Mix Examples - Arithmetic Progressions Questions in English

(A) For the given $A.P.$,the $10^{th}$ term $T_{10} = 73$ and the $20^{th}$ term $T_{20} = 143$.
Using the formula for the common difference $d = \frac{T_m - T_n}{m - n}$,we get:
$d = \frac{T_{20} - T_{10}}{20 - 10}$
$d = \frac{143 - 73}{10} = \frac{70}{10} = 7$
Now,using the formula $T_n = a + (n - 1)d$ for the $10^{th}$ term:
$T_{10} = a + 9d$
$73 = a + 9(7)$
$73 = a + 63$
$a = 10$
Finally,the $30^{th}$ term $T_{30}$ is given by:
$T_{30} = a + 29d$
$T_{30} = 10 + 29(7)$
$T_{30} = 10 + 203 = 213$
Thus,the first term is $10$,the common difference is $7$,and the $30^{th}$ term is $213$.

(B) For the given $A.P.$,the first term $a = 4$ and the common difference $d = 9 - 4 = 5$.
Suppose,the $n^{th}$ term $T_n$ of the $A.P.$ is $224$.
The formula for the $n^{th}$ term is $T_n = a + (n - 1)d$.
Substituting the values:
$224 = 4 + (n - 1)5$
Subtracting $4$ from both sides:
$220 = 5(n - 1)$
Dividing by $5$:
$44 = n - 1$
Adding $1$ to both sides:
$n = 45$
Thus,the $45^{th}$ term of the $A.P.$ is $224$.

(C) The amounts to be paid by Ajaybhai in monthly instalments are as follows:
Amount paid in the $1^{st}$ instalment $= Rs. 100$
Amount paid in the $2^{nd}$ instalment $= Rs. 100 + 50 = Rs. 150$
Amount paid in the $3^{rd}$ instalment $= Rs. 150 + 50 = Rs. 200$
Thus,the amounts (in $Rs.$) to be paid by Ajaybhai form a finite Arithmetic Progression ($A$.$P$.): $100, 150, 200, \dots$ up to $24$ terms.
For this $A$.$P$.,the first term $a = 100$ and the common difference $d = 150 - 100 = 50$.
To find the amount paid in the $24^{th}$ instalment,we use the formula for the $n^{th}$ term of an $A$.$P$.:
$T_n = a + (n - 1)d$
For $n = 24$:
$T_{24} = a + (24 - 1)d$
$T_{24} = 100 + 23(50)$
$T_{24} = 100 + 1150$
$T_{24} = 1250$
Thus,Ajaybhai will pay $Rs. 1250$ in the $24^{th}$ instalment.

(N/A) To determine if the sequence $1, 4, 9, 16, \ldots$ is an $A.P.$,we check the difference between consecutive terms.
Let the sequence be $a_1, a_2, a_3, a_4, \ldots$ where $a_1 = 1, a_2 = 4, a_3 = 9, a_4 = 16$.
The difference $d_1 = a_2 - a_1 = 4 - 1 = 3$.
The difference $d_2 = a_3 - a_2 = 9 - 4 = 5$.
Since $d_1 \neq d_2$,the common difference is not constant.
Therefore,the sequence $1, 4, 9, 16, \ldots$ is not an $A.P.$

(A) To determine if the sequence $5, 11, 17, 23, \ldots$ is an $A.P.$,we check the common difference $d$ between consecutive terms.
$d_1 = 11 - 5 = 6$
$d_2 = 17 - 11 = 6$
$d_3 = 23 - 17 = 6$
Since the common difference $d = 6$ is constant,the sequence is an $A.P.$
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$,where $a = 5$ and $d = 6$.
$T_n = 5 + (n - 1)6$
$T_n = 5 + 6n - 6$
$T_n = 6n - 1$

(D) To check if the sequence $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots$ is an $A.P.$,we calculate the difference between consecutive terms.
Let $a_1 = \frac{1}{2}$,$a_2 = \frac{2}{3}$,$a_3 = \frac{3}{4}$,$a_4 = \frac{4}{5}$.
Calculate the common difference $d_1 = a_2 - a_1 = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$.
Calculate the common difference $d_2 = a_3 - a_2 = \frac{3}{4} - \frac{2}{3} = \frac{9-8}{12} = \frac{1}{12}$.
Since $d_1 \neq d_2$,the difference between consecutive terms is not constant.
Therefore,the given sequence is not an $A.P.$

(A) To determine if the sequence is an $A.P.$,we check the common difference $d$ between consecutive terms.
$d_1 = 2.3 - 1.4 = 0.9$
$d_2 = 3.2 - 2.3 = 0.9$
$d_3 = 4.1 - 3.2 = 0.9$
Since the common difference $d = 0.9$ is constant,the sequence is an $A.P.$
The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$,where $a = 1.4$ and $d = 0.9$.
$T_n = 1.4 + (n - 1)0.9$
$T_n = 1.4 + 0.9n - 0.9$
$T_n = 0.9n + 0.5$

(A) To determine if the sequence is an $A.P.$,we check the common difference $d = a_{n} - a_{n-1}$.
Here,$a_{1} = 111, a_{2} = 107, a_{3} = 103, a_{4} = 99$.
$d_{1} = a_{2} - a_{1} = 107 - 111 = -4$.
$d_{2} = a_{3} - a_{2} = 103 - 107 = -4$.
$d_{3} = a_{4} - a_{3} = 99 - 103 = -4$.
Since the common difference is constant $(d = -4)$,the sequence is an $A.P$.
The $n^{th}$ term formula is $a_{n} = a + (n - 1)d$.
Substituting $a = 111$ and $d = -4$:
$a_{n} = 111 + (n - 1)(-4) = 111 - 4n + 4 = -4n + 115$.
Thus,the $n^{th}$ term is $T_{n} = -4n + 115$.

(A) The general form of an $A.P.$ is given by $a, a+d, a+2d, a+3d, \ldots$
Given $a = 8$ and $d = 5$.
Substituting these values:
First term $(T_1)$ = $a = 8$
Second term $(T_2)$ = $a + d = 8 + 5 = 13$
Third term $(T_3)$ = $a + 2d = 8 + 2(5) = 8 + 10 = 18$
Fourth term $(T_4)$ = $a + 3d = 8 + 3(5) = 8 + 15 = 23$
Thus,the $A.P.$ is $8, 13, 18, 23, \ldots$
The general term $T_n$ is given by $T_n = a + (n-1)d = 8 + (n-1)5 = 8 + 5n - 5 = 5n + 3$.

(A) The general form of an $A.P.$ is given by $a, a+d, a+2d, a+3d, \ldots$
Given $a = -12$ and $d = 3$.
First term $T_1 = a = -12$.
Second term $T_2 = a + d = -12 + 3 = -9$.
Third term $T_3 = a + 2d = -12 + 2(3) = -12 + 6 = -6$.
Fourth term $T_4 = a + 3d = -12 + 3(3) = -12 + 9 = -3$.
Thus,the $A.P.$ is $-12, -9, -6, -3, \ldots$
The $n^{th}$ term is given by $T_n = a + (n-1)d = -12 + (n-1)3 = -12 + 3n - 3 = 3n - 15$.

(A) The general form of an $A.P.$ is given by $a, a+d, a+2d, a+3d, \ldots$
Given $a = 21$ and $d = -7$.
First term $T_1 = a = 21$.
Second term $T_2 = a + d = 21 + (-7) = 14$.
Third term $T_3 = a + 2d = 21 + 2(-7) = 21 - 14 = 7$.
Fourth term $T_4 = a + 3d = 21 + 3(-7) = 21 - 21 = 0$.
Thus,the $A.P.$ is $21, 14, 7, 0, \ldots$
The general term $T_n$ is given by $T_n = a + (n-1)d = 21 + (n-1)(-7) = 21 - 7n + 7 = -7n + 28$.

(A) The general form of an $A.P.$ is given by $a, a+d, a+2d, a+3d, \ldots$
Given $a = -15$ and $d = -7$.
First term $a_1 = -15$.
Second term $a_2 = a + d = -15 + (-7) = -22$.
Third term $a_3 = a + 2d = -15 + 2(-7) = -15 - 14 = -29$.
Fourth term $a_4 = a + 3d = -15 + 3(-7) = -15 - 21 = -36$.
The $n^{th}$ term is given by $T_n = a + (n-1)d = -15 + (n-1)(-7) = -15 - 7n + 7 = -7n - 8$.
Thus,the $A.P.$ is $-15, -22, -29, -36, \ldots$ and the general term is $T_n = -7n - 8$.

(A) The general form of an $A.P.$ is $a, a+d, a+2d, a+3d, \ldots$
Given $a = \frac{15}{2}$ and $d = \frac{3}{2}$.
First term $T_1 = a = \frac{15}{2} = 7.5$.
Second term $T_2 = a + d = \frac{15}{2} + \frac{3}{2} = \frac{18}{2} = 9$.
Third term $T_3 = a + 2d = \frac{15}{2} + 2(\frac{3}{2}) = \frac{15}{2} + 3 = \frac{15+6}{2} = \frac{21}{2} = 10.5$.
Fourth term $T_4 = a + 3d = \frac{15}{2} + 3(\frac{3}{2}) = \frac{15+9}{2} = \frac{24}{2} = 12$.
Thus,the $A.P.$ is $\frac{15}{2}, 9, \frac{21}{2}, 12, \ldots$
The general term is $T_n = a + (n-1)d = \frac{15}{2} + (n-1)\frac{3}{2} = \frac{15 + 3n - 3}{2} = \frac{3n + 12}{2}$.

(N/A) The given arithmetic progression is $27, 22, 17, 12, \ldots$
Here,the first term $a = 27$.
The common difference $d = 22 - 27 = -5$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
Substituting the values,we get $T_n = 27 + (n - 1)(-5)$.
$T_n = 27 - 5n + 5$.
$T_n = -5n + 32$.

(A) The given sequence is $10.5, 11.7, 12.9, 14.1, \dots$
Here,the first term $a = 10.5$.
The common difference $d = 11.7 - 10.5 = 1.2$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
Substituting the values,we get $T_n = 10.5 + (n - 1)1.2$.
$T_n = 10.5 + 1.2n - 1.2$.
$T_n = 1.2n + 9.3$.

(A) The given $A.P.$ is $\frac{4}{3}, 2, \frac{8}{3}, \frac{10}{3}, \ldots$
Here,the first term $a = \frac{4}{3}$.
The common difference $d = 2 - \frac{4}{3} = \frac{6-4}{3} = \frac{2}{3}$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n-1)d$.
Substituting the values,we get $T_n = \frac{4}{3} + (n-1)\frac{2}{3}$.
$T_n = \frac{4}{3} + \frac{2n}{3} - \frac{2}{3}$.
$T_n = \frac{2n + 2}{3}$.

(A) The given $A.P.$ is $9, 13, 17, 21, \ldots$
Here,the first term $a = 9$.
The common difference $d = 13 - 9 = 4$.
We need to find the $12^{th}$ term $(a_{12})$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $a_{12} = 9 + (12 - 1) \times 4$.
$a_{12} = 9 + 11 \times 4$.
$a_{12} = 9 + 44$.
$a_{12} = 53$.
Therefore,the $12^{th}$ term is $53$.

(B) The given $A.P.$ is $50, 56, 62, 68, \ldots$
Here,the first term $a = 50$.
The common difference $d = 56 - 50 = 6$.
We need to find the $100^{th}$ term $(a_{100})$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $a_{100} = 50 + (100 - 1) \times 6$.
$a_{100} = 50 + 99 \times 6$.
$a_{100} = 50 + 594$.
$a_{100} = 644$.

(C) The given $A.P.$ is $13, 8, 3, -2, \ldots$
Here,the first term $a = 13$.
The common difference $d = 8 - 13 = -5$.
We need to find the $45^{th}$ term,denoted by $a_{45}$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values $a = 13$,$n = 45$,and $d = -5$:
$a_{45} = 13 + (45 - 1)(-5)$
$a_{45} = 13 + (44)(-5)$
$a_{45} = 13 - 220$
$a_{45} = -207$.
Therefore,the $45^{th}$ term is $-207$.

(A) The given $A.P.$ is $6.4, 7.6, 8.8, 10, \ldots$
Here,the first term $a = 6.4$.
The common difference $d = 7.6 - 6.4 = 1.2$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
Substituting the values: $T_n = 6.4 + (n - 1)1.2$.
$T_n = 6.4 + 1.2n - 1.2$.
$T_n = 1.2n + 5.2$.

(B) The given $A.P.$ is $18, 16.5, 15, 13.5, \ldots$
Here,the first term $a = 18$.
The common difference $d = 16.5 - 18 = -1.5$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
Substituting the values,we get:
$T_n = 18 + (n - 1)(-1.5)$
$T_n = 18 - 1.5n + 1.5$
$T_n = -1.5n + 19.5$.

(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.
For the $7^{th}$ term: $T_7 = a + 6d = 46$ --- $(1)$
For the $12^{th}$ term: $T_{12} = a + 11d = 71$ --- $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(a + 11d) - (a + 6d) = 71 - 46$
$5d = 25$
$d = 5$
Substituting $d = 5$ in equation $(1)$:
$a + 6(5) = 46$
$a + 30 = 46$
$a = 16$
The $n^{th}$ term is $T_n = a + (n - 1)d = 16 + (n - 1)5 = 16 + 5n - 5 = 5n + 11$.

(A) Let the first term be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.
For the $12^{th}$ term: $a + 11d = 4$ --- $(1)$
For the $20^{th}$ term: $a + 19d = -20$ --- $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(a + 19d) - (a + 11d) = -20 - 4$
$8d = -24$
$d = -3$
Substitute $d = -3$ into equation $(1)$:
$a + 11(-3) = 4$
$a - 33 = 4$
$a = 37$
The $n^{th}$ term is $T_n = a + (n - 1)d = 37 + (n - 1)(-3) = 37 - 3n + 3 = -3n + 40$.

(D) The given $A.P.$ is $7, 11, 15, \ldots, 107$.
Here,the first term $a = 7$.
The common difference $d = 11 - 7 = 4$.
The last term $a_n = 107$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $107 = 7 + (n - 1)4$.
$107 - 7 = (n - 1)4$.
$100 = (n - 1)4$.
$n - 1 = 100 / 4 = 25$.
$n = 25 + 1 = 26$.
Therefore,the number of terms in the $A.P.$ is $26$.

(A) Given $A.P.$ is $6, 5.5, 5, \ldots, -12$.
Here,the first term $a = 6$.
The common difference $d = 5.5 - 6 = -0.5$.
The last term $a_n = -12$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $-12 = 6 + (n - 1)(-0.5)$.
$-12 - 6 = (n - 1)(-0.5)$.
$-18 = (n - 1)(-0.5)$.
$n - 1 = \frac{-18}{-0.5} = 36$.
$n = 36 + 1 = 37$.
Thus,the number of terms in the $A.P.$ is $37$.

(B) The given $A.P.$ is $13, 20, 27, 34, \ldots$
Here,the first term $a = 13$ and the common difference $d = 20 - 13 = 7$.
Let the $n^{th}$ term be $a_n = 384$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $384 = 13 + (n - 1)7$.
$384 - 13 = (n - 1)7$.
$371 = (n - 1)7$.
$n - 1 = 371 / 7 = 53$.
$n = 53 + 1 = 54$.
Therefore,the $54^{th}$ term of the $A.P.$ is $384$.

(C) The given $A.P.$ is $34, 27, 20, 13, \ldots$
Here,the first term $a = 34$ and the common difference $d = 27 - 34 = -7$.
We need to find the term $n$ such that $a_n = -519$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $-519 = 34 + (n - 1)(-7)$.
$-519 - 34 = (n - 1)(-7)$.
$-553 = (n - 1)(-7)$.
$n - 1 = \frac{-553}{-7} = 79$.
$n = 79 + 1 = 80$.
Therefore,the $80^{th}$ term of the $A.P.$ is $-519$.

(B) The given $A.P.$ is $112, 107, 102, \ldots$
Here,the first term $a = 112$ and the common difference $d = 107 - 112 = -5$.
We want to find the first negative term,so we set the $n^{th}$ term $a_n < 0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $112 + (n - 1)(-5) < 0$.
$112 - 5n + 5 < 0$
$117 - 5n < 0$
$117 < 5n$
$n > \frac{117}{5}$
$n > 23.4$
Since $n$ must be an integer,the smallest integer greater than $23.4$ is $24$.
Therefore,the $24^{th}$ term is the first negative term.

(A) The given $A.P.$ is $83, 77, 71, \ldots$
Here,the first term $a = 83$ and the common difference $d = 77 - 83 = -6$.
We want to find the first negative term,so we set the $n^{th}$ term $a_n < 0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $83 + (n - 1)(-6) < 0$.
$83 - 6n + 6 < 0$
$89 - 6n < 0$
$89 < 6n$
$n > \frac{89}{6}$
$n > 14.833...$
Since $n$ must be an integer,the smallest integer greater than $14.833$ is $n = 15$.
Therefore,the $15^{th}$ term is the first negative term.

(B) The given $A.P.$ is $12, 21, 30, \ldots, 363$.
Here,the first term $a = 12$ and the common difference $d = 21 - 12 = 9$.
The last term $l = 363$.
To find the $n^{th}$ term from the end of an $A.P.$,we use the formula: $n^{th}$ term from end $= l - (n - 1)d$.
Here,$n = 12$,$l = 363$,and $d = 9$.
Substituting the values: $12^{th}$ term from end $= 363 - (12 - 1) \times 9$.
$= 363 - (11 \times 9) = 363 - 99 = 264$.
Thus,the $12^{th}$ term from the end is $264$.

(C) The given $A.P.$ is $85, 80, 75, \ldots, -30$.
Here,the first term $a = 85$ and the common difference $d = 80 - 85 = -5$.
The last term $l = -30$.
To find the $n^{th}$ term from the end of an $A.P.$,the formula is $l - (n - 1)d$.
Here,we need the $5^{th}$ term from the end,so $n = 5$.
$5^{th} \text{ term from the end} = l - (5 - 1)d$
$= -30 - (4)(-5)$
$= -30 + 20$
$= -10$.

(D) Given $A.P.$ is $5, 15, 25, \ldots$
Here,the first term $a = 5$ and the common difference $d = 15 - 5 = 10$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
The $31^{st}$ term is $a_{31} = a + 30d = 5 + 30(10) = 5 + 300 = 305$.
Let the $n^{th}$ term exceed the $31^{st}$ term by $130$.
So,$a_n = a_{31} + 130$.
$a_n = 305 + 130 = 435$.
Using the formula $a_n = a + (n - 1)d$:
$435 = 5 + (n - 1)10$.
$430 = (n - 1)10$.
$43 = n - 1$.
$n = 44$.
Therefore,the $44^{th}$ term exceeds the $31^{st}$ term by $130$.

(A) The given $A.P.$ is $14, 18, 22, \ldots$
Here,the first term $a = 14$ and the common difference $d = 18 - 14 = 4$.
Let the $n^{th}$ term of the $A.P.$ be $142$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $142 = 14 + (n - 1)4$.
$142 - 14 = (n - 1)4$.
$128 = (n - 1)4$.
$n - 1 = 128 / 4 = 32$.
$n = 32 + 1 = 33$.
Since $n$ is a positive integer,$142$ is the $33^{rd}$ term of the $A.P.$

(N/A) The given $A.P.$ is $242, 236, 230, \ldots$
Here,the first term $a = 242$ and the common difference $d = 236 - 242 = -6$.
Let the $n^{th}$ term of the $A.P.$ be $0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $0 = 242 + (n - 1)(-6)$.
$-242 = -6(n - 1)$.
$n - 1 = \frac{242}{6} = \frac{121}{3} = 40.33$.
$n = 41.33$.
Since $n$ must be a positive integer,$0$ cannot be a term of this $A.P.$

(A) The $n^{th}$ term of the $A.P.$ is given by $T_{n} = -4n + 15$.
To find the $15^{th}$ term $(T_{15})$,substitute $n = 15$ into the formula:
$T_{15} = -4(15) + 15 = -60 + 15 = -45$.
The common difference $(d)$ of an $A.P.$ given by $T_{n} = an + b$ is the coefficient of $n$.
Here,$d = -4$.
Alternatively,$T_{1} = -4(1) + 15 = 11$ and $T_{2} = -4(2) + 15 = 7$.
$d = T_{2} - T_{1} = 7 - 11 = -4$.

(A) The given $A.P.$ is $9, 13, 17, \ldots$
Here,the first term $a = 9$ and the common difference $d = 13 - 9 = 4$.
The $n^{th}$ term of an $A.P.$ is given by the formula $T_n = a + (n - 1)d$.
For the $12^{th}$ term $(n = 12)$:
$T_{12} = 9 + (12 - 1) \times 4 = 9 + 11 \times 4 = 9 + 44 = 53$.
For the $24^{th}$ term $(n = 24)$:
$T_{24} = 9 + (24 - 1) \times 4 = 9 + 23 \times 4 = 9 + 92 = 101$.
Thus,the $12^{th}$ term is $53$ and the $24^{th}$ term is $101$.

(A) The given $A.P.$ is $9, 12, 15, \ldots$
Here,the first term $a = 9$ and the common difference $d = 12 - 9 = 3$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
For the $n^{th}$ term: $T_n = 9 + (n - 1)3 = 9 + 3n - 3 = 3n + 6$.
For the $16^{th}$ term: $T_{16} = 3(16) + 6 = 48 + 6 = 54$.
Thus,the $16^{th}$ term is $54$ and the $n^{th}$ term is $3n + 6$.

(B) Given $A.P.$ is $-1, 3, 7, 11, \ldots$
Here,the first term $a = -1$.
The common difference $d = 3 - (-1) = 3 + 1 = 4$.
Let the $n^{th}$ term be $a_n = 95$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $95 = -1 + (n - 1)4$.
$95 + 1 = (n - 1)4$.
$96 = (n - 1)4$.
$n - 1 = 96 / 4 = 24$.
$n = 24 + 1 = 25$.
Therefore,the $25^{th}$ term of the $A.P.$ is $95$.

(N/A) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$.
Given $T_8 = 31$,so $a + 7d = 31$ (Equation $1$).
Given $T_{15} = T_{11} + 16$,so $(a + 14d) = (a + 10d) + 16$.
This simplifies to $4d = 16$,which gives $d = 4$.
Substituting $d = 4$ into Equation $1$: $a + 7(4) = 31 \implies a + 28 = 31 \implies a = 3$.
The $A.P.$ is $a, a+d, a+2d, \ldots$ which is $3, 7, 11, 15, \ldots$.
The $20^{th}$ term is $T_{20} = a + 19d = 3 + 19(4) = 3 + 76 = 79$.

(D) Given $A.P.$ is $5, 10, 15, \ldots$
Here,first term $a = 5$ and common difference $d = 10 - 5 = 5$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
The $31^{st}$ term is $a_{31} = 5 + (31 - 1)5 = 5 + 30 \times 5 = 5 + 150 = 155$.
Let the $n^{th}$ term exceed the $31^{st}$ term by $130$,so $a_n = a_{31} + 130$.
$a_n = 155 + 130 = 285$.
Using the formula $a_n = a + (n - 1)d$:
$285 = 5 + (n - 1)5$.
$280 = (n - 1)5$.
$n - 1 = 280 / 5 = 56$.
$n = 56 + 1 = 57$.
Therefore,the $57^{th}$ term exceeds the $31^{st}$ term by $130$.

(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$.
Given,$T_{10} = 52 \implies a + 9d = 52$ (Equation $1$).
Also,$T_{17} - T_{13} = 20$.
$(a + 16d) - (a + 12d) = 20 \implies 4d = 20 \implies d = 5$.
Substituting $d = 5$ in Equation $1$: $a + 9(5) = 52 \implies a + 45 = 52 \implies a = 7$.
The $A.P.$ is $a, a+d, a+2d, \ldots$ which is $7, 12, 17, \ldots$.
The $30^{th}$ term is $T_{30} = a + 29d = 7 + 29(5) = 7 + 145 = 152$.

(B) The given $A.P.$ is $3, 7, 11, \ldots$
Here,the first term $a = 3$ and the common difference $d = 7 - 3 = 4$.
Let the $n^{th}$ term of the $A.P.$ be $184$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values,we get: $184 = 3 + (n - 1)4$.
$184 - 3 = 4(n - 1)$.
$181 = 4(n - 1)$.
$n - 1 = 181 / 4 = 45.25$.
$n = 45.25 + 1 = 46.25$.
Since $n$ must be a positive integer,$184$ cannot be a term of this $A.P.$

(A) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.
Given $T_{10} = 52$,we have $a + 9d = 52$ (Equation $1$).
Given $T_{16} = 82$,we have $a + 15d = 82$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 15d) - (a + 9d) = 82 - 52$,which gives $6d = 30$,so $d = 5$.
Substituting $d = 5$ into Equation $1$: $a + 9(5) = 52$,so $a + 45 = 52$,which gives $a = 7$.
The $n^{th}$ term is $T_n = a + (n - 1)d = 7 + (n - 1)5 = 7 + 5n - 5 = 5n + 2$.
The $32^{nd}$ term is $T_{32} = 5(32) + 2 = 160 + 2 = 162$.

(A) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.
Given,$T_7 = a + 6d = -1$ (Equation $1$).
Given,$T_{16} = a + 15d = 17$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 15d) - (a + 6d) = 17 - (-1)$.
$9d = 18$,which gives $d = 2$.
Substituting $d = 2$ into Equation $1$: $a + 6(2) = -1$.
$a + 12 = -1$,so $a = -13$.
The general term is $T_n = a + (n - 1)d = -13 + (n - 1)2$.
$T_n = -13 + 2n - 2 = 2n - 15$.

(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
According to the problem,$5 \times a_5 = 8 \times a_8$.
Substituting the formula for the terms: $5(a + 4d) = 8(a + 7d)$.
Expanding the equation: $5a + 20d = 8a + 56d$.
Rearranging the terms: $5a - 8a = 56d - 20d$.
$-3a = 36d$,which simplifies to $a = -12d$.
We need to find the $13^{th}$ term,$a_{13} = a + 12d$.
Substituting $a = -12d$ into the expression for $a_{13}$:
$a_{13} = -12d + 12d = 0$.
Therefore,the $13^{th}$ term of the $A.P.$ is $0$.

(N/A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.
Given: $T_p = a + (p - 1)d = q$ --- $(1)$
Given: $T_q = a + (q - 1)d = p$ --- $(2)$
Subtracting equation $(2)$ from $(1)$:
$(a + (p - 1)d) - (a + (q - 1)d) = q - p$
$(p - 1 - q + 1)d = q - p$
$(p - q)d = -(p - q)$
$d = -1$
Substituting $d = -1$ in equation $(1)$:
$a + (p - 1)(-1) = q$
$a - p + 1 = q$
$a = p + q - 1$
The general term $T_n$ is given by:
$T_n = a + (n - 1)d$
$T_n = (p + q - 1) + (n - 1)(-1)$
$T_n = p + q - 1 - n + 1$
$T_n = p + q - n$

(B) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
Given,the sixth term $a_6 = 19$,so $a + 5d = 19$ (Equation $1$).
Given,the seventeenth term $a_{17} = 41$,so $a + 16d = 41$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(a + 16d) - (a + 5d) = 41 - 19$
$11d = 22$
$d = 2$.
Substituting $d = 2$ in Equation $1$:
$a + 5(2) = 19$
$a + 10 = 19$
$a = 9$.
Now,find the $50^{th}$ term $a_{50}$:
$a_{50} = a + (50 - 1)d$
$a_{50} = 9 + 49(2)$
$a_{50} = 9 + 98 = 107$.

(N/A) Let the first term of the $A.P.$ be $a$ and the common difference be $d.$
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d.$
Given that the $9^{th}$ term is $0$,we have:
$a_9 = a + (9 - 1)d = 0$
$a + 8d = 0$
$a = -8d$ --- $(1)$
Now,we need to find the $29^{th}$ term:
$a_{29} = a + (29 - 1)d = a + 28d$
Substituting $a = -8d$ from $(1)$:
$a_{29} = -8d + 28d = 20d$ --- $(2)$
Next,we find the $19^{th}$ term:
$a_{19} = a + (19 - 1)d = a + 18d$
Substituting $a = -8d$ from $(1)$:
$a_{19} = -8d + 18d = 10d$ --- $(3)$
Comparing $(2)$ and $(3)$,we see that $a_{29} = 20d$ and $2 \times a_{19} = 2 \times 10d = 20d.$
Therefore,$a_{29} = 2 \times a_{19}.$ Hence,the $29^{th}$ term is two times its $19^{th}$ term.

(A) For the first $A.P.$: $a_1 = 9$,$d_1 = 7 - 9 = -2$. The $n^{th}$ term is $a_n = a_1 + (n - 1)d_1 = 9 + (n - 1)(-2) = 9 - 2n + 2 = 11 - 2n$.
For the second $A.P.$: $a_2 = 15$,$d_2 = 12 - 15 = -3$. The $n^{th}$ term is $a_n = a_2 + (n - 1)d_2 = 15 + (n - 1)(-3) = 15 - 3n + 3 = 18 - 3n$.
Since the $n^{th}$ terms are equal: $11 - 2n = 18 - 3n$.
Rearranging the terms: $3n - 2n = 18 - 11$.
Therefore,$n = 7$.

(N/A) The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
Given,$a_4 + a_8 = 24$.
$(a + 3d) + (a + 7d) = 24 \implies 2a + 10d = 24 \implies a + 5d = 12$ (Equation $1$).
Also,$a_6 + a_{10} = 34$.
$(a + 5d) + (a + 9d) = 34 \implies 2a + 14d = 34 \implies a + 7d = 17$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(a + 7d) - (a + 5d) = 17 - 12 \implies 2d = 5 \implies d = 2.5$.
Substituting $d = 2.5$ in Equation $1$:
$a + 5(2.5) = 12 \implies a + 12.5 = 12 \implies a = -0.5$.
Thus,the first term $a = -0.5$ and the common difference $d = 2.5$.