Math Notes For Class 11 Chapter 4 Ex 4.1 Solved & Quiz

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 $8, 12, 16$.