Your Worksheet is Ready
CBSE - Class 10 Mathematics Arithmetic Progression Worksheet
EXERCISE 5.1
Worksheet Answers
Solution:
Given:
Principal amount ($P$) = ₹ $10000$
Rate of interest ($r$) = $8\%$ per annum
The interest is compounded annually.
To Find:
Whether the sequence of amounts at the end of each year forms an Arithmetic Progression (AP).
Step 1: Understanding the Formula for Compound Interest
The amount ($A$) after $n$ years with compound interest is given by the formula:
$A = P \left(1 + \frac{r}{100}\right)^n$
Where:
Step 2: Calculating the amount for consecutive years
Let $a_1, a_2, a_3, \dots$ be the amount in the account at the end of the 1st, 2nd, and 3rd year respectively.
For $n = 1$ (Amount at the end of 1st year):
$a_1 = 10000 \left(1 + \frac{8}{100}\right)^1 = 10000(1.08) = 10800$
For $n = 2$ (Amount at the end of 2nd year):
$a_2 = 10000 \left(1 + \frac{8}{100}\right)^2 = 10000(1.08)^2 = 10000(1.1664) = 11664$
For $n = 3$ (Amount at the end of 3rd year):
$a_3 = 10000 \left(1 + \frac{8}{100}\right)^3 = 10000(1.259712) = 12597.12$
Step 3: Checking for Arithmetic Progression
A sequence is an Arithmetic Progression if the difference between consecutive terms is constant (i.e., $a_{n+1} - a_n = d$, where $d$ is the common difference).
Calculate the first difference ($d_1$):
$d_1 = a_2 - a_1 = 11664 - 10800 = 864$
Calculate the second difference ($d_2$):
$d_2 = a_3 - a_2 = 12597.12 - 11664 = 933.12$
Step 4: Conclusion
Since $d_1 \neq d_2$ ($864 \neq 933.12$), the difference between consecutive terms is not constant.
[By definition, a sequence is an AP if and only if the common difference is constant for all terms.]
Final Answer: The list of numbers does not form an Arithmetic Progression because the difference between consecutive terms is not constant.
Solution:
Given: A sequence of numbers $1^2, 3^2, 5^2, 7^2, \dots$
To Find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and the next three terms.
Step 1: Evaluating the terms of the sequence
Let the sequence be denoted by $a_1, a_2, a_3, a_4, \dots$
Calculating the values of the given terms:
$a_1 = 1^2 = 1$
$a_2 = 3^2 = 9$
$a_3 = 5^2 = 25$
$a_4 = 7^2 = 49$
The sequence is $1, 9, 25, 49, \dots$
Step 2: Checking for a common difference
A sequence is an AP if the difference between consecutive terms is constant, i.e., $a_{n+1} - a_n = d$ for all $n$.
Calculate the difference between the first and second terms ($d_1$):
$d_1 = a_2 - a_1 = 9 - 1 = 8$
Calculate the difference between the second and third terms ($d_2$):
$d_2 = a_3 - a_2 = 25 - 9 = 16$
Calculate the difference between the third and fourth terms ($d_3$):
$d_3 = a_4 - a_3 = 49 - 25 = 24$
Step 3: Conclusion on the nature of the sequence
Since $d_1 \neq d_2 \neq d_3$ (specifically, $8 \neq 16 \neq 24$), the difference between consecutive terms is not constant.
[Definition of an Arithmetic Progression: A sequence is an AP if and only if the difference between any two consecutive terms is constant.]
Final Answer: The given sequence $1^2, 3^2, 5^2, 7^2, \dots$ does not form an Arithmetic Progression because the common difference is not constant.
Solution:
Given: A sequence of numbers: $0, -4, -8, -12, \dots$
To find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference ($d$).
3. Write the next three terms of the sequence.
Step 1: Definition of an Arithmetic Progression
A sequence $a_1, a_2, a_3, \dots, a_n$ is said to be an Arithmetic Progression if the difference between consecutive terms remains constant. This constant difference is called the common difference ($d$), defined as:
$d = a_{n} - a_{n-1}$ for all $n > 1$.
Step 2: Calculating differences between consecutive terms
Let the given terms be:
$a_1 = 0$
$a_2 = -4$
$a_3 = -8$
$a_4 = -12$
Calculate the differences:
Difference 1 ($d_1$): $a_2 - a_1 = -4 - 0 = -4$
Difference 2 ($d_2$): $a_3 - a_2 = -8 - (-4) = -8 + 4 = -4$
Difference 3 ($d_3$): $a_4 - a_3 = -12 - (-8) = -12 + 8 = -4$
Step 3: Verification of AP
[Since $d_1 = d_2 = d_3 = -4$, the difference between consecutive terms is constant.]
Therefore, the given sequence forms an Arithmetic Progression with a common difference $d = -4$.
Step 4: Finding the next three terms
To find the next terms, we add the common difference ($d = -4$) to the last known term ($a_4 = -12$).
Let the next three terms be $a_5, a_6,$ and $a_7$.
Calculation for $a_5$:
$a_5 = a_4 + d = -12 + (-4) = -16$
Calculation for $a_6$:
$a_6 = a_5 + d = -16 + (-4) = -20$
Calculation for $a_7$:
$a_7 = a_6 + d = -20 + (-4) = -24$
Final Answer:
The sequence forms an AP.
The common difference $d = -4$.
The next three terms are -16, -20, -24.
Solution:
Given:
To Find:
Determine whether the sequence of taxi fares for $1, 2, 3, 4, \dots$ kilometers forms an Arithmetic Progression (AP) and provide the justification.
Definition:
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$). If the terms are $a_1, a_2, a_3, \dots, a_n$, then the sequence is an AP if $(a_2 - a_1) = (a_3 - a_2) = (a_4 - a_3) = d$.
Step 1: Calculating the terms of the sequence
Let $a_n$ be the fare for $n$ kilometers.
The sequence of fares is: $15, 23, 31, 39, \dots$
Step 2: Verifying the common difference
To check if the sequence is an AP, we calculate the difference between consecutive terms:
Step 3: Conclusion
Since the difference between consecutive terms is constant ($d = 8$), the sequence satisfies the definition of an Arithmetic Progression.
Final Answer: Yes, the situation forms an Arithmetic Progression because the fare increases by a constant amount of ₹ 8 for each additional kilometer, resulting in a common difference of 8.
Solution:
Given: A sequence of numbers: $1^2, 5^2, 7^2, 73, \dots$
To Find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and the next three terms.
Step 1: Simplifying the terms of the sequence
Let the given sequence be denoted by $a_1, a_2, a_3, a_4, \dots$
Calculating the values of each term:
$a_1 = 1^2 = 1$
$a_2 = 5^2 = 25$
$a_3 = 7^2 = 49$
$a_4 = 73$
The sequence is: $1, 25, 49, 73, \dots$
Step 2: Checking for a common difference
A sequence is an AP if the difference between consecutive terms ($a_{n+1} - a_n$) is constant. This constant is called the common difference $d$.
Calculate the difference between the first and second terms ($d_1$):
$d_1 = a_2 - a_1 = 25 - 1 = 24$
Calculate the difference between the second and third terms ($d_2$):
$d_2 = a_3 - a_2 = 49 - 25 = 24$
Calculate the difference between the third and fourth terms ($d_3$):
$d_3 = a_4 - a_3 = 73 - 49 = 24$
[Since $d_1 = d_2 = d_3 = 24$, the difference between consecutive terms is constant.]
Step 3: Conclusion on AP status
Because the difference between consecutive terms is constant ($d = 24$), the given sequence forms an Arithmetic Progression.
Step 4: Finding the next three terms
To find the next three terms ($a_5, a_6, a_7$), we add the common difference $d = 24$ to the preceding term.
Fifth term ($a_5$):
$a_5 = a_4 + d = 73 + 24 = 97$
Sixth term ($a_6$):
$a_6 = a_5 + d = 97 + 24 = 121$
Seventh term ($a_7$):
$a_7 = a_6 + d = 121 + 24 = 145$
Final Answer: The sequence forms an AP with a common difference $d = 24$. The next three terms are 97, 121, and 145.
Solution:
Given: A sequence of numbers: $0.2, 0.22, 0.222, 0.2222, \dots$
To Find: Determine if the given sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and write the next three terms.
Step 1: Definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).
For a sequence $a_1, a_2, a_3, a_4, \dots$, the sequence is an AP if and only if:
$a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = d$
Step 2: Calculating the differences between consecutive terms
Let the terms be:
$a_1 = 0.2$
$a_2 = 0.22$
$a_3 = 0.222$
$a_4 = 0.2222$
Calculate the difference between the first and second term ($d_1$):
$d_1 = a_2 - a_1 = 0.22 - 0.2 = 0.02$
Calculate the difference between the second and third term ($d_2$):
$d_2 = a_3 - a_2 = 0.222 - 0.22 = 0.002$
Calculate the difference between the third and fourth term ($d_3$):
$d_3 = a_4 - a_3 = 0.2222 - 0.222 = 0.0002$
Step 3: Comparing the differences
We observe that:
$d_1 = 0.02$
$d_2 = 0.002$
$d_3 = 0.0002$
Since $d_1 \neq d_2 \neq d_3$, the difference between consecutive terms is not constant.
Step 4: Conclusion
Because the common difference is not constant, the given sequence $0.2, 0.22, 0.222, 0.2222, \dots$ does not satisfy the condition for an Arithmetic Progression.
Final Answer: The given sequence does not form an AP because the difference between consecutive terms is not constant.
Solution:
Given: An Arithmetic Progression (AP) sequence: $-5, -1, 3, 7, \dots$
To find: The first term ($a$) and the common difference ($d$) of the given AP.
Step 1: Identifying the first term ($a$)
In an Arithmetic Progression, the first term is denoted by the variable $a$ (or $a_1$). By observing the given sequence:
Sequence: $-5, -1, 3, 7, \dots$
The first term is the element at the first position in the sequence.
Therefore, $a = -5$.
Step 2: Defining the common difference ($d$)
The common difference ($d$) of an Arithmetic Progression is defined as the difference between any two consecutive terms. Mathematically, it is expressed as:
$d = a_{n} - a_{n-1}$
where $a_n$ is the $n^{th}$ term and $a_{n-1}$ is the preceding term.
Step 3: Calculating the common difference ($d$)
We can calculate $d$ by subtracting the first term from the second term:
$d = a_2 - a_1$
Given $a_1 = -5$ and $a_2 = -1$:
$d = (-1) - (-5)$
[Applying the rule of signs: subtracting a negative number is equivalent to adding its positive counterpart]
$d = -1 + 5$
$d = 4$
Step 4: Verification of the common difference
To ensure the sequence is indeed an AP, we verify the difference between other consecutive terms:
For the third and second terms: $d = a_3 - a_2 = 3 - (-1) = 3 + 1 = 4$
For the fourth and third terms: $d = a_4 - a_3 = 7 - 3 = 4$
[Since the difference is constant throughout the sequence, the value $d = 4$ is confirmed.]
Final Answer: The first term ($a$) is $-5$ and the common difference ($d$) is $4$.
Solution:
Given: An Arithmetic Progression (AP) sequence: $3, 1, -1, -3, \dots$
To Find: The first term ($a$) and the common difference ($d$) of the given AP.
Step 1: Defining the variables of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is known as the common difference ($d$), and the first number in the sequence is denoted as the first term ($a$).
The sequence is represented as: $a_1, a_2, a_3, a_4, \dots, a_n$
Step 2: Identifying the first term ($a$)
By definition, the first term ($a$) is the first element of the sequence.
Given sequence: $3, 1, -1, -3, \dots$
Therefore, $a = a_1 = 3$.
Step 3: Calculating the common difference ($d$)
The common difference ($d$) is calculated by subtracting any term from the term that immediately follows it. The formula is given by:
$d = a_{n} - a_{n-1}$
Let us calculate $d$ using the first two terms ($a_1$ and $a_2$):
$d = a_2 - a_1$
$d = 1 - 3$
$d = -2$
Step 4: Verification of the common difference
To ensure the sequence is indeed an AP, we verify the difference across other consecutive terms:
For $a_3$ and $a_2$: $d = a_3 - a_2 = -1 - 1 = -2$
For $a_4$ and $a_3$: $d = a_4 - a_3 = -3 - (-1) = -3 + 1 = -2$
[Since the difference is constant across all tested terms, the common difference $d = -2$ is verified.]
Final Answer: The first term $a = 3$ and the common difference $d = -2$.
Solution:
Given: A sequence of numbers: $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$
To find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference $d$.
3. Write the next three terms of the sequence.
Step 1: Simplify the terms of the sequence
Let the terms of the sequence be represented by $a_1, a_2, a_3, a_4, \dots$
$a_1 = \sqrt{3}$
$a_2 = \sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$
$a_3 = \sqrt{9} = 3$
$a_4 = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
Step 2: Check for a common difference
A sequence is an AP if the difference between consecutive terms ($d = a_{n+1} - a_n$) is constant.
Calculate the difference between the first two terms ($d_1$):
$d_1 = a_2 - a_1 = \sqrt{6} - \sqrt{3}$
$d_1 = \sqrt{3}(\sqrt{2} - 1)$
Calculate the difference between the second and third terms ($d_2$):
$d_2 = a_3 - a_2 = 3 - \sqrt{6}$
$d_2 = \sqrt{3}(\sqrt{3} - \sqrt{2})$
Step 3: Compare the differences
We observe that $d_1 \neq d_2$:
$\sqrt{3}(\sqrt{2} - 1) \approx 1.732(1.414 - 1) = 1.732(0.414) \approx 0.717$
$\sqrt{3}(\sqrt{3} - \sqrt{2}) \approx 1.732(1.732 - 1.414) = 1.732(0.318) \approx 0.551$
[Since the difference between consecutive terms is not constant, the sequence does not satisfy the definition of an Arithmetic Progression.]
Step 4: Conclusion
Because the common difference $d$ is not constant throughout the sequence, the given list of numbers does not form an Arithmetic Progression.
Final Answer: The given sequence $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$ does not form an AP because the difference between consecutive terms is not constant.
Solution:
Given: A sequence of numbers: $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$
To find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and write the next three terms.
Step 1: Definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).
Let the terms be $a_1, a_2, a_3, a_4, \dots$
Here, $a_1 = 3$, $a_2 = 3+\sqrt{2}$, $a_3 = 3+2\sqrt{2}$, $a_4 = 3+3\sqrt{2}$.
Step 2: Calculating the differences between consecutive terms
We calculate the difference $d_n = a_{n+1} - a_n$ for consecutive terms:
Difference 1 ($d_1$):
$d_1 = a_2 - a_1 = (3 + \sqrt{2}) - 3$
$d_1 = 3 - 3 + \sqrt{2} = \sqrt{2}$
Difference 2 ($d_2$):
$d_2 = a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2})$
$d_2 = 3 + 2\sqrt{2} - 3 - \sqrt{2}$
$d_2 = (3 - 3) + (2\sqrt{2} - \sqrt{2}) = \sqrt{2}$
Difference 3 ($d_3$):
$d_3 = a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2})$
$d_3 = 3 + 3\sqrt{2} - 3 - 2\sqrt{2}$
$d_3 = (3 - 3) + (3\sqrt{2} - 2\sqrt{2}) = \sqrt{2}$
Step 3: Verification
Since $d_1 = d_2 = d_3 = \sqrt{2}$, the difference between consecutive terms is constant. Therefore, the given sequence is an Arithmetic Progression with common difference $d = \sqrt{2}$.
Step 4: Finding the next three terms
To find the next terms, we add the common difference $d = \sqrt{2}$ to the last known term ($a_4 = 3 + 3\sqrt{2}$):
Fifth term ($a_5$):
$a_5 = a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$
Sixth term ($a_6$):
$a_6 = a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$
Seventh term ($a_7$):
$a_7 = a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$
Final Answer: The sequence forms an AP with common difference $d = \sqrt{2}$. The next three terms are $3+4\sqrt{2}, 3+5\sqrt{2},$ and $3+6\sqrt{2}$.
Solution:
Given: The first term of the Arithmetic Progression (AP) is $a = -1.25$ and the common difference is $d = -0.25$.
To Find: The first four terms of the Arithmetic Progression.
Step 1: Understanding the definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is denoted by $d$. If the first term is $a$, the terms of the AP are given by the general form: $a, a+d, a+2d, a+3d, \dots$
Step 2: Calculating the first four terms
Let the terms be denoted as $a_1, a_2, a_3,$ and $a_4$.
Calculation for the first term ($a_1$):
$a_1 = a$
$a_1 = -1.25$
Calculation for the second term ($a_2$):
$a_2 = a + d$
$a_2 = -1.25 + (-0.25)$
$a_2 = -1.25 - 0.25$
$a_2 = -1.50$
Calculation for the third term ($a_3$):
$a_3 = a_2 + d$
$a_3 = -1.50 + (-0.25)$
$a_3 = -1.50 - 0.25$
$a_3 = -1.75$
Calculation for the fourth term ($a_4$):
$a_4 = a_3 + d$
$a_4 = -1.75 + (-0.25)$
$a_4 = -1.75 - 0.25$
$a_4 = -2.00$
Step 3: Summary of the sequence
The first four terms of the AP are $-1.25, -1.50, -1.75,$ and $-2.00$.
Final Answer: The first four terms of the AP are $-1.25, -1.50, -1.75, -2.00$.
Solution:
Given: A sequence of numbers $a, a^2, a^3, a^4, \dots$
To find: Determine if the given sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and the next three terms.
Step 1: Definition of an Arithmetic Progression
A sequence $a_1, a_2, a_3, \dots, a_n$ is said to be an Arithmetic Progression if the difference between consecutive terms is constant. That is, $a_{n+1} - a_n = d$ for all $n \geq 1$, where $d$ is the common difference.
Step 2: Calculating the differences between consecutive terms
Let the terms be:
$a_1 = a$
$a_2 = a^2$
$a_3 = a^3$
$a_4 = a^4$
Calculate the difference between the first and second terms ($d_1$):
$d_1 = a_2 - a_1 = a^2 - a = a(a - 1)$
Calculate the difference between the second and third terms ($d_2$):
$d_2 = a_3 - a_2 = a^3 - a^2 = a^2(a - 1)$
Step 3: Comparing the differences
For the sequence to be an AP, the condition $d_1 = d_2$ must hold true.
Comparing $d_1$ and $d_2$:
$a(a - 1) \neq a^2(a - 1)$ (assuming $a \neq 0$ and $a \neq 1$)
Since $d_1 \neq d_2$, the difference between consecutive terms is not constant.
Step 4: Conclusion on the nature of the sequence
Because the difference between consecutive terms is not constant, the sequence $a, a^2, a^3, a^4, \dots$ does not satisfy the definition of an Arithmetic Progression.
Final Answer: The given sequence $a, a^2, a^3, a^4, \dots$ does not form an Arithmetic Progression because the common difference is not constant.
Solution:
Given: A sequence of numbers: $2, 4, 8, 16, \dots$
To Find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference $d$ and the next three terms.
Step 1: Definition and Condition for an Arithmetic Progression
A sequence $a_1, a_2, a_3, a_4, \dots$ is said to be an Arithmetic Progression if the difference between consecutive terms remains constant. This constant difference is called the common difference ($d$).
Mathematically, the condition is:
$a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = d$
Step 2: Calculating the differences between consecutive terms
Let the given terms be:
$a_1 = 2$
$a_2 = 4$
$a_3 = 8$
$a_4 = 16$
Now, calculate the differences:
Difference 1 ($d_1$): $a_2 - a_1 = 4 - 2 = 2$
Difference 2 ($d_2$): $a_3 - a_2 = 8 - 4 = 4$
Difference 3 ($d_3$): $a_4 - a_3 = 16 - 8 = 8$
Step 3: Analyzing the results
Comparing the differences obtained:
$d_1 = 2$
$d_2 = 4$
$d_3 = 8$
Since $d_1 \neq d_2 \neq d_3$, the difference between consecutive terms is not constant.
[By the definition of an Arithmetic Progression, a sequence must have a constant common difference to be classified as an AP.]
Step 4: Conclusion
Because the common difference is not constant, the sequence $2, 4, 8, 16, \dots$ does not form an Arithmetic Progression.
Final Answer: The given sequence $2, 4, 8, 16, \dots$ does not form an AP because the difference between consecutive terms is not constant.
Solution:
Given: A sequence of numbers: $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$
To Find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and the next three terms.
Step 1: Simplify the given terms
To analyze the sequence, we first simplify each radical term by extracting perfect square factors:
Term 1 ($a_1$) = $\sqrt{2} = 1\sqrt{2}$
Term 2 ($a_2$) = $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$
Term 3 ($a_3$) = $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$
Term 4 ($a_4$) = $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$
The sequence is: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$
Step 2: Verify if the sequence is an AP
A sequence is an AP if the difference between consecutive terms is constant. This constant is called the common difference ($d$).
Calculate the differences between consecutive terms:
$d_1 = a_2 - a_1 = 2\sqrt{2} - 1\sqrt{2} = (2-1)\sqrt{2} = \sqrt{2}$
$d_2 = a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = (3-2)\sqrt{2} = \sqrt{2}$
$d_3 = a_4 - a_3 = 4\sqrt{2} - 3\sqrt{2} = (4-3)\sqrt{2} = \sqrt{2}$
[Since $d_1 = d_2 = d_3 = \sqrt{2}$, the difference between consecutive terms is constant.]
Therefore, the given sequence is an Arithmetic Progression with common difference $d = \sqrt{2}$.
Step 3: Find the next three terms
The sequence currently has four terms. We need to find the 5th, 6th, and 7th terms using the formula $a_n = a_{n-1} + d$.
Term 5 ($a_5$) = $a_4 + d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50}$
Term 6 ($a_6$) = $a_5 + d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72}$
Term 7 ($a_7$) = $a_6 + d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98}$
Final Answer: The sequence forms an AP with common difference $d = \sqrt{2}$. The next three terms are $\sqrt{50}, \sqrt{72}, \text{ and } \sqrt{98}$.
Solution:
Given:
The first term of the Arithmetic Progression (AP), denoted by $a = -2$.
The common difference of the AP, denoted by $d = 0$.
To Find:
The first four terms of the Arithmetic Progression ($a_1, a_2, a_3, a_4$).
Step 1: Understanding the definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference ($d$).
The general form of an AP is given by: $a, a+d, a+2d, a+3d, \dots$
Where:
Step 2: Calculating the first four terms
We substitute the given values $a = -2$ and $d = 0$ into the general formulas defined in Step 1.
Calculation for the first term ($a_1$):
$a_1 = a = -2$
Calculation for the second term ($a_2$):
$a_2 = a + d$
$a_2 = -2 + 0$
$a_2 = -2$
Calculation for the third term ($a_3$):
$a_3 = a + 2d$
$a_3 = -2 + 2(0)$
$a_3 = -2 + 0$
$a_3 = -2$
Calculation for the fourth term ($a_4$):
$a_4 = a + 3d$
$a_4 = -2 + 3(0)$
$a_4 = -2 + 0$
$a_4 = -2$
Step 3: Conclusion
Since the common difference $d$ is $0$, every term in the sequence remains identical to the first term $a$.
Final Answer: The first four terms of the AP are -2, -2, -2, -2.
Solution:
Given: A sequence of numbers: $-10, -6, -2, 2, \dots$
To Find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference ($d$).
3. Write the next three terms of the sequence.
Step 1: Defining the terms of the sequence
Let the given sequence be represented by $a_1, a_2, a_3, a_4, \dots$ where:
$a_1 = -10$
$a_2 = -6$
$a_3 = -2$
$a_4 = 2$
Step 2: Checking for a common difference
A sequence is an Arithmetic Progression if the difference between consecutive terms is constant. This constant is called the common difference ($d$).
We calculate the differences between consecutive terms:
$d_1 = a_2 - a_1 = -6 - (-10) = -6 + 10 = 4$
$d_2 = a_3 - a_2 = -2 - (-6) = -2 + 6 = 4$
$d_3 = a_4 - a_3 = 2 - (-2) = 2 + 2 = 4$
Step 3: Verification of AP
Since $d_1 = d_2 = d_3 = 4$, the difference between consecutive terms is constant.
[Definition: An Arithmetic Progression is a sequence of numbers such that the difference between the consecutive terms is constant.]
Therefore, the given sequence is an AP with a common difference $d = 4$.
Step 4: Finding the next three terms
To find the next terms, we add the common difference $d = 4$ to the preceding term.
Let the next three terms be $a_5, a_6,$ and $a_7$.
Calculation for $a_5$:
$a_5 = a_4 + d = 2 + 4 = 6$
Calculation for $a_6$:
$a_6 = a_5 + d = 6 + 4 = 10$
Calculation for $a_7$:
$a_7 = a_6 + d = 10 + 4 = 14$
Final Answer:
The sequence forms an AP.
The common difference $d = 4$.
The next three terms are 6, 10, 14.
Solution:
Given: A sequence of numbers: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$
To Find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference ($d$).
3. Write the next three terms of the sequence.
Step 1: Definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).
Let the terms of the sequence be represented as $a_1, a_2, a_3, a_4, \dots$
The sequence is an AP if $(a_2 - a_1) = (a_3 - a_2) = (a_4 - a_3) = d$.
Step 2: Calculating the differences between consecutive terms
Given terms: $a_1 = -\frac{1}{2}$, $a_2 = -\frac{1}{2}$, $a_3 = -\frac{1}{2}$, $a_4 = -\frac{1}{2}$.
Difference $d_1 = a_2 - a_1 = (-\frac{1}{2}) - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$
Difference $d_2 = a_3 - a_2 = (-\frac{1}{2}) - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$
Difference $d_3 = a_4 - a_3 = (-\frac{1}{2}) - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$
Step 3: Verification and Conclusion on AP
Since $d_1 = d_2 = d_3 = 0$, the difference between consecutive terms is constant.
[Since the common difference is constant, the sequence forms an Arithmetic Progression.]
The common difference $d = 0$.
Step 4: Finding the next three terms
To find the next terms, we add the common difference $d$ to the last known term.
Let the next three terms be $a_5, a_6,$ and $a_7$.
$a_5 = a_4 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$
$a_6 = a_5 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$
$a_7 = a_6 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$
Final Answer: The sequence forms an AP with common difference $d = 0$. The next three terms are $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$.
Solution:
Given:
The cost of digging the first metre of the well = ₹ $150$.
The cost of digging each subsequent metre increases by a fixed amount of ₹ $50$.
To Find:
Determine whether the sequence of costs for digging $1, 2, 3, \dots, n$ metres forms an Arithmetic Progression (AP) and provide the justification.
Step 1: Defining the sequence of costs
Let $a_n$ represent the cost of digging the well up to $n$ metres.
According to the problem statement:
Cost for the first metre ($a_1$) = ₹ $150$.
Cost for the first two metres ($a_2$) = Cost of first metre + Cost of second metre = $150 + 50 = ₹ 200$.
Cost for the first three metres ($a_3$) = Cost of first two metres + Cost of third metre = $200 + 50 = ₹ 250$.
Cost for the first four metres ($a_4$) = Cost of first three metres + Cost of fourth metre = $250 + 50 = ₹ 300$.
Step 2: Analyzing the sequence
The list of numbers representing the costs is: $150, 200, 250, 300, \dots$
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 ($d$).
Let us calculate the differences between consecutive terms:
$d_1 = a_2 - a_1 = 200 - 150 = 50$
$d_2 = a_3 - a_2 = 250 - 200 = 50$
$d_3 = a_4 - a_3 = 300 - 250 = 50$
Step 3: Conclusion based on the definition of an AP
[Since the difference between any two consecutive terms is constant, i.e., $a_{n} - a_{n-1} = 50$ for all $n > 1$].
Because the common difference $d = 50$ is constant throughout the sequence, the situation represents an Arithmetic Progression.
Final Answer: Yes, the list of numbers forms an Arithmetic Progression because each term increases by a constant value of ₹ 50, which serves as the common difference.
Solution:
Given: A cylinder contains an initial amount of air. A vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at each stroke.
To Find: Determine whether the sequence of the amounts of air remaining in the cylinder after each stroke forms an Arithmetic Progression (AP).
Step 1: Defining the Variables
Let the initial amount of air present in the cylinder be $V$ units.
Let $a_1$ be the amount of air after the 0th stroke (initial state).
Let $a_2$ be the amount of air after the 1st stroke.
Let $a_3$ be the amount of air after the 2nd stroke.
Let $a_4$ be the amount of air after the 3rd stroke.
Step 2: Calculating the sequence of air amounts
The pump removes $\frac{1}{4}$ of the air present in the cylinder at each step.
Initial amount: $a_1 = V$
After the 1st stroke ($a_2$):
$a_2 = V - \frac{1}{4}V = \frac{3}{4}V$
After the 2nd stroke ($a_3$):
The pump removes $\frac{1}{4}$ of the air remaining, which is $a_2$.
$a_3 = a_2 - \frac{1}{4}a_2 = \frac{3}{4}a_2$
Substituting $a_2 = \frac{3}{4}V$:
$a_3 = \frac{3}{4} \times (\frac{3}{4}V) = \frac{9}{16}V$
After the 3rd stroke ($a_4$):
$a_4 = a_3 - \frac{1}{4}a_3 = \frac{3}{4}a_3$
Substituting $a_3 = \frac{9}{16}V$:
$a_4 = \frac{3}{4} \times (\frac{9}{16}V) = \frac{27}{64}V$
Step 3: Checking for Arithmetic Progression
A sequence is an Arithmetic Progression if the difference between consecutive terms is constant (i.e., $a_{n+1} - a_n = d$, where $d$ is the common difference).
Calculate the first difference ($d_1$):
$d_1 = a_2 - a_1 = \frac{3}{4}V - V = -\frac{1}{4}V$
Calculate the second difference ($d_2$):
$d_2 = a_3 - a_2 = \frac{9}{16}V - \frac{3}{4}V$
To subtract, find a common denominator (16):
$d_2 = \frac{9}{16}V - \frac{12}{16}V = -\frac{3}{16}V$
Step 4: Comparison and Conclusion
Since $d_1 \neq d_2$ (because $-\frac{1}{4}V \neq -\frac{3}{16}V$), the difference between consecutive terms is not constant.
[Definition of an Arithmetic Progression: A sequence of numbers is an AP if the difference between any two consecutive terms is constant.]
Final Answer: The list of numbers does not form an Arithmetic Progression because the difference between consecutive terms is not constant.
Solution:
Given: A sequence of numbers: $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
To find:
1. Determine if the given sequence forms an Arithmetic Progression (AP).
2. If it forms an AP, find the common difference ($d$).
3. Write the next three terms of the sequence.
Step 1: Identifying the terms of the sequence
Let the sequence be denoted by $a_1, a_2, a_3, a_4, \dots$
$a_1 = 2$
$a_2 = \frac{5}{2}$
$a_3 = 3$
$a_4 = \frac{7}{2}$
Step 2: Checking for a common difference
A sequence is an AP if the difference between consecutive terms is constant. This constant is called the common difference ($d$).
Calculate the difference between consecutive terms:
Difference $d_1 = a_2 - a_1 = \frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}$
Difference $d_2 = a_3 - a_2 = 3 - \frac{5}{2} = \frac{6}{2} - \frac{5}{2} = \frac{1}{2}$
Difference $d_3 = a_4 - a_3 = \frac{7}{2} - 3 = \frac{7}{2} - \frac{6}{2} = \frac{1}{2}$
Step 3: Conclusion on AP status
[Since $d_1 = d_2 = d_3 = \frac{1}{2}$, the difference between consecutive terms is constant.]
Therefore, the given sequence is an Arithmetic Progression with a common difference $d = \frac{1}{2}$.
Step 4: Finding the next three terms
To find the next terms, we add the common difference $d = \frac{1}{2}$ to the last known term ($a_4 = \frac{7}{2}$).
Fifth term ($a_5$) = $a_4 + d = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$
Sixth term ($a_6$) = $a_5 + d = 4 + \frac{1}{2} = \frac{8}{2} + \frac{1}{2} = \frac{9}{2}$
Seventh term ($a_7$) = $a_6 + d = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5$
Final Answer: The sequence forms an AP with common difference $d = \frac{1}{2}$. The next three terms are $4, \frac{9}{2}, 5$.