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 $a = -2$ 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) The given sequence is $-10, -6, -2, 2, \ldots$
Here,the first term $a_1 = -10$,the second term $a_2 = -6$,the third term $a_3 = -2$,and the fourth term $a_4 = 2$.
To check if the sequence is an $AP$,we calculate the common difference $d$ between consecutive terms:
$d_1 = a_2 - a_1 = -6 - (-10) = -6 + 10 = 4$
$d_2 = a_3 - a_2 = -2 - (-6) = -2 + 6 = 4$
$d_3 = a_4 - a_3 = 2 - (-2) = 2 + 2 = 4$
Since the difference between consecutive terms is constant $(d = 4)$,the given list of numbers forms an $AP$ with a common difference $d = 4$.

(C) The general form of an $AP$ is given by $a, a + d, a + 2d, a + 3d, \dots$
Given that the first term $a = -2$ and the common difference $d = -2$.
Substituting these values into the general form:
First term: $a = -2$
Second term: $a + d = -2 + (-2) = -4$
Third term: $a + 2d = -2 + 2(-2) = -2 - 4 = -6$
Fourth term: $a + 3d = -2 + 3(-2) = -2 - 6 = -8$
Thus,the first four terms are $-2, -4, -6, -8$.

(B) The common difference $d$ of an $AP$ is calculated as $d = a_{n} - a_{n-1}$.
For the given $AP: 10, 5, 0, -5, \ldots$:
$a_{2} - a_{1} = 5 - 10 = -5$
$a_{3} - a_{2} = 0 - 5 = -5$
$a_{4} - a_{3} = -5 - 0 = -5$
Since the difference between consecutive terms is constant,the given list of numbers forms an $AP$ with a common difference $d = -5$.
Therefore,the statement that the common difference $d$ is equal to $5$ is false.

(B) The formula for the amount $A$ under compound interest is $A = P(1 + r/100)^n$,where $P = 1000$,$r = 10$,and $n$ is the number of years.
Amount at the end of the $1^{st}$ year: $A_1 = 1000(1 + 10/100)^1 = 1000(1.1) = 1100$.
Amount at the end of the $2^{nd}$ year: $A_2 = 1000(1 + 10/100)^2 = 1000(1.21) = 1210$.
Amount at the end of the $3^{rd}$ year: $A_3 = 1000(1 + 10/100)^3 = 1000(1.331) = 1331$.
The sequence of amounts is $1100, 1210, 1331, \ldots$.
To check if it is an $AP$,we calculate the common difference:
$A_2 - A_1 = 1210 - 1100 = 110$.
$A_3 - A_2 = 1331 - 1210 = 121$.
Since $A_2 - A_1 \neq A_3 - A_2$,the difference between consecutive terms is not constant.
Therefore,the sequence does not form an $AP$.

(A) $(i)$ Here,$t_{1} = -1, t_{2} = -1, t_{3} = -1$ and $t_{4} = -1$.
$t_{2} - t_{1} = -1 - (-1) = 0$
$t_{3} - t_{2} = -1 - (-1) = 0$
$t_{4} - t_{3} = -1 - (-1) = 0$
Since the common difference $d = 0$ is constant,the given list of numbers forms an $AP$.
$(ii)$ Here,$t_{1} = 0, t_{2} = 2, t_{3} = 0$ and $t_{4} = 2$.
$t_{2} - t_{1} = 2 - 0 = 2$
$t_{3} - t_{2} = 0 - 2 = -2$
Since $t_{2} - t_{1} \neq t_{3} - t_{2}$,the given list of numbers does not form an $AP$.
$(iii)$ Here,$t_{1} = 1, t_{2} = 1, t_{3} = 2$ and $t_{4} = 2$.
$t_{2} - t_{1} = 1 - 1 = 0$
$t_{3} - t_{2} = 2 - 1 = 1$
Since $t_{2} - t_{1} \neq t_{3} - t_{2}$,the given list of numbers does not form an $AP$.

(I, IV) $(i)$ Here,$t_1 = 11, t_2 = 22, t_3 = 33$.
$t_2 - t_1 = 22 - 11 = 11$.
$t_3 - t_2 = 33 - 22 = 11$.
Since the common difference $d = 11$ is constant,this forms an $AP$.
$(ii)$ Here,$t_1 = \frac{1}{2}, t_2 = \frac{1}{3}, t_3 = \frac{1}{4}$.
$t_2 - t_1 = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$.
$t_3 - t_2 = \frac{1}{4} - \frac{1}{3} = -\frac{1}{12}$.
Since the differences are not equal,this does not form an $AP$.
$(iii)$ The sequence is $2, 4, 8, 16, \ldots$.
$t_2 - t_1 = 4 - 2 = 2$.
$t_3 - t_2 = 8 - 4 = 4$.
Since the differences are not equal,this does not form an $AP$.
$(iv)$ The sequence is $\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}, \ldots$.
$t_2 - t_1 = 2\sqrt{3} - \sqrt{3} = \sqrt{3}$.
$t_3 - t_2 = 3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$.
$t_4 - t_3 = 4\sqrt{3} - 3\sqrt{3} = \sqrt{3}$.
Since the common difference $d = \sqrt{3}$ is constant,this forms an $AP$.

(B) False.
Here,$a_{1} = -1, a_{2} = -\frac{3}{2}, a_{3} = -2$ and $a_{4} = \frac{5}{2}$.
Calculate the common differences:
$a_{2} - a_{1} = -\frac{3}{2} - (-1) = -\frac{3}{2} + 1 = -\frac{1}{2}$.
$a_{3} - a_{2} = -2 - (-\frac{3}{2}) = -2 + \frac{3}{2} = -\frac{1}{2}$.
$a_{4} - a_{3} = \frac{5}{2} - (-2) = \frac{5}{2} + 2 = \frac{9}{2}$.
For a sequence to be an $AP$,the difference between any two consecutive terms must be constant.
Although $a_{2} - a_{1} = a_{3} - a_{2} = -\frac{1}{2}$,the difference $a_{4} - a_{3} = \frac{9}{2}$.
Since the common difference is not constant throughout the sequence,it does not form an $AP$.

(N/A) Let the common difference of both $APs$ be $d$.
Given that the first term of the first $AP$ is $a_1 = 2$ and the first term of the second $AP$ is $b_1 = 7$.
The $n^{\text{th}}$ term of an $AP$ is given by $a_n = a + (n-1)d$.
For the first $AP$,the $n^{\text{th}}$ term is $a_n = 2 + (n-1)d$.
For the second $AP$,the $n^{\text{th}}$ term is $b_n = 7 + (n-1)d$.
The difference between the $n^{\text{th}}$ terms is $b_n - a_n = [7 + (n-1)d] - [2 + (n-1)d] = 7 - 2 = 5$.
Since the difference is independent of $n$,the difference between any two corresponding terms is always $5$.
Thus,the difference between the $10^{\text{th}}$ terms is $5$,and the difference between the $21^{\text{st}}$ terms is also $5$.

(B) No,the statement is false. The total fare (in $Rs.$) after each $km$ is calculated as follows:
For $1$ $km$: $15$
For $2$ $km$: $15 + 8 = 23$
For $3$ $km$: $15 + 2 \times 8 = 31$
For $4$ $km$: $15 + 3 \times 8 = 39$
Thus,the sequence of total fares is $15, 23, 31, 39, \ldots$.
Let $t_1 = 15, t_2 = 23, t_3 = 31, t_4 = 39$.
Now,calculating the common difference:
$t_2 - t_1 = 23 - 15 = 8$
$t_3 - t_2 = 31 - 23 = 8$
$t_4 - t_3 = 39 - 31 = 8$
Since the difference between consecutive terms is constant (common difference $d = 8$),the sequence forms an Arithmetic Progression $(AP)$.

(A) $(i)$ The fee charged from a student every month is $400, 400, 400, 400, \ldots$. This forms an $AP$ because the common difference $(d) = 400 - 400 = 0$ is constant.
$(ii)$ The fee charged for Classes $I$ to $XII$ is $250, (250+50), (250+2 \times 50), (250+3 \times 50), \ldots$,i.e.,$250, 300, 350, 400, \ldots$. This forms an $AP$ because the common difference $(d) = 300 - 250 = 50$ is constant.
$(iii)$ Simple interest for one year is $\frac{1000 \times 10 \times 1}{100} = 100$. The amount at the end of every year is $1000, (1000+100), (1000+200), (1000+300), \ldots$,i.e.,$1000, 1100, 1200, 1300, \ldots$. This forms an $AP$ because the common difference $(d) = 1100 - 1000 = 100$ is constant.
$(iv)$ Let the initial number of bacteria be $x$. Since they double every second,the sequence is $x, 2x, 4x, 8x, \ldots$. Here,$t_2 - t_1 = x$ and $t_3 - t_2 = 2x$. Since the difference is not constant,it does not form an $AP$.

(N/A) $(i)$ Yes,here $a_{n}=2n-3$.
Put $n=1, a_{1}=2(1)-3=-1$.
Put $n=2, a_{2}=2(2)-3=1$.
Put $n=3, a_{3}=2(3)-3=3$.
Put $n=4, a_{4}=2(4)-3=5$.
The list of numbers is $-1, 1, 3, 5, \ldots$.
Here,$a_{2}-a_{1}=1-(-1)=2$,$a_{3}-a_{2}=3-1=2$,and $a_{4}-a_{3}=5-3=2$.
Since the common difference $d=2$ is constant,$2n-3$ is the $n^{\text{th}}$ term of an $AP$.
$(ii)$ No,here $a_{n}=3n^{2}+5$.
Put $n=1, a_{1}=3(1)^{2}+5=8$.
Put $n=2, a_{2}=3(2)^{2}+5=17$.
Put $n=3, a_{3}=3(3)^{2}+5=32$.
The list of numbers is $8, 17, 32, \ldots$.
Here,$a_{2}-a_{1}=17-8=9$ and $a_{3}-a_{2}=32-17=15$.
Since $a_{2}-a_{1} \neq a_{3}-a_{2}$,it does not form an $AP$.
$(iii)$ No,here $a_{n}=1+n+n^{2}$.
Put $n=1, a_{1}=1+1+(1)^{2}=3$.
Put $n=2, a_{2}=1+2+(2)^{2}=7$.
Put $n=3, a_{3}=1+3+(3)^{2}=13$.
The list of numbers is $3, 7, 13, \ldots$.
Here,$a_{2}-a_{1}=7-3=4$ and $a_{3}-a_{2}=13-7=6$.
Since $a_{2}-a_{1} \neq a_{3}-a_{2}$,it does not form an $AP$.

(D) If three terms $a, b, c$ are in $AP$,then $b - a = c - b$,which implies $2b = a + c$.
Given terms are $n-2, 4n-1, 5n+2$.
Applying the condition $2(4n-1) = (n-2) + (5n+2)$.
$8n - 2 = 6n$.
$8n - 6n = 2$.
$2n = 2$.
Therefore,$n = 1$.

(N/A) Here,$a_{1}=0, a_{2}=\frac{1}{4}, a_{3}=\frac{1}{2}$ and $a_{4}=\frac{3}{4}$.
Calculate the common difference $d$:
$a_{2}-a_{1} = \frac{1}{4} - 0 = \frac{1}{4}$
$a_{3}-a_{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$
$a_{4}-a_{3} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$
Since the difference between consecutive terms is constant $(d = \frac{1}{4})$,the given sequence forms an $AP$.
The next three terms are:
$a_{5} = a_{4} + d = \frac{3}{4} + \frac{1}{4} = 1$
$a_{6} = a_{5} + d = 1 + \frac{1}{4} = \frac{5}{4}$
$a_{7} = a_{6} + d = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$

(A) Given sequence: $5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$
Here,$a_{1}=5, a_{2}=\frac{14}{3}, a_{3}=\frac{13}{3}$ and $a_{4}=4$.
Calculate the common differences:
$a_{2}-a_{1} = \frac{14}{3} - 5 = \frac{14-15}{3} = -\frac{1}{3}$
$a_{3}-a_{2} = \frac{13}{3} - \frac{14}{3} = -\frac{1}{3}$
$a_{4}-a_{3} = 4 - \frac{13}{3} = \frac{12-13}{3} = -\frac{1}{3}$
Since the difference between consecutive terms is constant $(d = -\frac{1}{3})$,the given sequence is an $AP$.
The next three terms are:
$a_{5} = a_{4} + d = 4 + (-\frac{1}{3}) = \frac{12-1}{3} = \frac{11}{3}$
$a_{6} = a_{5} + d = \frac{11}{3} + (-\frac{1}{3}) = \frac{10}{3}$
$a_{7} = a_{6} + d = \frac{10}{3} + (-\frac{1}{3}) = \frac{9}{3} = 3$
Thus,the next three terms are $\frac{11}{3}, \frac{10}{3}, 3$.

(N/A) Here,$a_{1}=\sqrt{3}, a_{2}=2 \sqrt{3}$ and $a_{3}=3 \sqrt{3}$.
Calculate the common difference:
$a_{2}-a_{1}=2 \sqrt{3}-\sqrt{3}=\sqrt{3}$
$a_{3}-a_{2}=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3}$
Since $a_{2}-a_{1}=a_{3}-a_{2}=\sqrt{3}$,which is the common difference $(d)$,the given sequence forms an $AP$.
The next three terms are:
$a_{4}=a_{3}+d=3 \sqrt{3}+\sqrt{3}=4 \sqrt{3}$
$a_{5}=a_{4}+d=4 \sqrt{3}+\sqrt{3}=5 \sqrt{3}$
$a_{6}=a_{5}+d=5 \sqrt{3}+\sqrt{3}=6 \sqrt{3}$

(N/A) Given sequence: $a_1 = a+b, a_2 = a+1+b, a_3 = a+1+b+1$.
Calculate the common difference $d$:
$d_1 = a_2 - a_1 = (a+1+b) - (a+b) = 1$.
$d_2 = a_3 - a_2 = (a+1+b+1) - (a+1+b) = 1$.
Since $d_1 = d_2 = 1$,the common difference is constant,so the sequence is an $AP$.
The next three terms are:
$a_4 = a_3 + d = (a+1+b+1) + 1 = a+b+3 = (a+2)+(b+1)$.
$a_5 = a_4 + d = (a+b+3) + 1 = a+b+4 = (a+2)+(b+2)$.
$a_6 = a_5 + d = (a+b+4) + 1 = a+b+5 = (a+3)+(b+2)$.

(A) Given sequence: $a_1 = a, a_2 = 2a+1, a_3 = 3a+2, a_4 = 4a+3$.
Calculate the common difference $d$:
$d_1 = a_2 - a_1 = (2a+1) - a = a+1$
$d_2 = a_3 - a_2 = (3a+2) - (2a+1) = a+1$
$d_3 = a_4 - a_3 = (4a+3) - (3a+2) = a+1$
Since $d_1 = d_2 = d_3 = a+1$,the common difference is constant. Therefore,the sequence is an $AP$.
The next three terms are:
$a_5 = a_4 + d = (4a+3) + (a+1) = 5a+4$
$a_6 = a_5 + d = (5a+4) + (a+1) = 6a+5$
$a_7 = a_6 + d = (6a+5) + (a+1) = 7a+6$

(B) From the figure,it can be observed that:
$1^{\text{st}}$ step is $\frac{1}{2} \, m$ wide,
$2^{\text{nd}}$ step is $1 \, m$ wide,
$3^{\text{rd}}$ step is $\frac{3}{2} \, m$ wide.
Therefore,the width of each step is increasing by $\frac{1}{2} \, m$ each time,whereas their height $\frac{1}{4} \, m$ and length $50 \, m$ remain the same.
Thus,the widths of these steps form an arithmetic progression: $\frac{1}{2}, 1, \frac{3}{2}, 2, \dots$
Volume of concrete in $1^{\text{st}}$ step $= \text{length} \times \text{width} \times \text{height} = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4} \, m^3$.
Volume of concrete in $2^{\text{nd}}$ step $= 50 \times 1 \times \frac{1}{4} = \frac{25}{2} \, m^3$.
Volume of concrete in $3^{\text{rd}}$ step $= 50 \times \frac{3}{2} \times \frac{1}{4} = \frac{75}{4} \, m^3$.
The volumes of concrete in these steps form an $A.P.$ with first term $a = \frac{25}{4}$ and common difference $d = \frac{25}{2} - \frac{25}{4} = \frac{25}{4}$.
The total volume for $n = 15$ steps is given by the sum $S_n = \frac{n}{2} [2a + (n - 1)d]$.
$S_{15} = \frac{15}{2} \left[ 2 \left( \frac{25}{4} \right) + (15 - 1) \left( \frac{25}{4} \right) \right]$
$S_{15} = \frac{15}{2} \left[ \frac{25}{2} + 14 \times \frac{25}{4} \right] = \frac{15}{2} \left[ \frac{25}{2} + \frac{175}{2} \right]$
$S_{15} = \frac{15}{2} \left[ \frac{200}{2} \right] = \frac{15}{2} \times 100 = 750$.
Thus,the total volume of concrete required is $750 \, m^3$.

Given that,the first term $(a) = \frac{1}{2}$ and the common difference $(d) = -\frac{1}{6}$.
The general form of an $AP$ is $a, a+d, a+2d, \dots$
First term $(T_1) = a = \frac{1}{2}$.
Second term $(T_2) = a + d = \frac{1}{2} + (-\frac{1}{6}) = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.
Third term $(T_3) = a + 2d = \frac{1}{2} + 2(-\frac{1}{6}) = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$.
Hence,the required three terms are $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$.