Textbook -Arithmetic Progressions Questions in English

(N/A) The given arithmetic progression $(AP)$ is $\frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \ldots$
The first term $a$ is the first number in the sequence,so $a = \frac{3}{2}$.
The common difference $d$ is calculated by subtracting any term from the term that follows it,i.e.,$d = a_2 - a_1$.
Substituting the values,$d = \frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2} = \frac{-2}{2} = -1$.
Thus,the first term $a = \frac{3}{2}$ and the common difference $d = -1$.

(A) We have $a_{2}-a_{1} = 10-4 = 6$.
$a_{3}-a_{2} = 16-10 = 6$.
$a_{4}-a_{3} = 22-16 = 6$.
Since the difference $a_{k+1}-a_{k}$ is constant for all $k$,the given list of numbers forms an $AP$ with the common difference $d = 6$.
The next two terms are:
$22 + 6 = 28$
$28 + 6 = 34$.

(A) To check if the sequence forms an $AP$,we calculate the common difference $d = a_{k+1} - a_k$ for consecutive terms:
$a_2 - a_1 = -1 - 1 = -2$
$a_3 - a_2 = -3 - (-1) = -3 + 1 = -2$
$a_4 - a_3 = -5 - (-3) = -5 + 3 = -2$
Since the difference $d = -2$ is constant,the given sequence forms an $AP$.
The next two terms are calculated by adding the common difference $d = -2$ to the last term:
Next term $1 = -5 + (-2) = -7$
Next term $2 = -7 + (-2) = -9$
Thus,the next two terms are $-7$ and $-9$.

(N/A) To check if the sequence forms an $AP$,we calculate the common difference $d$ between consecutive terms.
First,calculate the difference between the second and first term:
$a_{2} - a_{1} = 2 - (-2) = 2 + 2 = 4$
Next,calculate the difference between the third and second term:
$a_{3} - a_{2} = -2 - 2 = -4$
Since $a_{2} - a_{1} \neq a_{3} - a_{2}$ (i.e.,$4 \neq -4$),the common difference is not constant.
Therefore,the given list of numbers does not form an $AP$.

(NO) An arithmetic progression $(AP)$ is a sequence of numbers such that the difference between any two consecutive terms is constant.
Let the sequence be $a_1, a_2, a_3, a_4, \ldots = 1, 1, 1, 2, 2, 2, 3, 3, 3, \ldots$
Calculate the common difference $d$ between consecutive terms:
$d_1 = a_2 - a_1 = 1 - 1 = 0$
$d_2 = a_3 - a_2 = 1 - 1 = 0$
$d_3 = a_4 - a_3 = 2 - 1 = 1$
Since $d_1 = d_2 \neq d_3$,the difference between consecutive terms is not constant.
Therefore,the given sequence does not form an $AP$.

(A) It can be observed that:
Taxi fare for $1^{\text{st}} \ km = 15$
Taxi fare for first $2 \ km = 15 + 8 = 23$
Taxi fare for first $3 \ km = 23 + 8 = 31$
Taxi fare for first $4 \ km = 31 + 8 = 39$
Clearly,the sequence $15, 23, 31, 39, \dots$ forms an $A.P.$ because the difference between any two consecutive terms is constant,i.e.,$d = 8$.

(B) Let the initial volume of air in a cylinder be $V$ liters.
In each stroke,the vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder.
This means that after every stroke,the remaining air is $1 - \frac{1}{4} = \frac{3}{4}$ of the previous amount.
Therefore,the sequence of volumes after each stroke is $V, \frac{3}{4}V, \left(\frac{3}{4}\right)^2 V, \left(\frac{3}{4}\right)^3 V, \dots$.
To check if this is an arithmetic progression $(A.P.)$,we calculate the difference between consecutive terms:
$d_1 = \frac{3}{4}V - V = -\frac{1}{4}V$
$d_2 = \left(\frac{3}{4}\right)^2 V - \frac{3}{4}V = \frac{9}{16}V - \frac{12}{16}V = -\frac{3}{16}V$.
Since $d_1 \neq d_2$,the difference between consecutive terms is not constant.
Thus,this list of numbers does not form an $A.P.$

(A) Cost of digging for the first metre $= ₹ 150$.
Cost of digging for the first $2$ metres $= 150 + 50 = ₹ 200$.
Cost of digging for the first $3$ metres $= 200 + 50 = ₹ 250$.
Cost of digging for the first $4$ metres $= 250 + 50 = ₹ 300$.
Clearly, the sequence $150, 200, 250, 300, \dots$ forms an Arithmetic Progression $(A.P.)$ because the difference between any two consecutive terms is constant, i.e., $d = 50$.

(B) We know that if a principal amount $P$ is deposited at $r \%$ compound interest per annum for $n$ years,the amount after $n$ years is given by $A = P(1 + \frac{r}{100})^n$.
For the given problem,$P = 10000$ and $r = 8$.
The amount in the account after each year is:
Year $1$: $10000(1 + \frac{8}{100})^1 = 10800$
Year $2$: $10000(1 + \frac{8}{100})^2 = 11664$
Year $3$: $10000(1 + \frac{8}{100})^3 = 12597.12$
The sequence of amounts is $10800, 11664, 12597.12, \dots$
To check if this is an Arithmetic Progression $(A.P.)$,we calculate the common difference:
$d_1 = 11664 - 10800 = 864$
$d_2 = 12597.12 - 11664 = 933.12$
Since $d_1 \neq d_2$,the difference between consecutive terms is not constant. Therefore,this list of numbers does not form an $A.P.$

(A) Given: First term $a = 10$ and common difference $d = 10.$
The general form of an $AP$ is $a, a+d, a+2d, a+3d, \dots$
First term $a_1 = a = 10.$
Second term $a_2 = a + d = 10 + 10 = 20.$
Third term $a_3 = a + 2d = 10 + 2(10) = 10 + 20 = 30.$
Fourth term $a_4 = a + 3d = 10 + 3(10) = 10 + 30 = 40.$
Thus,the first four terms of the $AP$ are $10, 20, 30,$ and $40.$

(A) Given: First term $a = -2$ and common difference $d = 0.$
The general form of an $AP$ is $a, a+d, a+2d, a+3d, \dots$
First term $a_1 = a = -2.$
Second term $a_2 = a + d = -2 + 0 = -2.$
Third term $a_3 = a + 2d = -2 + 2(0) = -2.$
Fourth term $a_4 = a + 3d = -2 + 3(0) = -2.$
Therefore,the first four terms of the $AP$ are $-2, -2, -2, -2.$

(A) Given: First term $a = 4$ and common difference $d = -3.$
The general form of an $AP$ is $a, a+d, a+2d, a+3d, \dots$
First term $a_1 = a = 4.$
Second term $a_2 = a + d = 4 + (-3) = 4 - 3 = 1.$
Third term $a_3 = a + 2d = 4 + 2(-3) = 4 - 6 = -2.$
Fourth term $a_4 = a + 3d = 4 + 3(-3) = 4 - 9 = -5.$
Therefore,the first four terms of the $AP$ are $4, 1, -2, -5.$

(A) Given: $a = -1$ and $d = \frac{1}{2}$.
The general form of an $AP$ is $a_1, a_2, a_3, a_4, \dots$ where $a_n = a + (n-1)d$.
$a_1 = a = -1$
$a_2 = a + d = -1 + \frac{1}{2} = -\frac{1}{2}$
$a_3 = a + 2d = -1 + 2(\frac{1}{2}) = -1 + 1 = 0$
$a_4 = a + 3d = -1 + 3(\frac{1}{2}) = -1 + \frac{3}{2} = \frac{1}{2}$
Thus,the first four terms of the $AP$ are $-1, -\frac{1}{2}, 0, \frac{1}{2}$.

(A) Given: First term $a = -1.25$ and common difference $d = -0.25.$
The terms of an $AP$ are given by $a_n = a + (n-1)d.$
First term: $a_1 = a = -1.25$
Second term: $a_2 = a + d = -1.25 + (-0.25) = -1.50$
Third term: $a_3 = a + 2d = -1.25 + 2(-0.25) = -1.25 - 0.50 = -1.75$
Fourth term: $a_4 = a + 3d = -1.25 + 3(-0.25) = -1.25 - 0.75 = -2.00$
Thus,the first four terms of the $AP$ are $-1.25, -1.50, -1.75,$ and $-2.00.$

(C) The given arithmetic progression is $3, 1, -1, -3, \ldots$
The first term $(a)$ is the first number in the sequence,so $a = 3$.
The common difference $(d)$ is calculated by subtracting the first term from the second term:
$d = a_2 - a_1$
$d = 1 - 3$
$d = -2$
Thus,the first term is $3$ and the common difference is $-2$.

(A) The given arithmetic progression is $-5, -1, 3, 7, \ldots$.
The first term $(a)$ is the first number in the sequence,so $a = -5$.
The common difference $(d)$ is calculated by subtracting the first term from the second term:
$d = a_2 - a_1$
$d = (-1) - (-5)$
$d = -1 + 5$
$d = 4$.
Thus,the first term is $-5$ and the common difference is $4$.

(N/A) The given arithmetic progression is $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots$
The first term $(a)$ is the first number in the sequence,so $a = \frac{1}{3}$.
The common difference $(d)$ is calculated by subtracting the first term from the second term:
$d = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}$.
Thus,the first term is $\frac{1}{3}$ and the common difference is $\frac{4}{3}$.

(A) The given arithmetic progression is $0.6, 1.7, 2.8, 3.9, \ldots$
The first term $(a)$ is the first number in the sequence,so $a = 0.6$.
The common difference $(d)$ is calculated by subtracting the first term from the second term:
$d = a_2 - a_1$
$d = 1.7 - 0.6$
$d = 1.1$
Thus,the first term is $0.6$ and the common difference is $1.1$.

(N/A) Given sequence: $2, 4, 8, 16, \ldots$
To check if the sequence is an $AP$,we calculate the difference between consecutive terms:
$a_2 - a_1 = 4 - 2 = 2$
$a_3 - a_2 = 8 - 4 = 4$
$a_4 - a_3 = 16 - 8 = 8$
Since the difference between consecutive terms $(a_{k+1} - a_k)$ is not constant,the given sequence does not form an $AP$.

(A) Given sequence: $2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$
To check if it is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = \frac{5}{2} - 2 = \frac{1}{2}$
$a_{3} - a_{2} = 3 - \frac{5}{2} = \frac{1}{2}$
$a_{4} - a_{3} = \frac{7}{2} - 3 = \frac{1}{2}$
Since the difference $a_{k+1} - a_{k}$ is constant,the sequence is an $AP$ with common difference $d = \frac{1}{2}$.
The next three terms are:
$a_{5} = \frac{7}{2} + \frac{1}{2} = 4$
$a_{6} = 4 + \frac{1}{2} = \frac{9}{2}$
$a_{7} = \frac{9}{2} + \frac{1}{2} = 5$

(D) Given sequence: $-10, -6, -2, 2, \ldots$
To check if it is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = (-6) - (-10) = -6 + 10 = 4$
$a_{3} - a_{2} = (-2) - (-6) = -2 + 6 = 4$
$a_{4} - a_{3} = 2 - (-2) = 2 + 2 = 4$
Since the difference $a_{k+1} - a_{k}$ is constant $(d = 4)$,the sequence forms an $AP$.
The next three terms are:
$a_{5} = a_{4} + d = 2 + 4 = 6$
$a_{6} = a_{5} + d = 6 + 4 = 10$
$a_{7} = a_{6} + d = 10 + 4 = 14$
Thus,the common difference is $d = 4$ and the next three terms are $6, 10, 14$.

(A) The given sequence is $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots$
To check if the sequence is in $AP$,we calculate the difference between consecutive terms:
$a_2 - a_1 = (3 + \sqrt{2}) - 3 = \sqrt{2}$
$a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2}) = \sqrt{2}$
$a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = \sqrt{2}$
Since the difference $a_{k+1} - a_k$ is constant,the sequence is an $AP$ with common difference $d = \sqrt{2}$.
The next three terms are:
$a_5 = a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$
$a_6 = a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$
$a_7 = a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$

(B) Given sequence: $0.2, 0.22, 0.222, 0.2222, \ldots$
To check if the sequence is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = 0.22 - 0.2 = 0.02$
$a_{3} - a_{2} = 0.222 - 0.22 = 0.002$
$a_{4} - a_{3} = 0.2222 - 0.222 = 0.0002$
Since the difference $a_{k+1} - a_{k}$ is not constant,the given sequence does not form an $AP$.

(A) Given sequence: $0, -4, -8, -12, \ldots$
To check if the sequence is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = (-4) - 0 = -4$
$a_{3} - a_{2} = (-8) - (-4) = -4$
$a_{4} - a_{3} = (-12) - (-8) = -4$
Since the difference $a_{k+1} - a_{k}$ is constant,the sequence is an $AP$ with common difference $d = -4$.
The next three terms are:
$a_{5} = a_{4} + d = -12 + (-4) = -16$
$a_{6} = a_{5} + d = -16 + (-4) = -20$
$a_{7} = a_{6} + d = -20 + (-4) = -24$
Thus,the common difference is $-4$ and the next three terms are $-16, -20, -24$.

(A) Given sequence: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots$
To check if it is an $AP$,we calculate the difference between consecutive terms:
$a_2 - a_1 = (-\frac{1}{2}) - (-\frac{1}{2}) = 0$
$a_3 - a_2 = (-\frac{1}{2}) - (-\frac{1}{2}) = 0$
$a_4 - a_3 = (-\frac{1}{2}) - (-\frac{1}{2}) = 0$
Since the difference $a_{k+1} - a_k$ is constant $(0)$ for all $k$,the sequence is an $AP$ with common difference $d = 0$.
The next three terms are:
$a_5 = a_4 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$
$a_6 = a_5 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$
$a_7 = a_6 + d = -\frac{1}{2} + 0 = -\frac{1}{2}$

(NONE) Given sequence: $1, 3, 9, 27, \ldots$
To check if the sequence is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = 3 - 1 = 2$
$a_{3} - a_{2} = 9 - 3 = 6$
$a_{4} - a_{3} = 27 - 9 = 18$
Since the common difference $(a_{k+1} - a_{k})$ is not constant (i.e.,$2 \neq 6 \neq 18$),the given sequence does not form an $AP$.

(A) The given sequence is $a, 2a, 3a, 4a, \dots$
To check if it is an $AP,$ we calculate the difference between consecutive terms:
$a_2 - a_1 = 2a - a = a$
$a_3 - a_2 = 3a - 2a = a$
$a_4 - a_3 = 4a - 3a = a$
Since the difference $a_{k+1} - a_k$ is constant $(d = a)$,the sequence forms an $AP.$
The next three terms are:
$a_5 = 4a + a = 5a$
$a_6 = 5a + a = 6a$
$a_7 = 6a + a = 7a$

(NONE) The given sequence is $a, a^{2}, a^{3}, a^{4}, \ldots$
To check if the sequence is in an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = a^{2} - a = a(a - 1)$
$a_{3} - a_{2} = a^{3} - a^{2} = a^{2}(a - 1)$
$a_{4} - a_{3} = a^{4} - a^{3} = a^{3}(a - 1)$
Since the difference $a_{k+1} - a_{k}$ is not constant (i.e.,$a(a-1) \neq a^{2}(a-1)$ for $a \neq 0, 1$),the given sequence does not form an $AP$.

(A) The given sequence is $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$
We can simplify the terms as:
$a_1 = \sqrt{2} = 1\sqrt{2}$
$a_2 = \sqrt{8} = 2\sqrt{2}$
$a_3 = \sqrt{18} = 3\sqrt{2}$
$a_4 = \sqrt{32} = 4\sqrt{2}$
To check if it is an $AP$,we find the common difference $d = a_{k+1} - a_k$:
$a_2 - a_1 = 2\sqrt{2} - 1\sqrt{2} = \sqrt{2}$
$a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$
$a_4 - a_3 = 4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$
Since the difference $d = \sqrt{2}$ is constant,the sequence is an $AP$.
The next three terms are:
$a_5 = a_4 + d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50}$
$a_6 = a_5 + d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72}$
$a_7 = a_6 + d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98}$

(N/A) The given sequence is $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$
To check if the sequence is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2} - 1)$
$a_{3} - a_{2} = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6} = \sqrt{3}(\sqrt{3} - \sqrt{2})$
$a_{4} - a_{3} = \sqrt{12} - \sqrt{9} = 2\sqrt{3} - 3 = \sqrt{3}(2 - \sqrt{3})$
Since the difference $a_{k+1} - a_{k}$ is not constant for all $k$,the given sequence does not form an $AP$.

(N/A) The given sequence is $1^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots$
This can be written as $1, 9, 25, 49, \ldots$
To check if the sequence forms an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = 9 - 1 = 8$
$a_{3} - a_{2} = 25 - 9 = 16$
$a_{4} - a_{3} = 49 - 25 = 24$
Since the difference $a_{k+1} - a_{k}$ is not constant (i.e.,$8 \neq 16 \neq 24$),the given sequence does not form an $AP$.

(A) The given sequence is $1^{2}, 5^{2}, 7^{2}, 73, \ldots$
Calculating the values,we get $1, 25, 49, 73, \ldots$
To check if it is an $AP$,we find the difference between consecutive terms:
$a_{2} - a_{1} = 25 - 1 = 24$
$a_{3} - a_{2} = 49 - 25 = 24$
$a_{4} - a_{3} = 73 - 49 = 24$
Since the difference $a_{k+1} - a_{k}$ is constant $(d = 24)$,the given sequence forms an $AP$.
The common difference is $d = 24$.
The next three terms are:
$a_{5} = 73 + 24 = 97$
$a_{6} = 97 + 24 = 121$
$a_{7} = 121 + 24 = 145$

(A) Given the arithmetic progression $(AP)$ is $2, 7, 12, \ldots$
Here,the first term $a = 2$.
The common difference $d = 7 - 2 = 5$.
We need to find the $10^{th}$ term,so $n = 10$.
The formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n - 1)d$.
Substituting the values,we get $a_{10} = 2 + (10 - 1) \times 5$.
$a_{10} = 2 + 9 \times 5$.
$a_{10} = 2 + 45 = 47$.
Therefore,the $10^{th}$ term of the given $AP$ is $47$.

(N/A) For the given $AP$,the first term $a = 21$ and the common difference $d = 18 - 21 = -3$.
We use the formula for the $n$-th term: $a_n = a + (n - 1)d$.
To find which term is $-81$,we set $a_n = -81$:
$-81 = 21 + (n - 1)(-3)$
$-81 = 21 - 3n + 3$
$-81 = 24 - 3n$
$3n = 24 + 81$
$3n = 105$
$n = 35$.
Thus,the $35$th term of the $AP$ is $-81$.
To check if any term is $0$,we set $a_n = 0$:
$0 = 21 + (n - 1)(-3)$
$0 = 21 - 3n + 3$
$3n = 24$
$n = 8$.
Since $n = 8$ is a positive integer,the $8$th term of the $AP$ is $0$.

(A) The general term of an $AP$ is given by $a_n = a + (n - 1)d$.
Given that the $3^{rd}$ term is $5$,we have:
$a_3 = a + 2d = 5$ $...(1)$
Given that the $7^{th}$ term is $9$,we have:
$a_7 = a + 6d = 9$ $...(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(a + 6d) - (a + 2d) = 9 - 5$
$4d = 4$
$d = 1$
Substituting $d = 1$ in equation $(1)$:
$a + 2(1) = 5$
$a = 3$
The $AP$ is given by $a, a+d, a+2d, a+3d, \dots$
Substituting the values,we get $3, 3+1, 3+2, 3+3, \dots$,which is $3, 4, 5, 6, \dots$

(B) We have the sequence $5, 11, 17, 23, \ldots$
First,we check if it is an Arithmetic Progression $(AP)$:
$a_{2}-a_{1} = 11-5 = 6$
$a_{3}-a_{2} = 17-11 = 6$
$a_{4}-a_{3} = 23-17 = 6$
Since the common difference $d = 6$ is constant,the sequence is an $AP$ with first term $a = 5$.
Let $301$ be the $n$th term of this $AP$. The formula for the $n$th term is:
$a_{n} = a + (n-1)d$
Substituting the values:
$301 = 5 + (n-1)6$
$301 = 5 + 6n - 6$
$301 = 6n - 1$
$302 = 6n$
$n = \frac{302}{6} = \frac{151}{3} = 50.33$
Since $n$ must be a positive integer for any term in an $AP$,and $50.33$ is not an integer,$301$ is not a term of the given list of numbers.

(A) The list of two-digit numbers divisible by $3$ is $12, 15, 18, \ldots, 99$.
This sequence forms an Arithmetic Progression $(AP)$ where the first term $a = 12$,the common difference $d = 3$,and the last term $a_n = 99$.
The formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n - 1)d$.
Substituting the values: $99 = 12 + (n - 1) \times 3$.
Subtracting $12$ from both sides: $87 = (n - 1) \times 3$.
Dividing by $3$: $n - 1 = 29$.
Therefore,$n = 29 + 1 = 30$.
Thus,there are $30$ two-digit numbers divisible by $3$.

(B) The given $AP$ is $10, 7, 4, \ldots, -62$.
Here,the first term $a = 10$,common difference $d = 7 - 10 = -3$,and the last term $l = -62$.
To find the $n^{th}$ term from the end,we can use the formula: $n^{th}$ term from the end $= l - (n - 1)d$.
Here,$n = 11$,$l = -62$,and $d = -3$.
Substituting these values into the formula:
$11^{th}$ term from the end $= -62 - (11 - 1)(-3)$
$= -62 - (10)(-3)$
$= -62 + 30$
$= -32$.
Thus,the $11^{th}$ term from the last term is $-32$.

(C) The formula for simple interest is $\text{Simple Interest} = \frac{P \times R \times T}{100}$.
Interest at the end of the $1^{st}$ year $= \frac{1000 \times 8 \times 1}{100} = ₹ 80$.
Interest at the end of the $2^{nd}$ year $= \frac{1000 \times 8 \times 2}{100} = ₹ 160$.
Interest at the end of the $3^{rd}$ year $= \frac{1000 \times 8 \times 3}{100} = ₹ 240$.
The sequence of interests is $80, 160, 240, \dots$.
Since the difference between consecutive terms is constant $(d = 80)$,this sequence forms an $AP$ with first term $a = 80$ and common difference $d = 80$.
To find the interest at the end of $30 \text{ years}$,we calculate the $30^{th}$ term $(a_{30})$:
$a_{30} = a + (30 - 1)d = 80 + 29 \times 80 = 80 + 2320 = 2400$.
Thus,the interest at the end of $30 \text{ years}$ is ₹ $2400$.

(D) The number of rose plants in the $1^{st}, 2^{nd}, 3^{rd}, \dots$ rows are $23, 21, 19, \dots, 5$.
This sequence forms an Arithmetic Progression $(AP)$ because the difference between consecutive terms is constant.
Here,the first term $a = 23$ and the common difference $d = 21 - 23 = -2$.
The last term (or $n^{th}$ term) is $a_n = 5$.
Using the formula for the $n^{th}$ term of an $AP$: $a_n = a + (n - 1)d$.
Substituting the values: $5 = 23 + (n - 1)(-2)$.
Subtracting $23$ from both sides: $5 - 23 = (n - 1)(-2) \implies -18 = (n - 1)(-2)$.
Dividing by $-2$: $9 = n - 1$.
Therefore,$n = 10$.
There are $10$ rows in the flower bed.

(N/A) The general term of an $AP$ is given by the formula: $a_{n} = a + (n - 1)d$.
$(i)$ Given $a = 7, d = 3, n = 8$.
$a_{8} = 7 + (8 - 1)3 = 7 + (7)(3) = 7 + 21 = 28$.
$(ii)$ Given $a = -18, n = 10, a_{n} = 0$.
$0 = -18 + (10 - 1)d \rightarrow 18 = 9d \rightarrow d = 2$.
$(iii)$ Given $d = -3, n = 18, a_{n} = -5$.
$-5 = a + (18 - 1)(-3) \rightarrow -5 = a + (17)(-3) \rightarrow -5 = a - 51 \rightarrow a = 46$.
$(iv)$ Given $a = -18.9, d = 2.5, a_{n} = 3.6$.
$3.6 = -18.9 + (n - 1)2.5 \rightarrow 3.6 + 18.9 = (n - 1)2.5 \rightarrow 22.5 = (n - 1)2.5 \rightarrow n - 1 = 9 \rightarrow n = 10$.
$(v)$ Given $a = 3.5, d = 0, n = 105$.
$a_{105} = 3.5 + (105 - 1)0 = 3.5 + 0 = 3.5$.

(B) Given that the $A.P.$ is $10, 7, 4, \ldots$
The first term,$a = 10$.
The common difference,$d = a_2 - a_1 = 7 - 10 = -3$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
To find the $30^{th}$ term $(n = 30)$:
$a_{30} = 10 + (30 - 1)(-3)$
$a_{30} = 10 + (29)(-3)$
$a_{30} = 10 - 87$
$a_{30} = -77$
Hence,the correct answer is $B$.

(C) Given the $AP: -3, -\frac{1}{2}, 2, \ldots$
The first term $a = -3$.
The common difference $d = a_2 - a_1 = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = \frac{5}{2}$.
The formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n - 1)d$.
To find the $11^{th}$ term $(n = 11)$:
$a_{11} = -3 + (11 - 1) \left( \frac{5}{2} \right)$
$a_{11} = -3 + 10 \left( \frac{5}{2} \right)$
$a_{11} = -3 + 25$
$a_{11} = 22$.
Therefore,the correct choice is $C$.

(D) Given the $A.P.$ as $2, \square, 26$.
Here,the first term $a = 2$ and the third term $a_3 = 26$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
For the third term $(n = 3)$:
$a_3 = a + (3 - 1)d$
$26 = 2 + 2d$
$24 = 2d$
$d = 12$.
Now,to find the missing second term $(a_2)$:
$a_2 = a + (2 - 1)d$
$a_2 = 2 + 12 = 14$.
Therefore,the missing term is $14$.

(18, 8) The given sequence is $\square, 13, \square, 3$.
For this $AP$,the second term is $a_2 = 13$ and the fourth term is $a_4 = 3$.
We know that the $n^{th}$ term of an $AP$ is given by $a_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.
For $a_2 = 13$:
$a + (2 - 1)d = 13 \implies a + d = 13$ $...(i)$
For $a_4 = 3$:
$a + (4 - 1)d = 3 \implies a + 3d = 3$ $...(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 3d) - (a + d) = 3 - 13$
$2d = -10$
$d = -5$
Substituting $d = -5$ into equation $(i)$:
$a + (-5) = 13$
$a = 13 + 5 = 18$
Now,the first term is $a = 18$ and the third term is $a_3 = a + 2d$:
$a_3 = 18 + 2(-5) = 18 - 10 = 8$.
Therefore,the missing terms are $18$ and $8$ respectively.

(N/A) Given the $A.P.$ as $5, \square, \square, 9 \frac{1}{2}$.
Here,the first term $a = 5$.
The fourth term $a_4 = 9 \frac{1}{2} = \frac{19}{2}$.
We know the formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
For $n = 4$,$a_4 = a + 3d$.
Substituting the values: $\frac{19}{2} = 5 + 3d$.
Subtracting $5$ from both sides: $\frac{19}{2} - 5 = 3d \implies \frac{19 - 10}{2} = 3d \implies \frac{9}{2} = 3d$.
Dividing by $3$,we get the common difference $d = \frac{3}{2}$.
Now,find the missing terms:
The second term $a_2 = a + d = 5 + \frac{3}{2} = \frac{10 + 3}{2} = \frac{13}{2}$.
The third term $a_3 = a + 2d = 5 + 2(\frac{3}{2}) = 5 + 3 = 8$.
Thus,the missing terms are $\frac{13}{2}$ and $8$.

(N/A) For this $A.P.,$ the first term $a = -4$ and the sixth term $a_{6} = 6$.
We know the formula for the $n^{th}$ term of an $A.P.$ is $a_{n} = a + (n - 1)d$.
Substituting the values for the sixth term:
$a_{6} = a + (6 - 1)d$
$6 = -4 + 5d$
$10 = 5d$
$d = 2$
Now,we find the missing terms:
$a_{2} = a + d = -4 + 2 = -2$
$a_{3} = a + 2d = -4 + 2(2) = 0$
$a_{4} = a + 3d = -4 + 3(2) = 2$
$a_{5} = a + 4d = -4 + 4(2) = 4$
Therefore,the missing terms are $-2, 0, 2,$ and $4$ respectively.

(A) Let the $A.P.$ be $a_1, a_2, a_3, a_4, a_5, a_6.$
Given: $a_2 = 38$ and $a_6 = -22.$
We know the general term formula: $a_n = a + (n-1)d.$
For $n=2$: $a + d = 38$ $...(1)$
For $n=6$: $a + 5d = -22$ $...(2)$
Subtracting equation $(1)$ from $(2)$:
$(a + 5d) - (a + d) = -22 - 38$
$4d = -60$
$d = -15$
Substituting $d = -15$ in equation $(1)$:
$a + (-15) = 38$
$a = 38 + 15 = 53$
Now,finding the missing terms:
$a_1 = a = 53$
$a_3 = a + 2d = 53 + 2(-15) = 53 - 30 = 23$
$a_4 = a + 3d = 53 + 3(-15) = 53 - 45 = 8$
$a_5 = a + 4d = 53 + 4(-15) = 53 - 60 = -7$
Thus,the missing terms are $53, 23, 8,$ and $-7$.

(B) The given $AP$ is $3, 8, 13, 18, \ldots$
For this $AP$,the first term $a = 3$.
The common difference $d = a_2 - a_1 = 8 - 3 = 5$.
Let the $n^{th}$ term of this $AP$ be $78$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values,we get $78 = 3 + (n - 1)5$.
Subtracting $3$ from both sides,$75 = (n - 1)5$.
Dividing by $5$,we get $n - 1 = 15$.
Therefore,$n = 16$.
Hence,the $16^{th}$ term of this $AP$ is $78$.

(B) For the given $AP$: $7, 13, 19, \ldots, 205$.
Here,the first term $a = 7$.
The common difference $d = a_2 - a_1 = 13 - 7 = 6$.
Let the number of terms in the $AP$ be $n$.
The last term $a_n = 205$.
Using the formula for the $n^{th}$ term of an $AP$: $a_n = a + (n - 1)d$.
Substituting the values: $205 = 7 + (n - 1)6$.
Subtracting $7$ from both sides: $198 = (n - 1)6$.
Dividing by $6$: $33 = n - 1$.
Therefore,$n = 34$.
Thus,the number of terms in the given $AP$ is $34$.