Given: An Arithmetic Progression (AP) with the second term $a_2 = 38$ and the sixth term $a_6 = -22$.

To find: The missing terms $a_1$, $a_3$, $a_4$, and $a_5$.

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

Step 2: Formulate equations based on the given terms.
For the second term ($n=2$):
$a_2 = a + (2 - 1)d = 38$
$a + d = 38$ --- (Equation 1)

For the sixth term ($n=6$):
$a_6 = a + (6 - 1)d = -22$
$a + 5d = -22$ --- (Equation 2)

Step 3: Solve the system of linear equations for $a$ and $d$.
Subtract Equation 1 from Equation 2:
$(a + 5d) - (a + d) = -22 - 38$
$a - a + 5d - d = -60$
$4d = -60$
$d = \frac{-60}{4}$
$d = -15$ [Since the common difference is constant]

Substitute $d = -15$ into Equation 1:
$a + (-15) = 38$
$a = 38 + 15$
$a = 53$

Step 4: Calculate the missing terms.
The first term $a_1 = a = 53$.
The third term $a_3 = a + 2d = 53 + 2(-15) = 53 - 30 = 23$.
The fourth term $a_4 = a + 3d = 53 + 3(-15) = 53 - 45 = 8$.
The fifth term $a_5 = a + 4d = 53 + 4(-15) = 53 - 60 = -7$.

Summary of terms:
$a_1 = 53$
$a_2 = 38$
$a_3 = 23$
$a_4 = 8$
$a_5 = -7$
$a_6 = -22$

Final Answer: The missing terms are 53, 23, 8, and -7. The complete AP is 53, 38, 23, 8, -7, -22.

Given: An Arithmetic Progression (AP) with the first term $a = -4$ and the sixth term $a_6 = 6$. The sequence is $-4, \square, \square, \square, \square, 6$.

To find: The four missing terms in the sequence, which correspond to $a_2, a_3, a_4,$ and $a_5$.

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

Step 2: Formulate equations based on the given information.
We are given:
$a_1 = a = -4$
$a_6 = 6$
Using the formula $a_n = a + (n - 1)d$ for $n = 6$:
$a_6 = a + (6 - 1)d$
$6 = -4 + 5d$ [Substituting the known values of $a_6$ and $a$]

Step 3: Solve for the common difference ($d$).
$6 + 4 = 5d$ [Adding 4 to both sides]
$10 = 5d$
$d = \frac{10}{5}$ [Dividing both sides by 5]
$d = 2$

Step 4: Calculate the missing terms.
Now that we have $a = -4$ and $d = 2$, we can find the missing terms using $a_n = a + (n - 1)d$:

For the second term ($a_2$):
$a_2 = a + d = -4 + 2 = -2$

For the third term ($a_3$):
$a_3 = a + 2d = -4 + 2(2) = -4 + 4 = 0$

For the fourth term ($a_4$):
$a_4 = a + 3d = -4 + 3(2) = -4 + 6 = 2$

For the fifth term ($a_5$):
$a_5 = a + 4d = -4 + 4(2) = -4 + 8 = 4$

Step 5: Verification.
The sequence is $-4, -2, 0, 2, 4, 6$.
Checking the common difference: $-2 - (-4) = 2$; $0 - (-2) = 2$; $2 - 0 = 2$; $4 - 2 = 2$; $6 - 4 = 2$.
Since the common difference is constant, the values are correct.

Final Answer: The missing terms are -2, 0, 2, and 4. The complete AP is -4, -2, 0, 2, 4, 6.

Given:

The first term of the Arithmetic Progression (AP), $a = -18.9$

The common difference of the AP, $d = 2.5$

The $n^{th}$ term of the AP, $a_n = 3.6$

To find:

The number of terms, $n$.

Formula:

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

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

Step 1: Substitution of given values into the formula

Substitute $a = -18.9$, $d = 2.5$, and $a_n = 3.6$ into the formula $a_n = a + (n - 1)d$:

$3.6 = -18.9 + (n - 1)(2.5)$

Step 2: Isolating the term containing $n$

Add $18.9$ to both sides of the equation to isolate the term with $(n-1)$:

$3.6 + 18.9 = (n - 1)(2.5)$

$22.5 = (n - 1)(2.5)$

Step 3: Solving for $(n - 1)$

Divide both sides of the equation by $2.5$:

$\frac{22.5}{2.5} = n - 1$

[Since $\frac{22.5}{2.5} = \frac{225}{25} = 9$]

$9 = n - 1$

Step 4: Solving for $n$

Add $1$ to both sides of the equation:

$n = 9 + 1$

$n = 10$

Conclusion:

By substituting the given values into the standard formula for the $n^{th}$ term of an AP and performing algebraic simplification, we determine that the number of terms $n$ is $10$.

Final Answer: n = 10

Given: An Arithmetic Progression (AP) sequence: $10, 7, 4, \dots$

To Find: The $30^{th}$ term of the given AP.

Step 1: Identify the parameters of the Arithmetic Progression.

An Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. This constant is known as the common difference ($d$).

The first term ($a$) is the first number in the sequence:
$a = 10$

The common difference ($d$) is calculated by subtracting the first term from the second term:
$d = a_2 - a_1$
$d = 7 - 10$
$d = -3$

Step 2: State the formula for the $n^{th}$ term of an AP.

The general formula to find the $n^{th}$ term ($a_n$) of an Arithmetic Progression is:
$a_n = a + (n - 1)d$
Where:
$a_n$ = the $n^{th}$ term to be found
$a$ = the first term
$n$ = the position of the term
$d$ = the common difference

Step 3: Substitute the known values into the formula.

We are looking for the $30^{th}$ term, so $n = 30$.
Substituting $a = 10$, $d = -3$, and $n = 30$ into the formula:

$a_{30} = 10 + (30 - 1)(-3)$

Step 4: Perform the arithmetic calculations.

First, solve the expression inside the parentheses:
$a_{30} = 10 + (29)(-3)$

Next, perform the multiplication:
$29 \times -3 = -87$

Finally, perform the addition:
$a_{30} = 10 - 87$
$a_{30} = -77$

Justification: By applying the standard formula for the $n^{th}$ term of an arithmetic progression, we have determined that the sequence decreases by $3$ at each step. Starting from $10$, after $29$ steps of decreasing by $3$, the value reaches $-77$.

Final Answer: The 30th term of the AP is -77.

Given:

The first term of the Arithmetic Progression (AP), $a = 3.5$.

The common difference of the AP, $d = 0$.

The number of terms in the AP, $n = 105$.

To find:

The $n$th term of the AP, denoted by $a_n$.

Step 1: Stating the relevant formula

The $n$th term of an Arithmetic Progression is given by the formula:

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

[Where $a$ is the first term, $d$ is the common difference, and $n$ is the position of the term in the sequence.]

Step 2: Substituting the given values into the formula

Substitute $a = 3.5$, $d = 0$, and $n = 105$ into the formula:

$a_{105} = 3.5 + (105 - 1) \times 0$

Step 3: Performing the arithmetic calculations

First, calculate the value inside the parentheses:

$105 - 1 = 104$

Now, substitute this back into the equation:

$a_{105} = 3.5 + (104) \times 0$

[Since any number multiplied by zero is zero, $104 \times 0 = 0$]

$a_{105} = 3.5 + 0$

$a_{105} = 3.5$

Justification:

In an Arithmetic Progression where the common difference $d = 0$, every term in the sequence is identical to the first term $a$. Since $a = 3.5$, all terms $a_1, a_2, ..., a_{105}$ are equal to $3.5$.

Final Answer: 3.5

Given: An Arithmetic Progression (AP) with the first term $a = 5$ and the fourth term $a_4 = 9\frac{1}{2}$.

To find: The missing terms in the boxes, which correspond to the second term ($a_2$) and the third term ($a_3$).

Step 1: Expressing the given terms using the general formula for an AP.

The general term of an Arithmetic Progression is given by the formula: $a_n = a + (n - 1)d$, where $a$ is the first term, $d$ is the common difference, and $n$ is the position of the term.

Given $a = 5$.

Given $a_4 = 9\frac{1}{2} = \frac{19}{2}$.

Step 2: Formulating the equation for the common difference ($d$).

Using the formula for the fourth term ($n=4$):

$a_4 = a + (4 - 1)d$

$\frac{19}{2} = 5 + 3d$ [Substituting the known values]

Subtract 5 from both sides:

$\frac{19}{2} - 5 = 3d$

$\frac{19 - 10}{2} = 3d$ [Finding a common denominator]

$\frac{9}{2} = 3d$

Divide both sides by 3:

$d = \frac{9}{2 \times 3}$

$d = \frac{3}{2}$ or $1.5$

Step 3: Calculating the missing terms.

Now that we have $a = 5$ and $d = \frac{3}{2}$, we can find the missing terms $a_2$ and $a_3$.

For the second term ($a_2$):

$a_2 = a + d$

$a_2 = 5 + \frac{3}{2}$

$a_2 = \frac{10 + 3}{2} = \frac{13}{2} = 6\frac{1}{2}$

For the third term ($a_3$):

$a_3 = a + 2d$

$a_3 = 5 + 2(\frac{3}{2})$

$a_3 = 5 + 3 = 8$

Verification:

The sequence is $5, 6\frac{1}{2}, 8, 9\frac{1}{2}$.

Common difference check: $6.5 - 5 = 1.5$; $8 - 6.5 = 1.5$; $9.5 - 8 = 1.5$. Since the common difference is constant, the values are correct.

Final Answer: The missing terms are $6\frac{1}{2}$ and $8$.

Given:

We are looking for three-digit numbers. The range of three-digit numbers is from $100$ to $999$. We are specifically interested in those numbers that are divisible by $7$.

To Find:

The total count of three-digit numbers divisible by $7$.

Step 1: Identifying the first three-digit number divisible by 7

The smallest three-digit number is $100$. Let us divide $100$ by $7$:
$100 \div 7 = 14$ with a remainder of $2$.
To find the next multiple of $7$, we calculate: $100 + (7 - 2) = 105$.
Thus, the first three-digit number divisible by $7$ is $a = 105$.

Step 2: Identifying the last three-digit number divisible by 7

The largest three-digit number is $999$. Let us divide $999$ by $7$:
$999 \div 7 = 142$ with a remainder of $5$.
To find the largest multiple of $7$ less than or equal to $999$, we subtract the remainder from $999$:
$999 - 5 = 994$.
Thus, the last three-digit number divisible by $7$ is $l = 994$.

Step 3: Formulating the Arithmetic Progression (AP)

The sequence of three-digit numbers divisible by $7$ forms an Arithmetic Progression where:
First term ($a$) = $105$
Common difference ($d$) = $7$
Last term ($a_n$) = $994$

Step 4: Applying the formula for the $n^{th}$ term of an AP

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

Step 5: Solving for $n$

Substitute the known values into the formula:
$994 = 105 + (n - 1)7$
Subtract $105$ from both sides:
$994 - 105 = (n - 1)7$
$889 = (n - 1)7$
Divide both sides by $7$:
$889 / 7 = n - 1$
$127 = n - 1$
Add $1$ to both sides:
$n = 127 + 1$
$n = 128$

Final Answer:

There are 128 three-digit numbers that are divisible by 7.

Given:

The common difference of the Arithmetic Progression (AP), $d = -3$.

The number of terms in the AP, $n = 18$.

The $n$th term of the AP, $a_n = -5$.

To find:

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

Step 1: Stating the General Formula for an Arithmetic Progression

The $n$th term of an Arithmetic Progression is given by the standard formula:

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

[Where $a_n$ is the $n$th term, $a$ is the first term, $n$ is the position of the term, and $d$ is the common difference.]

Step 2: Substituting the Given Values into the Formula

Substitute $a_n = -5$, $n = 18$, and $d = -3$ into the equation:

$-5 = a + (18 - 1)(-3)$

Step 3: Simplifying the Expression

First, perform the subtraction inside the parentheses:

$-5 = a + (17)(-3)$

[Since $18 - 1 = 17$]

Next, perform the multiplication:

$-5 = a - 51$

[Since $17 \times -3 = -51$]

Step 4: Solving for $a$

To isolate $a$, add $51$ to both sides of the equation:

$-5 + 51 = a$

$a = 46$

Final Answer: The first term of the Arithmetic Progression is 46.

Given: An Arithmetic Progression (AP) with missing terms represented as: $\square, 13, \square, 3$.

To find: The missing terms in the first and third positions of the sequence.

Step 1: Defining the variables of the Arithmetic Progression

Let the terms of the Arithmetic Progression be denoted by $a_1, a_2, a_3, a_4$.
The general form of an AP is given by $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
From the given sequence:
$a_1 = \square$
$a_2 = 13$
$a_3 = \square$
$a_4 = 3$

Step 2: Formulating the equations based on the general term formula

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

For the second term ($n=2$):
$a_2 = a + (2-1)d = a + d = 13$ --- (Equation 1)

For the fourth term ($n=4$):
$a_4 = a + (4-1)d = a + 3d = 3$ --- (Equation 2)

Step 3: Solving the system of linear equations

To find the common difference $d$, subtract Equation 1 from Equation 2:
$(a + 3d) - (a + d) = 3 - 13$
$a - a + 3d - d = -10$
$2d = -10$
$d = \frac{-10}{2}$
$d = -5$

Now, substitute the value of $d$ into Equation 1 to find the first term $a$:
$a + (-5) = 13$
$a = 13 + 5$
$a = 18$

Step 4: Calculating the missing terms

The first term ($a_1$) is $a = 18$.

The third term ($a_3$) is calculated as:
$a_3 = a + 2d$
$a_3 = 18 + 2(-5)$ [Substituting $a=18$ and $d=-5$]
$a_3 = 18 - 10$
$a_3 = 8$

Verification:
The sequence is $18, 13, 8, 3$.
Common difference check: $13 - 18 = -5$; $8 - 13 = -5$; $3 - 8 = -5$.
Since the common difference is constant, the values are correct.

Final Answer: The missing terms are 18 and 8. The complete AP is 18, 13, 8, 3.

Given:

The 11th term of an Arithmetic Progression ($a_{11}$) = $38$.

The 16th term of an Arithmetic Progression ($a_{16}$) = $73$.

To Find:

The 31st term of the Arithmetic Progression ($a_{31}$).

Step 1: Defining the General Formula for an Arithmetic Progression

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

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

Where:

• $a$ is the first term of the AP.
• $d$ is the common difference.
• $n$ is the position of the term.

Step 2: Formulating Equations based on the Given Information

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

For the 11th term ($n=11$):

$a_{11} = a + (11 - 1)d = 38$

$a + 10d = 38$ --- (Equation 1)

For the 16th term ($n=16$):

$a_{16} = a + (16 - 1)d = 73$

$a + 15d = 73$ --- (Equation 2)

Step 3: Solving the System of Linear Equations

To find the values of $a$ and $d$, we subtract Equation 1 from Equation 2:

$(a + 15d) - (a + 10d) = 73 - 38$

$a - a + 15d - 10d = 35$

$5d = 35$

$d = \frac{35}{5}$

$d = 7$

Now, substitute the value of $d = 7$ into Equation 1 to find $a$:

$a + 10(7) = 38$

$a + 70 = 38$

$a = 38 - 70$

$a = -32$

Step 4: Calculating the 31st Term

Now that we have $a = -32$ and $d = 7$, we find $a_{31}$ using the general formula:

$a_{31} = a + (31 - 1)d$

$a_{31} = -32 + (30)d$

$a_{31} = -32 + 30(7)$

$a_{31} = -32 + 210$

$a_{31} = 178$

Final Answer: The 31st term of the Arithmetic Progression is 178.

Given:

Ramkali's savings in the first week ($a_1$) = ₹ $5$.

The weekly increase in savings ($d$) = ₹ $1.75$.

The savings in the $n$th week ($a_n$) = ₹ $20.75$.

To Find:

The value of $n$ (the number of weeks).

Step 1: Identifying the Arithmetic Progression (AP)

Since the savings increase by a constant amount each week, the sequence of savings forms an Arithmetic Progression.

Let the first term be $a = 5$.

Let the common difference be $d = 1.75$.

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

Step 2: Formulating the Equation

Substitute the given values into the formula:

$20.75 = 5 + (n - 1) \times 1.75$

Step 3: Solving for $n$

Subtract $5$ from both sides of the equation:

$20.75 - 5 = (n - 1) \times 1.75$

$15.75 = (n - 1) \times 1.75$

Divide both sides by $1.75$ to isolate the term $(n - 1)$:

$\frac{15.75}{1.75} = n - 1$

To simplify the division, multiply both numerator and denominator by $100$:

$\frac{1575}{175} = n - 1$

Perform the division:

$1575 \div 175 = 9$

$9 = n - 1$

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

$n = 9 + 1$

$n = 10$

Step 4: Verification

If $n = 10$, then $a_{10} = 5 + (10 - 1) \times 1.75$

$a_{10} = 5 + 9 \times 1.75$

$a_{10} = 5 + 15.75$

$a_{10} = 20.75$

[Since the calculated $a_{10}$ matches the given $a_n$, the value of $n$ is correct.]

Final Answer: The value of $n$ is $10$.

Given:

An Arithmetic Progression (AP) where:

• The sum of the 4th term ($a_4$) and the 8th term ($a_8$) is $24$. That is, $a_4 + a_8 = 24$.
• The sum of the 6th term ($a_6$) and the 10th term ($a_{10}$) is $44$. That is, $a_6 + a_{10} = 44$.

To Find:

The first three terms of the AP, which are $a$, $a+d$, and $a+2d$.

Step 1: Define the general term of an AP

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

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

where $a$ is the first term and $d$ is the common difference.

Step 2: Formulate equations based on the given conditions

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

For the first condition ($a_4 + a_8 = 24$):

$a_4 = a + 3d$

$a_8 = a + 7d$

$(a + 3d) + (a + 7d) = 24$

$2a + 10d = 24$

Dividing the entire equation by 2 to simplify:

$a + 5d = 12$ --- (Equation 1)

For the second condition ($a_6 + a_{10} = 44$):

$a_6 = a + 5d$

$a_{10} = a + 9d$

$(a + 5d) + (a + 9d) = 44$

$2a + 14d = 44$

Dividing the entire equation by 2 to simplify:

$a + 7d = 22$ --- (Equation 2)

Step 3: Solve the system of linear equations

Subtract Equation 1 from Equation 2:

$(a + 7d) - (a + 5d) = 22 - 12$

$2d = 10$

$d = 5$ [Dividing both sides by 2]

Now, substitute $d = 5$ into Equation 1 to find $a$:

$a + 5(5) = 12$

$a + 25 = 12$

$a = 12 - 25$

$a = -13$

Step 4: Determine the first three terms

The first three terms are defined as $a_1$, $a_2$, and $a_3$:

$a_1 = a = -13$

$a_2 = a + d = -13 + 5 = -8$

$a_3 = a + 2d = -13 + 2(5) = -13 + 10 = -3$

Final Answer: The first three terms of the AP are -13, -8, and -3.

Given: An Arithmetic Progression (AP) sequence: $11, 8, 5, 2, \dots$ and a specific number $-150$.

To Find: Whether $-150$ is a term of the given Arithmetic Progression.

Step 1: Identifying the parameters of the Arithmetic Progression

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

The given sequence is: $11, 8, 5, 2, \dots$

Let the first term be $a = 11$.

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

$d = a_2 - a_1 = 8 - 11 = -3$

$d = a_3 - a_2 = 5 - 8 = -3$

Thus, $a = 11$ and $d = -3$.

Step 2: Applying the General Term Formula

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

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

Where:

• $a_n$ is the $n^{th}$ term.
• $a$ is the first term.
• $n$ is the position of the term (which must be a positive integer).
• $d$ is the common difference.

Step 3: Setting up the equation

Assume that $-150$ is the $n^{th}$ term of the AP. Therefore, we set $a_n = -150$.

Substituting the known values into the formula:

$-150 = 11 + (n - 1)(-3)$

Step 4: Solving for $n$

Subtract $11$ from both sides of the equation:

$-150 - 11 = (n - 1)(-3)$

$-161 = (n - 1)(-3)$

Divide both sides by $-3$:

$\frac{-161}{-3} = n - 1$

$\frac{161}{3} = n - 1$

Add $1$ to both sides to isolate $n$:

$n = \frac{161}{3} + 1$

$n = \frac{161 + 3}{3}$

$n = \frac{164}{3}$

$n = 54.66\dots$

Step 5: Logical Conclusion

In an Arithmetic Progression, the position of a term ($n$) must always be a natural number (i.e., a positive integer: $1, 2, 3, \dots$).

[Since $n = 54.66\dots$ is not an integer, it implies that $-150$ cannot be a term in this sequence.]

Final Answer: -150 is not a term of the given AP because the value of $n$ is not a positive integer.

Given: An Arithmetic Progression (AP) sequence: $18, 15\frac{1}{2}, 13, \dots, -47$.

To find: The number of terms ($n$) in the given AP.

Step 1: Identify the parameters of the Arithmetic Progression.

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

The first term is $a = 18$.

The second term is $a_2 = 15\frac{1}{2} = \frac{31}{2}$.

The common difference ($d$) is calculated as $d = a_2 - a_1$:

$d = \frac{31}{2} - 18$

$d = \frac{31}{2} - \frac{36}{2}$ [Converting 18 to a fraction with denominator 2]

$d = -\frac{5}{2}$

Step 2: State the General Term Formula.

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

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

Where:

• $a_n$ is the last term (given as $-47$)
• $a$ is the first term ($18$)
• $n$ is the number of terms
• $d$ is the common difference ($-\frac{5}{2}$)

Step 3: Substitute the values into the formula and solve for $n$.

$-47 = 18 + (n - 1)\left(-\frac{5}{2}\right)$

Subtract 18 from both sides:

$-47 - 18 = (n - 1)\left(-\frac{5}{2}\right)$

$-65 = (n - 1)\left(-\frac{5}{2}\right)$

Multiply both sides by $-\frac{2}{5}$ to isolate $(n - 1)$:

$-65 \times \left(-\frac{2}{5}\right) = n - 1$

$\frac{130}{5} = n - 1$ [Since $65 \div 5 = 13$]

$26 = n - 1$

Add 1 to both sides:

$n = 26 + 1$

$n = 27$

Step 4: Conclusion.

Since $n$ represents the count of terms, it must be a positive integer. Our result $n = 27$ satisfies this condition.

Final Answer: The number of terms in the given AP is 27.

Given: An Arithmetic Progression (AP) series: $7, 13, 19, \dots, 205$.

To find: The number of terms ($n$) in the given AP.

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 ($a$) is the first number in the sequence: $a = 7$.
- The second term ($a_2$) is $13$.
- The common difference ($d$) is calculated as the difference between any two consecutive terms: $d = a_2 - a_1$.
$d = 13 - 7 = 6$.
- The last term ($a_n$ or $l$) is given as $205$.

Step 2: State the relevant formula.
The formula for the $n^{th}$ term of an Arithmetic Progression is given by:
$a_n = a + (n - 1)d$
Where:
$a_n$ = the $n^{th}$ term
$a$ = the first term
$n$ = the number of terms
$d$ = the common difference

Step 3: Substitute the known values into the formula.
Substituting $a_n = 205$, $a = 7$, and $d = 6$ into the formula:
$205 = 7 + (n - 1)6$

Step 4: Solve for $n$.
Subtract $7$ from both sides of the equation:
$205 - 7 = (n - 1)6$
$198 = (n - 1)6$

Divide both sides by $6$:
$\frac{198}{6} = n - 1$
$33 = n - 1$

Add $1$ to both sides to isolate $n$:
$n = 33 + 1$
$n = 34$

Justification:
Since the number of terms must be a positive integer, and our result $n = 34$ satisfies this condition, the calculation is consistent with the properties of an Arithmetic Progression.

Final Answer: The number of terms in the given AP is 34.

Given:

An Arithmetic Progression (AP) consists of $n = 50$ terms.

The 3rd term ($a_3$) is $12$.

The last term ($a_{50}$) is $106$.

To Find:

The 29th term ($a_{29}$) of the AP.

Step 1: Defining the General Formula for an AP

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

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

Where:

• $a$ is the first term.
• $d$ is the common difference.
• $n$ is the position of the term.

Step 2: Formulating Equations based on Given Information

Using the formula for the 3rd term ($n=3$):

$a_3 = a + (3 - 1)d = 12$

$a + 2d = 12$ --- (Equation 1)

Using the formula for the 50th term ($n=50$):

$a_{50} = a + (50 - 1)d = 106$

$a + 49d = 106$ --- (Equation 2)

Step 3: Solving the System of Linear Equations

To find the values of $a$ and $d$, subtract Equation 1 from Equation 2:

$(a + 49d) - (a + 2d) = 106 - 12$

$a - a + 49d - 2d = 94$

$47d = 94$

$d = \frac{94}{47}$

$d = 2$ [Since $47 \times 2 = 94$]

Step 4: Finding the First Term ($a$)

Substitute the value of $d = 2$ into Equation 1:

$a + 2(2) = 12$

$a + 4 = 12$

$a = 12 - 4$

$a = 8$

Step 5: Calculating the 29th Term ($a_{29}$)

Now, use the general formula $a_n = a + (n - 1)d$ for $n = 29$:

$a_{29} = a + (29 - 1)d$

$a_{29} = 8 + (28)(2)$

$a_{29} = 8 + 56$

$a_{29} = 64$

Final Answer: The 29th term of the Arithmetic Progression is 64.

Given: An Arithmetic Progression (AP) sequence: $3, 8, 13, \dots, 253$.

To Find: The $20^{th}$ term from the end (last term) of the given AP.

Step 1: Identify the parameters of the given AP.

The sequence is $3, 8, 13, \dots, 253$.

First term ($a$) = $3$.

Common difference ($d$) = $a_2 - a_1 = 8 - 3 = 5$.

Last term ($l$) = $253$.

Step 2: Formulate the strategy to find the $n^{th}$ term from the end.

To find the $n^{th}$ term from the end of an AP, we can reverse the sequence. The new AP will have the last term of the original AP as its first term, and the common difference will be the negative of the original common difference ($-d$).

Let the reversed AP be: $253, 248, 243, \dots, 3$.

For this reversed AP:

New first term ($a'$) = $253$.

New common difference ($d'$) = $-5$.

We need to find the $20^{th}$ term ($n = 20$).

Step 3: Apply the formula for the $n^{th}$ term of an AP.

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

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

[Where $a_n$ is the $n^{th}$ term, $a'$ is the first term, $n$ is the position, and $d'$ is the common difference.]

Step 4: Substitute the values into the formula.

$a_{20} = 253 + (20 - 1)(-5)$

$a_{20} = 253 + (19)(-5)$

[Performing the multiplication: $19 \times -5 = -95$]

$a_{20} = 253 - 95$

Step 5: Perform the final subtraction.

$a_{20} = 158$

Final Answer: The 20th term from the last term of the AP is 158.

Given: An Arithmetic Progression (AP) sequence: $3, 8, 13, 18, \dots$

To Find: The position of the term $78$ in the given AP (i.e., the value of $n$ such that $a_n = 78$).

Step 1: Identify the parameters of the Arithmetic Progression.

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

The first term is given by: $a = 3$.

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

$d = a_2 - a_1 = 8 - 3 = 5$

$d = a_3 - a_2 = 13 - 8 = 5$

Thus, $a = 3$ and $d = 5$.

Step 2: State the formula for the $n^{th}$ term of an AP.

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

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

[Where $a_n$ is the $n^{th}$ term, $a$ is the first term, $n$ is the position of the term, and $d$ is the common difference.]

Step 3: Substitute the known values into the formula.

We are given that $a_n = 78$. Substituting $a_n = 78$, $a = 3$, and $d = 5$ into the formula:

$78 = 3 + (n - 1)5$

Step 4: Solve for $n$.

Subtract $3$ from both sides of the equation:

$78 - 3 = (n - 1)5$

$75 = (n - 1)5$

Divide both sides by $5$:

$\frac{75}{5} = n - 1$

$15 = n - 1$

Add $1$ to both sides to isolate $n$:

$n = 15 + 1$

$n = 16$

Step 5: Conclusion.

Since $n$ represents the position of the term in the sequence, it must be a positive integer. Our result $n = 16$ satisfies this condition.

Final Answer: The 16th term of the AP is 78.

Given:

1. The third term of the Arithmetic Progression ($a_3$) = $16$.

2. The seventh term ($a_7$) exceeds the fifth term ($a_5$) by $12$, which can be written as: $a_7 = a_5 + 12$.

To Find:

The Arithmetic Progression (AP), which is defined by its first term ($a$) and common difference ($d$).

Step 1: Establishing the General Formula

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

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

where $a$ is the first term and $d$ is the common difference.

Step 2: Formulating Equations based on Given Conditions

For the third term ($n=3$):

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

$16 = a + 2d$ --- (Equation 1)

For the relationship between the seventh and fifth terms:

$a_7 = a + (7 - 1)d = a + 6d$

$a_5 = a + (5 - 1)d = a + 4d$

Given $a_7 = a_5 + 12$, substitute the expressions:

$(a + 6d) = (a + 4d) + 12$

Step 3: Solving for the Common Difference ($d$)

Subtract $a$ from both sides:

$6d = 4d + 12$

Subtract $4d$ from both sides:

$6d - 4d = 12$

$2d = 12$

$d = \frac{12}{2}$

$d = 6$

Step 4: Solving for the First Term ($a$)

Substitute $d = 6$ into Equation 1:

$16 = a + 2(6)$

$16 = a + 12$

$a = 16 - 12$

$a = 4$

Step 5: Constructing the Arithmetic Progression

The general form of an AP is $a, a+d, a+2d, a+3d, \dots$

Term 1 ($a_1$) = $4$

Term 2 ($a_2$) = $a + d = 4 + 6 = 10$

Term 3 ($a_3$) = $a + 2d = 4 + 2(6) = 4 + 12 = 16$

Term 4 ($a_4$) = $a + 3d = 4 + 3(6) = 4 + 18 = 22$

Final Answer: The Arithmetic Progression is 4, 10, 16, 22, ...

Given:

Two Arithmetic Progressions (APs):

AP 1: $63, 65, 67, \dots$

AP 2: $3, 10, 17, \dots$

To Find:

The value of $n$ for which the $n$th term of AP 1 is equal to the $n$th term of AP 2.

Step 1: Identify the parameters of the first AP.

For the first AP: $63, 65, 67, \dots$

Let the first term be $a_1$ and the common difference be $d_1$.

$a_1 = 63$

$d_1 = 65 - 63 = 2$

The formula for the $n$th term of an AP is $a_n = a + (n - 1)d$.

Therefore, the $n$th term of the first AP ($A_n$) is:

$A_n = 63 + (n - 1)2$

$A_n = 63 + 2n - 2$

$A_n = 61 + 2n$ --- (Equation 1)

Step 2: Identify the parameters of the second AP.

For the second AP: $3, 10, 17, \dots$

Let the first term be $a_2$ and the common difference be $d_2$.

$a_2 = 3$

$d_2 = 10 - 3 = 7$

Therefore, the $n$th term of the second AP ($B_n$) is:

$B_n = 3 + (n - 1)7$

$B_n = 3 + 7n - 7$

$B_n = 7n - 4$ --- (Equation 2)

Step 3: Equate the $n$th terms and solve for $n$.

According to the problem, the $n$th terms are equal, so $A_n = B_n$.

Substituting Equation 1 and Equation 2:

$61 + 2n = 7n - 4$

Rearranging the terms to isolate $n$:

$61 + 4 = 7n - 2n$ [Transposing $-4$ to the left and $2n$ to the right]

$65 = 5n$

$n = \frac{65}{5}$ [Dividing both sides by 5]

$n = 13$

Step 4: Verification (Optional but recommended).

For $n = 13$ in AP 1: $A_{13} = 63 + (13 - 1)2 = 63 + 12(2) = 63 + 24 = 87$.

For $n = 13$ in AP 2: $B_{13} = 3 + (13 - 1)7 = 3 + 12(7) = 3 + 84 = 87$.

Since $87 = 87$, the value is verified.

Final Answer: The value of $n$ for which the $n$th terms of the two APs are equal is 13.