Generating Secure PDF Link
Math Notes For Class 11 Chapter 4 Ex 4.1 Solved & Quiz
Please Wait
10
seconds remaining...
Math Notes For Class 11 Chapter 4 Ex 4.1 Solved & Quiz
An Arithmetic Progression (A.P.) or Arithmetic Sequence 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 ($d$).
For a sequence $a_1, a_2, a_3, \dots, a_n$, the sequence is an A.P. if and only if:
$$d = a_{k} - a_{k-1} \quad \text{for all } k > 1$$If $a_1$ represents the first term and $d$ is the common difference, the sequence takes the standard form:
$$a_1, \; a_1 + d, \; a_1 + 2d, \; a_1 + 3d, \; \dots, \; a_1 + (n-1)d$$| Concept / Quantity | Mathematical Formula | Description / Key Notes |
|---|---|---|
| $n$-th General Term ($a_n$) | $$a_n = a_1 + (n-1)d$$ | Finds any specific term where $a_1$ is first term and $d$ is common difference. |
| Common Difference ($d$) | $$d = a_n - a_{n-1}$$ | Difference between any term and its immediate predecessor. |
| Sum of First $n$ Terms ($S_n$) | $$S_n = \frac{n}{2}\left[2a_1 + (n-1)d\right]$$ | Used when first term $a_1$, common difference $d$, and term number $n$ are known. |
| Alternative Sum Formula ($S_n$) | $$S_n = \frac{n}{2}\left(a_1 + a_n\right)$$ | Used when both the first term $a_1$ and final term $a_n$ are known. |
| Arithmetic Mean (A.M.) | $$A = \frac{a + b}{2}$$ | Single arithmetic mean between two numbers $a$ and $b$. |
Question 1 (Part i): Find the $12^{\text{th}}$ term of the arithmetic sequence $2, 6, 10, 14, \dots$
Solution:
Given arithmetic sequence: $2, 6, 10, 14, \dots$
First term ($a_1$) = $2$
Common difference ($d$) = $6 - 2 = 4$
Number of terms ($n$) = $12$
Using the general term formula for an A.P.:
$$a_n = a_1 + (n - 1)d$$
Substitute the values $a_1 = 2$, $d = 4$, and $n = 12$:
$$a_{12} = 2 + (12 - 1)(4)$$
$$a_{12} = 2 + (11)(4)$$
$$a_{12} = 2 + 44 = 46$$
Answer: $$a_{12} = 46$$
Question 1 (Part ii): Find the $20^{\text{th}}$ term of the arithmetic sequence $17, 13, 9, 5, \dots$
Solution:
Given arithmetic sequence: $17, 13, 9, 5, \dots$
First term ($a_1$) = $17$
Common difference ($d$) = $13 - 17 = -4$
Number of terms ($n$) = $20$
Using the $n$-th term formula:
$$a_n = a_1 + (n - 1)d$$
$$a_{20} = 17 + (20 - 1)(-4)$$
$$a_{20} = 17 + (19)(-4)$$
$$a_{20} = 17 - 76 = -59$$
Answer: $$a_{20} = -59$$
Question 2 (Part i): Find the $n$-th term formula for the arithmetic sequence where $a_1 = 5$ and $d = 4$.
Solution:
First term ($a_1$) = $5$
Common difference ($d$) = $4$
Using the general term formula:
$$a_n = a_1 + (n - 1)d$$
Substitute $a_1 = 5$ and $d = 4$:
$$a_n = 5 + (n - 1)(4)$$
$$a_n = 5 + 4n - 4$$
$$a_n = 4n + 1$$
Answer: $$a_n = 4n + 1$$
Question 2 (Part ii): Find the $n$-th term formula for the sequence $3, \frac{7}{2}, 4, \frac{9}{2}, \dots$
Solution:
Given sequence: $3, \frac{7}{2}, 4, \frac{9}{2}, \dots$
First term ($a_1$) = $3$
Common difference ($d$) = $\frac{7}{2} - 3 = \frac{7 - 6}{2} = \frac{1}{2}$
Using general term formula:
$$a_n = a_1 + (n - 1)d$$
$$a_n = 3 + (n - 1)\left(\frac{1}{2}\right)$$
$$a_n = 3 + \frac{n}{2} - \frac{1}{2}$$
$$a_n = \left(3 - \frac{1}{2}\right) + \frac{n}{2}$$
$$a_n = \frac{5}{2} + \frac{n}{2} = \frac{n + 5}{2}$$
Answer: $$a_n = \frac{n + 5}{2}$$
Question 3: If the $5^{\text{th}}$ term of an A.P. is $16$ and the $11^{\text{th}}$ term is $34$, find the first term $a_1$, common difference $d$, and the $20^{\text{th}}$ term $a_{20}$.
Solution:
We are given:
$$a_5 = 16 \implies a_1 + (5 - 1)d = 16 \implies a_1 + 4d = 16 \quad \text{--- (Equation 1)}$$
$$a_{11} = 34 \implies a_1 + (11 - 1)d = 34 \implies a_1 + 10d = 34 \quad \text{--- (Equation 2)}$$
Subtract Equation 1 from Equation 2:
$$(a_1 + 10d) - (a_1 + 4d) = 34 - 16$$
$$6d = 18 \implies d = 3$$
Substitute $d = 3$ into Equation 1:
$$a_1 + 4(3) = 16$$
$$a_1 + 12 = 16 \implies a_1 = 4$$
Now, calculate $a_{20}$:
$$a_{20} = a_1 + (20 - 1)d = 4 + (19)(3)$$
$$a_{20} = 4 + 57 = 61$$
Answer: $$a_1 = 4, \quad d = 3, \quad a_{20} = 61$$
Question 4: Which term of the arithmetic progression $5, 9, 13, 17, \dots$ is equal to $81$?
Solution:
First term ($a_1$) = $5$
Common difference ($d$) = $9 - 5 = 4$
Let the $n$-th term be $a_n = 81$.
Using formula $a_n = a_1 + (n - 1)d$:
$$81 = 5 + (n - 1)4$$
$$81 - 5 = 4(n - 1)$$
$$76 = 4(n - 1)$$
$$n - 1 = \frac{76}{4} = 19$$
$$n = 19 + 1 = 20$$
Answer: $$n = 20 \quad \text{(The } 20^{\text{th}} \text{ term is } 81\text{)}$$
Question 5 (Part i): Find the sum of the first $15$ terms ($S_{15}$) of the arithmetic series $3 + 7 + 11 + 15 + \dots$
Solution:
Given series: $3 + 7 + 11 + 15 + \dots$
First term ($a_1$) = $3$
Common difference ($d$) = $7 - 3 = 4$
Number of terms ($n$) = $15$
Using the sum formula:
$$S_n = \frac{n}{2}\left[2a_1 + (n - 1)d\right]$$
$$S_{15} = \frac{15}{2}\left[2(3) + (15 - 1)(4)\right]$$
$$S_{15} = \frac{15}{2}\left[6 + (14)(4)\right]$$
$$S_{15} = \frac{15}{2}\left[6 + 56\right]$$
$$S_{15} = \frac{15}{2}\left[62\right] = 15 \times 31 = 465$$
Answer: $$S_{15} = 465$$
Question 5 (Part ii): Find the sum of the first $20$ terms ($S_{20}$) of the arithmetic series $-8 + (-3) + 2 + 7 + \dots$
Solution:
Given series: $-8 + (-3) + 2 + 7 + \dots$
First term ($a_1$) = $-8$
Common difference ($d$) = $-3 - (-8) = 5$
Number of terms ($n$) = $20$
Using the sum formula:
$$S_n = \frac{n}{2}\left[2a_1 + (n - 1)d\right]$$
$$S_{20} = \frac{20}{2}\left[2(-8) + (20 - 1)(5)\right]$$
$$S_{20} = 10\left[-16 + (19)(5)\right]$$
$$S_{20} = 10\left[-16 + 95\right]$$
$$S_{20} = 10\left[79\right] = 790$$
Answer: $$S_{20} = 790$$
Question 6: Insert three arithmetic means (A.Ms) between $4$ and $20$.
Solution:
Let $A_1, A_2, A_3$ be the three arithmetic means between $4$ and $20$.
Then the sequence $4, A_1, A_2, A_3, 20$ forms an Arithmetic Progression.
Here:
First term ($a_1$) = $4$
Total number of terms ($n$) = $3 + 2 = 5$
Fifth term ($a_5$) = $20$
Using general term formula $a_n = a_1 + (n - 1)d$:
$$a_5 = a_1 + (5 - 1)d$$
$$20 = 4 + 4d$$
$$20 - 4 = 4d \implies 16 = 4d \implies d = 4$$
Now compute the three arithmetic means:
$$A_1 = a_1 + d = 4 + 4 = 8$$
$$A_2 = a_1 + 2d = 4 + 2(4) = 12$$
$$A_3 = a_1 + 3d = 4 + 3(4) = 16$$
Answer: The three arithmetic means are $8, 12, 16$.
Q1: What is the common difference $d$ of the A.P. $12, 7, 2, -3, \dots$?
Q2: If the $n$-th term of an A.P. is $a_n = 3n - 1$, what is the first term $a_1$?
Q3: What is the single Arithmetic Mean (A.M.) between $10$ and $30$?
Q4: If $a_1 = 2$ and $a_n = 50$ for $n = 10$, what is the sum $S_{10}$?
Q5: What is the $10^{\text{th}}$ term of the sequence $1, 4, 7, 10, \dots$?