between the 30th terms
Now, we find the difference between the 30th terms of the two A.P.s:

$$\text{Difference} = T_{30}^{(1)} – T_{30}^{(2)} = (8 + 29d) – (3 + 29d)$$

Simplifying:

$$\text{Difference} = 8 + 29d – 3 – 29d = 8 – 3 = 5$$

Final Answer:
The difference between their 30th terms is $\boxed{5}$.

Answer: Option B
Solution:
We are given that the $n$-th term of an arithmetic progression (A.P.) is:

$$T_n = 2n + 1$$

We need to find the sum of the first $n$ terms of this A.P.

Step 1: Express the sum of the first $n$ terms
The sum of the first $n$ terms of an A.P. is given by:

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

Where:
– $S_n$ is the sum of the first $n$ terms,
– $a$ is the first term,
– $l$ is the last term (which is the $n$-th term in this case).

Step 2: Find the first and $n$-th terms
From the formula for the $n$-th term $T_n = 2n + 1$:
– The first term $T_1$ is:

$$T_1 = 2(1) + 1 = 3$$

– The $n$-th term is $T_n = 2n + 1$.

Step 3: Use the sum formula
We know the sum formula:

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

Substitute $a = 3$ and $l = 2n + 1$:

$$S_n = \frac{n}{2} \cdot \left(3 + (2n + 1)\right)$$ $$S_n = \frac{n}{2} \cdot (2n + 4)$$ $$S_n = \frac{n}{2} \cdot 2(n + 2)$$ $$S_n = n(n + 2)$$

Final Answer:
The sum of the first $n$ terms of the A.P. is $\boxed{n(n + 2)}$.

Answer: Option D
Solution:
We are given that the sum of the first $n$ terms of an arithmetic progression (A.P.) is:

$$S_n = 3n^2 + n$$

and the common difference $d = 6$. We need to find the first term $a$ of the A.P.

Step 1: Recall the formula for the sum of the first $n$ terms of an A.P.
The sum of the first $n$ terms of an A.P. is given by the formula:

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

Where:
– $S_n$ is the sum of the first $n$ terms,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the number of terms.

Step 2: Set up the equation for the sum
We know that $S_n = 3n^2 + n$ and $d = 6$. Substituting these values into the sum formula:

$$3n^2 + n = \frac{n}{2} \cdot [2a + (n – 1) \cdot 6]$$

Step 3: Simplify the equation
First, simplify the right-hand side:

$$3n^2 + n = \frac{n}{2} \cdot [2a + 6(n – 1)]$$ $$3n^2 + n = \frac{n}{2} \cdot [2a + 6n – 6]$$ $$3n^2 + n = \frac{n}{2} \cdot (2a + 6n – 6)$$

Now, multiply both sides of the equation by 2 to eliminate the fraction:

$$2 \cdot (3n^2 + n) = n \cdot (2a + 6n – 6)$$ $$6n^2 + 2n = n(2a + 6n – 6)$$

Step 4: Simplify further
Divide both sides by $n$ (since $n \neq 0$):

$$6n + 2 = 2a + 6n – 6$$

Step 5: Solve for $a$
Now, subtract $6n$ from both sides:

$$2 = 2a – 6$$

Add 6 to both sides:

$$8 = 2a$$

Now, divide by 2:

$$a = 4$$

Final Answer:
The first term of the A.P. is $\boxed{4}$.

Answer: Option A
Solution:
We are given that four numbers are in an arithmetic progression (A.P.), and the sum of these numbers is 50. Additionally, the greatest number is 4 times the least number. We need to find the numbers.

Step 1: Let the four numbers be in terms of the first term and common difference
Let the four numbers in the A.P. be:
– The first term $a$,
– The second term $a + d$,
– The third term $a + 2d$,
– The fourth term $a + 3d$.

Step 2: Use the information about the sum of the numbers
The sum of these four numbers is 50:

$$a + (a + d) + (a + 2d) + (a + 3d) = 50$$

Simplifying the equation:

$$4a + 6d = 50$$

$$2a + 3d = 25 \quad \text{(Equation 1)}$$

Step 3: Use the information about the greatest and least numbers
We are also told that the greatest number is 4 times the least number. The greatest number is $a + 3d$, and the least number is $a$. Thus, we have:

$$a + 3d = 4a$$

Simplifying this equation:

$$3d = 3a$$

$$d = a \quad \text{(Equation 2)}$$

Step 4: Solve the system of equations
Now, substitute $d = a$ from Equation 2 into Equation 1:

$$2a + 3a = 25$$ $$5a = 25$$ $$a = 5$$

Step 5: Find $d$
Since $d = a$, we have:

$$d = 5$$

Step 6: Find the four numbers
Now that we know $a = 5$ and $d = 5$, the four numbers in the A.P. are:
– First number: $a = 5$,
– Second number: $a + d = 5 + 5 = 10$,
– Third number: $a + 2d = 5 + 2(5) = 15$,
– Fourth number: $a + 3d = 5 + 3(5) = 20$.

Final Answer:
The four numbers in the A.P. are $\boxed{5, 10, 15, 20}$.

Answer: Option A
Solution:
We are given that the first term of an arithmetic progression (A.P.) is $a = 2$, the common difference is $d = 4$, and we need to find the sum of the first 40 terms.

Step 1: Use the formula for the sum of the first $n$ terms of an A.P.
The sum of the first $n$ terms of an A.P. is given by the formula:

$$S_n = \frac{n}{2} \cdot \left( 2a + (n – 1) \cdot d \right)$$

Where:
– $S_n$ is the sum of the first $n$ terms,
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the number of terms.

Step 2: Substitute the known values
We are given:
– $a = 2$,
– $d = 4$,
– $n = 40$.

Substitute these values into the sum formula:

$$S_{40} = \frac{40}{2} \cdot \left( 2(2) + (40 – 1) \cdot 4 \right)$$ $$S_{40} = 20 \cdot \left( 4 + 39 \cdot 4 \right)$$ $$S_{40} = 20 \cdot \left( 4 + 156 \right)$$ $$S_{40} = 20 \cdot 160$$ $$S_{40} = 3200$$

Final Answer:
The sum of the first 40 terms of the A.P. is $\boxed{3200}$.

Answer: Option C
Solution:
We are given that the $n$-th term of an arithmetic progression (A.P.) is denoted by $T_n$, and the sum of the first $n$ terms is $S_n$. We are asked to express the $n$-th term $T_n$ in terms of the sum $S_n$.

Step 1: Recall the formulas for the sum of the first $n$ terms and the $n$-th term

1. The sum of the first $n$ terms of an A.P. is given by:

$$S_n = \frac{n}{2} \cdot \left(2a + (n – 1)d\right)$$

Where:
– $a$ is the first term,
– $d$ is the common difference,
– $n$ is the number of terms.

2. The $n$-th term of the A.P. is given by:

$$T_n = a + (n – 1) \cdot d$$

Step 2: Find the relation between $S_n$ and $T_n$
To express the $n$-th term $T_n$ in terms of the sum $S_n$, we can use the relationship between the sum of the first $n$ terms and the sum of the first $n – 1$ terms. Specifically, we know that:

$$S_n = S_{n-1} + T_n$$

Where $S_{n-1}$ is the sum of the first $n-1$ terms. Therefore, the $n$-th term $T_n$ can be written as:

$$T_n = S_n – S_{n-1}$$

Step 3: Substitute the formula for $S_n$ and $S_{n-1}$
Now, let’s substitute the sum formula for both $S_n$ and $S_{n-1}$.

– The sum of the first $n$ terms is:

$$S_n = \frac{n}{2} \cdot \left(2a + (n – 1)d\right)$$

– The sum of the first $n – 1$ terms is:

$$S_{n-1} = \frac{n-1}{2} \cdot \left(2a + (n – 2)d\right)$$

Step 4: Find the expression for $T_n$
Now, subtract $S_{n-1}$ from $S_n$ to find $T_n$:

$$T_n = \frac{n}{2} \cdot \left(2a + (n – 1)d\right) – \frac{n – 1}{2} \cdot \left(2a + (n – 2)d\right)$$

Simplifying this expression will yield the formula for $T_n$. However, an alternative approach, as shown earlier, is to use the general formula for the $n$-th term of an A.P.:

$$T_n = a + (n – 1) \cdot d$$

Conclusion:
The $n$-th term of an A.P. in terms of the sum of the first $n$ terms $S_n$ is:

$$T_n = S_n – S_{n-1}$$

Alternatively, you can use the standard formula for the $n$-th term:

$$T_n = a + (n – 1) \cdot d$$

Answer: Option B
Solution:
We are given that the sum of the first $n$ terms of an arithmetic progression (A.P.) is:

$$S_n = 3n^2 + 5n$$

We need to find the term in the A.P. that is equal to 164.

Step 1: Recall the relationship between the $n$-th term and the sum of the first $n$ terms
The $n$-th term $T_n$ of an A.P. can be found by subtracting the sum of the first $n-1$ terms $S_{n-1}$ from the sum of the first $n$ terms $S_n$:

$$T_n = S_n – S_{n-1}$$

Step 2: Find the expression for $T_n$
We know that $S_n = 3n^2 + 5n$. Now, we need to find $S_{n-1}$, which is the sum of the first $n-1$ terms:

$$S_{n-1} = 3(n-1)^2 + 5(n-1)$$

Expanding this expression:

$$S_{n-1} = 3(n^2 – 2n + 1) + 5(n – 1)$$ $$S_{n-1} = 3n^2 – 6n + 3 + 5n – 5$$ $$S_{n-1} = 3n^2 – n – 2$$

Now, subtract $S_{n-1}$ from $S_n$ to get $T_n$:

$$T_n = S_n – S_{n-1} = (3n^2 + 5n) – (3n^2 – n – 2)$$ $$T_n = 3n^2 + 5n – 3n^2 + n + 2$$ $$T_n = 6n + 2$$

Step 3: Set $T_n = 164$ and solve for $n$
We want to find the value of $n$ such that $T_n = 164$:

$$6n + 2 = 164$$

Subtract 2 from both sides:

$$6n = 162$$

Now, divide by 6:

$$n = 27$$

Final Answer:
The 27th term of the A.P. is 164. Therefore, $\boxed{27}$ is the answer.

Answer: Option C
Solution:
We are given that the first three terms of an arithmetic progression (A.P.) are:

1. First term: $3y – 1$
2. Second term: $3y + 5$
3. Third term: $5y + 1$

In an arithmetic progression, the difference between consecutive terms is constant. Therefore, the difference between the second term and the first term must be equal to the difference between the third term and the second term.

So, we have:

$$(3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)$$

Simplifying both sides:

On the left side:

$$(3y + 5) – (3y – 1) = 3y + 5 – 3y + 1 = 6$$

On the right side:

$$(5y + 1) – (3y + 5) = 5y + 1 – 3y – 5 = 2y – 4$$

So we have the equation:

$$6 = 2y – 4$$

Now, solve for $y$:

$$6 + 4 = 2y$$

$$10 = 2y$$

$$y = \frac{10}{2} = 5$$

So, $y = 5$.

Answer: Option C
Solution:
We are given that the 7th term and 13th term of an arithmetic progression (A.P.) are 34 and 64, respectively. We are asked to find the 18th term of the A.P.

In an A.P., the $n$-th term is given by the formula:

$$T_n = a + (n – 1) \cdot d$$

where:
– $T_n$ is the $n$-th term,
– $a$ is the first term, and
– $d$ is the common difference.

Step 1: Set up equations for the 7th and 13th terms

For the 7th term, we have:

$$T_7 = a + (7 – 1) \cdot d = a + 6d$$

We are given that $T_7 = 34$, so:

$$a + 6d = 34 \quad \text{(Equation 1)}$$

For the 13th term, we have:

$$T_{13} = a + (13 – 1) \cdot d = a + 12d$$

We are given that $T_{13} = 64$, so:

$$a + 12d = 64 \quad \text{(Equation 2)}$$

Step 2: Solve the system of equations

We now have the system of equations:

$$a + 6d = 34 \quad \text{(Equation 1)}$$ $$a + 12d = 64 \quad \text{(Equation 2)}$$

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

$$(a + 12d) – (a + 6d) = 64 – 34$$ $$a + 12d – a – 6d = 30$$ $$6d = 30$$ $$d = \frac{30}{6} = 5$$

Step 3: Find $a$

Now that we know $d = 5$, substitute this value into Equation 1:

$$a + 6 \cdot 5 = 34$$ $$a + 30 = 34$$ $$a = 34 – 30 = 4$$ Step 4: Find the 18th term

Now that we know $a = 4$ and $d = 5$, we can find the 18th term using the formula for the $n$-th term:

$$T_{18} = a + (18 – 1) \cdot d = 4 + 17 \cdot 5$$ $$T_{18} = 4 + 85 = 89$$

Thus, the 18th term of the A.P. is 89.

Answer: Option C
Solution:
We are given that the sum of the first $n$ terms of an arithmetic progression (A.P.) is:

$$S_n = 2n^2 + 5n$$

We are asked to find the $n$-th term of the A.P., denoted as $T_n$.

Step 1: Use the relation between sum and $n$-th term

The sum of the first $n$ terms of an A.P. is related to the $n$-th term by the following formula:

$$S_n = T_1 + T_2 + \cdots + T_n$$

The $n$-th term $T_n$ can be found using the relation:

$$T_n = S_n – S_{n-1}$$

Step 2: Calculate $S_{n-1}$

We know the formula for $S_n$, so now let’s find the formula for $S_{n-1}$:

$$S_{n-1} = 2(n-1)^2 + 5(n-1)$$

Expand the expression:

$$S_{n-1} = 2(n^2 – 2n + 1) + 5(n – 1)$$ $$S_{n-1} = 2n^2 – 4n + 2 + 5n – 5$$ $$S_{n-1} = 2n^2 + n – 3$$

Step 3: Find $T_n$

Now, substitute $S_n$ and $S_{n-1}$ into the formula for $T_n$:

$$T_n = S_n – S_{n-1}$$ $$T_n = (2n^2 + 5n) – (2n^2 + n – 3)$$ $$T_n = 2n^2 + 5n – 2n^2 – n + 3$$ $$T_n = (2n^2 – 2n^2) + (5n – n) + 3$$ $$T_n = 4n + 3$$

Thus, the $n$-th term of the A.P. is:

$$T_n = 4n + 3$$