Mix Examples - Arithmetic Progressions Questions in English

(A) The given $AP$ is $5, 8, 11, 14, \ldots$
Here,the first term $a = 5$.
The common difference $d = 8 - 5 = 3$.
We need to find the $10^{\text{th}}$ term $(a_{10})$.
The formula for the $n^{\text{th}}$ term of an $AP$ is $a_n = a + (n - 1)d$.
Substituting the values $a = 5$,$n = 10$,and $d = 3$:
$a_{10} = 5 + (10 - 1) \times 3$
$a_{10} = 5 + 9 \times 3$
$a_{10} = 5 + 27$
$a_{10} = 32$.
Therefore,the $10^{\text{th}}$ term is $32$.

(B) The formula for the $n^{th}$ term of an $AP$ is $a_{n} = a + (n - 1)d$.
Given values are $a = -7.2$,$d = 3.6$,and $a_{n} = 7.2$.
Substituting these values into the formula:
$7.2 = -7.2 + (n - 1)(3.6)$
Adding $7.2$ to both sides:
$7.2 + 7.2 = (n - 1)(3.6)$
$14.4 = (n - 1)(3.6)$
Dividing both sides by $3.6$:
$\frac{14.4}{3.6} = n - 1$
$4 = n - 1$
$n = 4 + 1$
$n = 5$

(C) In an $AP$,the $n^{th}$ term formula is given by $a_{n} = a + (n - 1)d$.
Given values are $d = -4$,$n = 7$,and $a_{n} = 4$.
Substituting these values into the formula:
$4 = a + (7 - 1)(-4)$
$4 = a + (6)(-4)$
$4 = a - 24$
$a = 4 + 24$
$a = 28$.

(D) The formula for the $n^{th}$ term of an Arithmetic Progression $(AP)$ is given by $a_{n} = a + (n - 1)d$.
Given values are $a = 3.5$,$d = 0$,and $n = 101$.
Substituting these values into the formula:
$a_{101} = 3.5 + (101 - 1) \times 0$
$a_{101} = 3.5 + 100 \times 0$
$a_{101} = 3.5 + 0$
$a_{101} = 3.5$
Therefore,the $101^{st}$ term of the $AP$ is $3.5$.

(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 $d = 4$.

(B) Given $AP$ is $-5, \frac{-5}{2}, 0, \frac{5}{2}, \dots$
Here,the first term $a = -5$.
The common difference $d = \frac{-5}{2} - (-5) = \frac{-5}{2} + 5 = \frac{-5 + 10}{2} = \frac{5}{2}$.
The formula for the $n^{\text{th}}$ term of an $AP$ is $a_n = a + (n - 1)d$.
To find the $11^{\text{th}}$ term $(n = 11)$:
$a_{11} = -5 + (11 - 1) \times \frac{5}{2}$
$a_{11} = -5 + 10 \times \frac{5}{2}$
$a_{11} = -5 + 5 \times 5$
$a_{11} = -5 + 25$
$a_{11} = 20$.

(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 $-2, -4, -6, -8$.

(A) Given,the first term $a = -3$.
The second term $a + d = 4$.
Substituting the value of $a$ in the second term equation: $-3 + d = 4$.
Therefore,the common difference $d = 4 + 3 = 7$.
The formula for the $n^{\text{th}}$ term of an $AP$ is $a_n = a + (n - 1)d$.
For the $21^{\text{st}}$ term $(n = 21)$:
$a_{21} = -3 + (21 - 1) \times 7$.
$a_{21} = -3 + 20 \times 7$.
$a_{21} = -3 + 140$.
$a_{21} = 137$.

(A) Given,the $2^{\text{nd}}$ term $a_2 = 13$ and the $5^{\text{th}}$ term $a_5 = 25$.
Using the formula for the $n^{\text{th}}$ term of an $AP$,$a_n = a + (n - 1)d$,we have:
$a + d = 13$ .....$(i)$
$a + 4d = 25$ .....$(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 4d) - (a + d) = 25 - 13$
$3d = 12$
$d = 4$
Substituting $d = 4$ into equation $(i)$:
$a + 4 = 13$
$a = 9$
Now,finding the $7^{\text{th}}$ term $a_7$:
$a_7 = a + (7 - 1)d$
$a_7 = 9 + 6(4)$
$a_7 = 9 + 24 = 33$
Thus,the $7^{\text{th}}$ term is $33$.

(B) Let the $n^{th}$ term of the given $AP$ be $210$.
Here,the first term $a = 21$.
The common difference $d = 42 - 21 = 21$.
We know the formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n - 1)d$.
Substituting the values,we get $210 = 21 + (n - 1)21$.
$210 = 21 + 21n - 21$.
$210 = 21n$.
$n = 210 / 21 = 10$.
Therefore,the $10^{th}$ term of the $AP$ is $210$.

(C) Given,the common difference of the $AP$ is $d = 5$.
The $n^{th}$ term of an $AP$ is given by the formula $a_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.
Now,we need to find $a_{18} - a_{13}$.
$a_{18} = a + (18 - 1)d = a + 17d$
$a_{13} = a + (13 - 1)d = a + 12d$
Subtracting the two terms:
$a_{18} - a_{13} = (a + 17d) - (a + 12d)$
$a_{18} - a_{13} = 17d - 12d = 5d$
Substituting the value of $d = 5$:
$a_{18} - a_{13} = 5 \times 5 = 25$.

(D) Given,$a_{18}-a_{14}=32$.
Using the formula for the $n^{th}$ term of an $AP$,$a_n = a + (n-1)d$,we have:
$a + (18-1)d - [a + (14-1)d] = 32$
$a + 17d - a - 13d = 32$
$4d = 32$
$d = 32 / 4 = 8$
Thus,the common difference $d$ is $8$.

(A) Let the first term of the first $AP$ be $a_1 = -1$ and the first term of the second $AP$ be $a_2 = -8$.
Let the common difference for both $AP$s be $d$.
The $n^{\text{th}}$ term of an $AP$ is given by $T_n = a + (n - 1)d$.
The $4^{\text{th}}$ term of the first $AP$ is $T_4 = a_1 + (4 - 1)d = -1 + 3d$.
The $4^{\text{th}}$ term of the second $AP$ is $T'_4 = a_2 + (4 - 1)d = -8 + 3d$.
The difference between their $4^{\text{th}}$ terms is $(a_1 + 3d) - (a_2 + 3d) = a_1 - a_2$.
Substituting the values,we get $(-1) - (-8) = -1 + 8 = 7$.
Thus,the difference between their $4^{\text{th}}$ terms is $7$.

(B) Given that $7$ times the $7^{\text{th}}$ term is equal to $11$ times the $11^{\text{th}}$ term.
$7 a_{7} = 11 a_{11}$
Using the formula for the $n^{\text{th}}$ term of an $AP$,$a_{n} = a + (n - 1)d$:
$7[a + (7 - 1)d] = 11[a + (11 - 1)d]$
$7(a + 6d) = 11(a + 10d)$
$7a + 42d = 11a + 110d$
Rearranging the terms to one side:
$11a - 7a + 110d - 42d = 0$
$4a + 68d = 0$
Dividing by $4$:
$a + 17d = 0$
We need to find the $18^{\text{th}}$ term $(a_{18})$:
$a_{18} = a + (18 - 1)d = a + 17d$
Since $a + 17d = 0$,therefore $a_{18} = 0$.

(C) The $n^{\text{th}}$ term of an $AP$ from the end is given by the formula:
$a_n = l - (n - 1)d$
Where $l$ is the last term,$n$ is the position from the end,and $d$ is the common difference.
Given the $AP: -11, -8, -5, \dots, 49$,the last term $l = 49$.
The common difference $d = -8 - (-11) = -8 + 11 = 3$.
We need to find the $4^{\text{th}}$ term from the end,so $n = 4$.
Substituting the values into the formula:
$a_4 = 49 - (4 - 1) \times 3$
$a_4 = 49 - (3 \times 3)$
$a_4 = 49 - 9 = 40$.
Thus,the $4^{\text{th}}$ term from the end is $40$.

(D) The famous mathematician associated with finding the sum of the first $100$ natural numbers is Carl Friedrich Gauss.
According to the anecdote,when he was a young student,he quickly calculated the sum of the first $100$ natural numbers $(1 + 2 + 3 + \dots + 100)$ by pairing the numbers: $(1 + 100) + (2 + 99) + \dots + (50 + 51) = 101 \times 50 = 5050$.

(A) Given,the first term $a = -5$ and the common difference $d = 2$.
We need to find the sum of the first $n = 6$ terms.
The formula for the sum of the first $n$ terms of an $AP$ is $S_{n} = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values $n = 6$,$a = -5$,and $d = 2$ into the formula:
$S_{6} = \frac{6}{2} [2(-5) + (6 - 1)(2)]$
$S_{6} = 3 [-10 + 5(2)]$
$S_{6} = 3 [-10 + 10]$
$S_{6} = 3(0) = 0$.
Thus,the sum of the first $6$ terms is $0$.

(B) Given,the $AP$ is $10, 6, 2, \ldots$
Here,the first term $a = 10$ and the common difference $d = 6 - 10 = -4$.
The formula for the sum of the first $n$ terms of an $AP$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 16$:
$S_{16} = \frac{16}{2}[2(10) + (16 - 1)(-4)]$
$S_{16} = 8[20 + 15(-4)]$
$S_{16} = 8[20 - 60]$
$S_{16} = 8(-40) = -320$.

(C) The sum of $n$ terms of an $AP$ is given by the formula: $S_{n} = \frac{n}{2}(a + a_{n})$.
Given $a = 1$,$a_{n} = 20$,and $S_{n} = 399$.
Substituting these values into the formula:
$399 = \frac{n}{2}(1 + 20)$
$399 = \frac{n}{2}(21)$
$399 = 10.5n$
$n = \frac{399}{10.5}$
$n = 38$.
Thus,the value of $n$ is $38$.

(D) The first five multiples of $3$ are $3, 6, 9, 12,$ and $15$.
This forms an Arithmetic Progression $(AP)$ where the first term $a = 3$,the common difference $d = 3$,and the number of terms $n = 5$.
The sum of the first $n$ terms of an $AP$ is given by the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values:
$S_5 = \frac{5}{2} [2(3) + (5 - 1)3]$
$S_5 = \frac{5}{2} [6 + 4(3)]$
$S_5 = \frac{5}{2} [6 + 12]$
$S_5 = \frac{5}{2} [18]$
$S_5 = 5 \times 9 = 45$.

(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 $AP$.

(A) Yes,it is possible to find the value directly.
The $n$-th term of an $AP$ is given by $a_n = a + (n - 1)d$.
Therefore,$a_{30} = a + (30 - 1)d = a + 29d$ and $a_{20} = a + (20 - 1)d = a + 19d$.
Subtracting the two terms: $a_{30} - a_{20} = (a + 29d) - (a + 19d) = 10d$.
From the given $AP$,the common difference $d = -7 - (-3) = -7 + 3 = -4$.
Substituting the value of $d$ into the expression: $a_{30} - a_{20} = 10(-4) = -40$.

(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 $5$.

(N/A) Let $0$ be the $n^{th}$ term of the given $AP$,i.e.,$a_n = 0$.
Given that,the first term $a = 31$ and the common difference $d = 28 - 31 = -3$.
The $n^{th}$ term of an $AP$ is given by the formula:
$a_n = a + (n - 1)d$
Substituting the values:
$0 = 31 + (n - 1)(-3)$
Rearranging the terms:
$3(n - 1) = 31$
$n - 1 = \frac{31}{3}$
$n = \frac{31}{3} + 1 = \frac{34}{3} = 11 \frac{1}{3}$
Since $n$ must be a positive integer (representing the position of a term),and $11 \frac{1}{3}$ is not an integer,$0$ is not a term of the given $AP$.

(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,$n = 1$.

(A) Given $AP$ is $-11, -7, -3, \ldots, 49$.
Here,the first term $a = -11$ and the common difference $d = -7 - (-11) = 4$.
The last term $a_n = 49$.
Using the formula $a_n = a + (n - 1)d$:
$49 = -11 + (n - 1) \times 4$
$60 = (n - 1) \times 4$
$n - 1 = 15 \implies n = 16$.
Since $n = 16$ is an even number,there are two middle terms: the $(\frac{n}{2})^{th}$ term and the $(\frac{n}{2} + 1)^{th}$ term.
These are the $8^{th}$ term and the $9^{th}$ term.
$a_8 = a + 7d = -11 + 7(4) = -11 + 28 = 17$.
$a_9 = a + 8d = -11 + 8(4) = -11 + 32 = 21$.
Thus,the middle terms are $17$ and $21$.

(A) Let the three terms in $AP$ be $a-d, a, a+d$.
According to the problem,the sum of these terms is $33$:
$(a-d) + a + (a+d) = 33$
$3a = 33$
$a = 11$
Also,the product of the first and the third term exceeds the second term by $29$:
$(a-d)(a+d) = a + 29$
$a^2 - d^2 = a + 29$
Substituting $a = 11$:
$11^2 - d^2 = 11 + 29$
$121 - d^2 = 40$
$d^2 = 121 - 40 = 81$
$d = \pm 9$
Case $1$: If $d = 9$,the terms are $11-9, 11, 11+9$,which are $2, 11, 20$.
Case $2$: If $d = -9$,the terms are $11-(-9), 11, 11+(-9)$,which are $20, 11, 2$.
Thus,the $AP$ is $2, 11, 20, \dots$ or $20, 11, 2, \dots$.

(A-B) $A_{1}: 2, -2, -6, -10, \ldots$
Here,the common difference $d = a_{2} - a_{1} = -2 - 2 = -4$. Thus,$(A_{1}) \rightarrow (B_{4})$.
$A_{2}: a = -18, n = 10, a_{n} = 0$. Using $a_{n} = a + (n - 1)d$:
$0 = -18 + (10 - 1)d \Rightarrow 18 = 9d \Rightarrow d = 2$. Thus,$(A_{2}) \rightarrow (B_{5})$.
$A_{3}: a = 0, a_{10} = 6$. Using $a_{n} = a + (n - 1)d$:
$a_{10} = a + 9d \Rightarrow 6 = 0 + 9d \Rightarrow d = \frac{6}{9} = \frac{2}{3}$. Thus,$(A_{3}) \rightarrow (B_{1})$.
$A_{4}: a_{2} = 13, a_{4} = 3$. Using $a_{n} = a + (n - 1)d$:
$a + d = 13$ $(i)$
$a + 3d = 3$ $(ii)$
Subtracting $(i)$ from $(ii)$: $(a + 3d) - (a + d) = 3 - 13 \Rightarrow 2d = -10 \Rightarrow d = -5$. Thus,$(A_{4}) \rightarrow (B_{2})$.
Final matching: $(A_{1}) \rightarrow (B_{4}), (A_{2}) \rightarrow (B_{5}), (A_{3}) \rightarrow (B_{1}), (A_{4}) \rightarrow (B_{2})$.

(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:
$a_5 = a_4 + d = (4a+3) + (a+1) = 5a+4$
$a_6 = a_5 + d = (5a+4) + (a+1) = 6a+5$
$a_7 = a_6 + d = (6a+5) + (a+1) = 7a+6$

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 $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$.

(A) Given that,the first term $(a) = -5$ and the common difference $(d) = -3$.
The general form of an $AP$ is $a, a+d, a+2d, \dots$
First term $(T_1) = a = -5$.
Second term $(T_2) = a + d = -5 + (-3) = -8$.
Third term $(T_3) = a + 2d = -5 + 2(-3) = -5 - 6 = -11$.
Hence,the first three terms of the $AP$ are $-5, -8, -11$.

(A) Given that,the first term $(a) = \sqrt{2}$ and the common difference $(d) = \frac{1}{\sqrt{2}}$.
The general form of an $AP$ is $a, a+d, a+2d, \dots$
First term $(T_{1}) = a = \sqrt{2}$.
Second term $(T_{2}) = a + d = \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{(\sqrt{2} \times \sqrt{2}) + 1}{\sqrt{2}} = \frac{2 + 1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
Third term $(T_{3}) = a + 2d = \sqrt{2} + 2 \times \left(\frac{1}{\sqrt{2}}\right) = \sqrt{2} + \sqrt{2} = 2\sqrt{2} = \frac{2 \times 2}{\sqrt{2}} = \frac{4}{\sqrt{2}}$.
Hence,the first three terms are $\sqrt{2}, \frac{3}{\sqrt{2}}, \frac{4}{\sqrt{2}}$.

(A) Given that,the first term $(a) = \sqrt{2}$ and the common difference $(d) = \frac{1}{\sqrt{2}}$.
The general form of an $AP$ is $a, a+d, a+2d, \dots$
First term $(T_1) = a = \sqrt{2}$.
Second term $(T_2) = a + d = \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{(\sqrt{2} \times \sqrt{2}) + 1}{\sqrt{2}} = \frac{2+1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
Third term $(T_3) = a + 2d = \sqrt{2} + 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} + \frac{2}{\sqrt{2}} = \frac{(\sqrt{2} \times \sqrt{2}) + 2}{\sqrt{2}} = \frac{2+2}{\sqrt{2}} = \frac{4}{\sqrt{2}}$.
Thus,the first three terms are $\sqrt{2}, \frac{3}{\sqrt{2}}, \frac{4}{\sqrt{2}}$.

(A) Since $a, 7, b, 23, c$ are in $AP$,the common difference $d$ is constant.
$d = 7 - a = b - 7 = 23 - b = c - 23$.
From $b - 7 = 23 - b$,we get $2b = 30$,so $b = 15$.
Using $7 - a = b - 7$,substitute $b = 15$:
$7 - a = 15 - 7 = 8$,which gives $a = 7 - 8 = -1$.
Using $c - 23 = 23 - b$,substitute $b = 15$:
$c - 23 = 23 - 15 = 8$,which gives $c = 8 + 23 = 31$.
Thus,$a = -1, b = 15, c = 31$.

(N/A) Let the first term of an $AP$ be $a$ and the common difference be $d$.
Given,$a_{5} = 19$ and $a_{13} - a_{8} = 20$.
Using the formula $a_{n} = a + (n - 1)d$:
For the fifth term: $a + 4d = 19$ ........$(i)$
For the difference between the thirteenth and eighth terms: $(a + 12d) - (a + 7d) = 20$
$5d = 20$
$d = 4$
Substituting $d = 4$ into equation $(i)$:
$a + 4(4) = 19$
$a + 16 = 19$
$a = 3$
The $AP$ is given by $a, a+d, a+2d, a+3d, \dots$
Substituting the values: $3, 3+4, 3+2(4), 3+3(4), \dots$
Thus,the $AP$ is $3, 7, 11, 15, \dots$

(C) Let the first term be $a$,common difference be $d$,and the number of terms be $n$.
The $n^{\text{th}}$ term of an $AP$ is given by $T_n = a + (n-1)d$.
Given that the $26^{\text{th}}$ term is $0$:
$a + 25d = 0$ .... $(i)$
Given that the $11^{\text{th}}$ term is $3$:
$a + 10d = 3$ .... $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(a + 25d) - (a + 10d) = 0 - 3$
$15d = -3$
$d = -\frac{3}{15} = -\frac{1}{5}$
Substituting $d = -\frac{1}{5}$ into equation $(i)$:
$a + 25(-\frac{1}{5}) = 0$
$a - 5 = 0$
$a = 5$
Given that the last term $l$ is $-\frac{1}{5}$:
$l = a + (n-1)d$
$-\frac{1}{5} = 5 + (n-1)(-\frac{1}{5})$
Multiply the entire equation by $5$:
$-1 = 25 - (n-1)$
$-1 = 25 - n + 1$
$-1 = 26 - n$
$n = 27$
Thus,the common difference is $-\frac{1}{5}$ and the number of terms is $27$.

(D) Let the first term be $a$ and the common difference be $d$ of the $AP$.
According to the question:
$a_5 + a_7 = 52$ and $a_{10} = 46$.
Using the formula $a_n = a + (n-1)d$:
$(a + 4d) + (a + 6d) = 52$
$2a + 10d = 52$
$a + 5d = 26$ ..... $(i)$
Also,$a_{10} = 46$:
$a + 9d = 46$ ..... $(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 9d) - (a + 5d) = 46 - 26$
$4d = 20$
$d = 5$
Substituting $d = 5$ in equation $(i)$:
$a + 5(5) = 26$
$a + 25 = 26$
$a = 1$
The $AP$ is given by $a, a+d, a+2d, a+3d, \ldots$
Substituting the values: $1, 1+5, 1+10, 1+15, \ldots$
Thus,the $AP$ is $1, 6, 11, 16, \ldots$

(A) Let the first term be $a$ and the common difference be $d$ for an $AP$.
Given that,the first term $a = 12$.
According to the problem,the $7^{\text{th}}$ term $(T_7)$ is $24$ less than the $11^{\text{th}}$ term $(T_{11})$.
So,$T_7 = T_{11} - 24$.
Using the formula for the $n^{\text{th}}$ term of an $AP$,$T_n = a + (n-1)d$,we have:
$a + 6d = (a + 10d) - 24$.
Subtracting $a$ from both sides,we get $6d = 10d - 24$.
Rearranging the terms,$4d = 24$,which gives $d = 6$.
Now,we need to find the $20^{\text{th}}$ term $(T_{20})$:
$T_{20} = a + (20-1)d = a + 19d$.
Substituting the values $a = 12$ and $d = 6$:
$T_{20} = 12 + 19(6) = 12 + 114 = 126$.
Thus,the $20^{\text{th}}$ term of the $AP$ is $126$.

(N/A) Let the first term be $a$ and the common difference be $d$ for an $AP$.
The $n^{\text{th}}$ term of an $AP$ is given by the formula $T_n = a + (n - 1)d$.
Given that the $9^{\text{th}}$ term is zero,we have $T_9 = 0$.
$a + (9 - 1)d = 0$
$a + 8d = 0$
$a = -8d$ ....$(i)$
Now,calculate the $19^{\text{th}}$ term $(T_{19})$:
$T_{19} = a + (19 - 1)d$
$T_{19} = a + 18d$
Substituting $a = -8d$ from equation $(i)$:
$T_{19} = -8d + 18d = 10d$ ....$(ii)$
Next,calculate the $29^{\text{th}}$ term $(T_{29})$:
$T_{29} = a + (29 - 1)d$
$T_{29} = a + 28d$
Substituting $a = -8d$ from equation $(i)$:
$T_{29} = -8d + 28d = 20d$ ....$(iii)$
From equations $(ii)$ and $(iii)$:
$T_{29} = 20d = 2 \times (10d) = 2 \times T_{19}$.
Thus,the $29^{\text{th}}$ term is twice its $19^{\text{th}}$ term.