Textbook -Arithmetic Progressions Questions in English

(C) Given $AP$ is $18, 15 \frac{1}{2}, 13, \ldots, -47$.
Here,the first term $a = 18$.
The common difference $d = a_2 - a_1 = 15 \frac{1}{2} - 18 = \frac{31}{2} - 18 = \frac{31 - 36}{2} = -\frac{5}{2}$.
Let the number of terms in this $AP$ be $n$. The $n^{th}$ term is $a_n = -47$.
Using the formula $a_n = a + (n - 1)d$:
$-47 = 18 + (n - 1)\left(-\frac{5}{2}\right)$
Subtract $18$ from both sides:
$-47 - 18 = (n - 1)\left(-\frac{5}{2}\right)$
$-65 = (n - 1)\left(-\frac{5}{2}\right)$
Multiply both sides by $-\frac{2}{5}$:
$n - 1 = -65 \times \left(-\frac{2}{5}\right)$
$n - 1 = 13 \times 2$
$n - 1 = 26$
$n = 27$.
Thus,the given $AP$ has $27$ terms.

(B) For this $AP$,the first term $a = 11$ and the common difference $d = a_2 - a_1 = 8 - 11 = -3$.
Let $-150$ be the $n^{\text{th}}$ term of this $AP$.
Using the formula for the $n^{\text{th}}$ term: $a_n = a + (n - 1)d$.
Substituting the values: $-150 = 11 + (n - 1)(-3)$.
$-150 = 11 - 3n + 3$.
$-150 = 14 - 3n$.
$3n = 14 + 150$.
$3n = 164$.
$n = \frac{164}{3} = 54.66$.
Since $n$ must be a positive integer for any term in an $AP$,and here $n$ is not an integer,$-150$ is not a term of this $AP$.

(A) Given that,
$a_{11} = 38$
$a_{16} = 73$
We know that the $n^{th}$ term of an $AP$ is given by $a_n = a + (n - 1)d$.
For the $11^{th}$ term: $a + 10d = 38$ $...(1)$
For the $16^{th}$ term: $a + 15d = 73$ $...(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(a + 15d) - (a + 10d) = 73 - 38$
$5d = 35$
$d = 7$
Substituting $d = 7$ in equation $(1)$:
$a + 10(7) = 38$
$a + 70 = 38$
$a = 38 - 70 = -32$
Now,find the $31^{st}$ term:
$a_{31} = a + (31 - 1)d$
$a_{31} = -32 + 30(7)$
$a_{31} = -32 + 210$
$a_{31} = 178$
Thus,the $31^{st}$ term is $178$.

(B) Given that the $3^{rd}$ term $a_3 = 12$ and the total number of terms $n = 50$,with the last term $a_{50} = 106$.
Using the formula for the $n^{th}$ term of an $AP$,$a_n = a + (n - 1)d$:
For the $3^{rd}$ term: $a + 2d = 12$ $(i)$
For the $50^{th}$ term: $a + 49d = 106$ $(ii)$
Subtracting equation $(i)$ from $(ii)$:
$(a + 49d) - (a + 2d) = 106 - 12$
$47d = 94$
$d = 2$
Substituting $d = 2$ in equation $(i)$:
$a + 2(2) = 12$
$a + 4 = 12$
$a = 8$
Now,find the $29^{th}$ term $a_{29}$:
$a_{29} = a + (29 - 1)d$
$a_{29} = 8 + 28(2)$
$a_{29} = 8 + 56 = 64$
Thus,the $29^{th}$ term is $64$.

(C) Given that,the $3^{rd}$ term $a_3 = 4$ and the $9^{th}$ term $a_9 = -8$.
We know the formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n - 1)d$.
For the $3^{rd}$ term: $a + 2d = 4$ $...(i)$
For the $9^{th}$ term: $a + 8d = -8$ $...(ii)$
Subtracting equation $(i)$ from $(ii)$:
$(a + 8d) - (a + 2d) = -8 - 4$
$6d = -12$
$d = -2$
Substituting $d = -2$ in equation $(i)$:
$a + 2(-2) = 4$
$a - 4 = 4$
$a = 8$
Now,let the $n^{th}$ term be zero,so $a_n = 0$:
$0 = a + (n - 1)d$
$0 = 8 + (n - 1)(-2)$
$0 = 8 - 2n + 2$
$2n = 10$
$n = 5$
Thus,the $5^{th}$ term of this $AP$ is zero.

(D) We know that for an $AP$,the $n^{th}$ term is given by the formula: $a_n = a + (n - 1)d$.
For the $17^{th}$ term:
$a_{17} = a + (17 - 1)d = a + 16d$.
For the $10^{th}$ term:
$a_{10} = a + (10 - 1)d = a + 9d$.
According to the problem,the $17^{th}$ term exceeds the $10^{th}$ term by $7$:
$a_{17} - a_{10} = 7$.
Substituting the expressions:
$(a + 16d) - (a + 9d) = 7$.
Simplifying the equation:
$a + 16d - a - 9d = 7$.
$7d = 7$.
Dividing both sides by $7$:
$d = 1$.
Therefore,the common difference is $1$.

(65) Given $A.P.$ is $3, 15, 27, 39, \ldots$
First term $a = 3$.
Common difference $d = 15 - 3 = 12$.
The $54^{th}$ term is given by $a_{54} = a + (54 - 1)d$.
$a_{54} = 3 + 53 \times 12 = 3 + 636 = 639$.
We need to find the term $a_n$ such that $a_n = a_{54} + 132$.
$a_n = 639 + 132 = 771$.
Using the formula $a_n = a + (n - 1)d$:
$771 = 3 + (n - 1)12$.
$768 = (n - 1)12$.
$n - 1 = \frac{768}{12} = 64$.
$n = 65$.
Alternatively,the difference between the $n^{th}$ term and the $54^{th}$ term is $(n - 54)d = 132$.
$(n - 54)12 = 132$.
$n - 54 = 11$.
$n = 65$.

(B) Let the first terms of the two $APs$ be $a_1$ and $a_2$ respectively,and let the common difference be $d$.
For the first $AP$,the $n^{th}$ term is $a_n = a_1 + (n-1)d$.
For the second $AP$,the $n^{th}$ term is $b_n = a_2 + (n-1)d$.
The $100^{th}$ terms are $a_{100} = a_1 + 99d$ and $b_{100} = a_2 + 99d$.
Given that the difference between the $100^{th}$ terms is $100$:
$(a_1 + 99d) - (a_2 + 99d) = 100$
$a_1 - a_2 = 100$ (Equation $1$)
Now,the $1000^{th}$ terms are $a_{1000} = a_1 + 999d$ and $b_{1000} = a_2 + 999d$.
The difference between the $1000^{th}$ terms is:
$(a_1 + 999d) - (a_2 + 999d) = a_1 - a_2$
Substituting the value from Equation $1$,the difference is $100$.

(C) The first three-digit number divisible by $7$ is $105$.
The next numbers are $105 + 7 = 112, 112 + 7 = 119, \ldots$
These numbers form an Arithmetic Progression $(A.P.)$ where the first term $a = 105$ and the common difference $d = 7$.
The largest three-digit number is $999$. Dividing $999$ by $7$,we get $999 = 7 \times 142 + 5$. The remainder is $5$.
Thus,the largest three-digit number divisible by $7$ is $999 - 5 = 994$.
Let $994$ be the $n^{th}$ term of the $A.P.$ sequence.
Using the formula $a_n = a + (n - 1)d$:
$994 = 105 + (n - 1)7$
$994 - 105 = (n - 1)7$
$889 = (n - 1)7$
$n - 1 = 889 / 7$
$n - 1 = 127$
$n = 128$
Therefore,there are $128$ three-digit numbers divisible by $7$.

(D) The first multiple of $4$ greater than $10$ is $12$. The next multiples are $16, 20, 24, \ldots$
These terms form an Arithmetic Progression $(A.P.)$ where the first term $a = 12$ and the common difference $d = 4$.
To find the last term,we divide $250$ by $4$. Since $250 = 4 \times 62 + 2$,the remainder is $2$. Therefore,the largest multiple of $4$ less than or equal to $250$ is $250 - 2 = 248$.
We have the $A.P.$ series: $12, 16, 20, 24, \ldots, 248$.
Using the formula for the $n^{\text{th}}$ term of an $A.P.$: $a_n = a + (n - 1)d$
Substituting the values: $248 = 12 + (n - 1)4$
$248 - 12 = (n - 1)4$
$236 = (n - 1)4$
$n - 1 = \frac{236}{4}$
$n - 1 = 59$
$n = 60$
Thus,there are $60$ multiples of $4$ between $10$ and $250$.

(A) For the first $AP: 63, 65, 67, \ldots$
Here,first term $a = 63$ and common difference $d = 65 - 63 = 2$.
The $n^{th}$ term is $a_n = a + (n - 1)d = 63 + (n - 1)2 = 63 + 2n - 2 = 61 + 2n$.
For the second $AP: 3, 10, 17, \ldots$
Here,first term $a = 3$ and common difference $d = 10 - 3 = 7$.
The $n^{th}$ term is $a_n = a + (n - 1)d = 3 + (n - 1)7 = 3 + 7n - 7 = 7n - 4$.
Given that the $n^{th}$ terms are equal:
$61 + 2n = 7n - 4$
$61 + 4 = 7n - 2n$
$65 = 5n$
$n = 13$.
Thus,the $13^{th}$ terms of both $APs$ are equal.

(A) Let the first term of the $AP$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $AP$ is given by $a_n = a + (n - 1)d$.
Given that the third term is $16$:
$a_3 = a + (3 - 1)d = 16$
$a + 2d = 16$ --- $(1)$
Given that the $7^{th}$ term exceeds the $5^{th}$ term by $12$:
$a_7 - a_5 = 12$
$(a + 6d) - (a + 4d) = 12$
$2d = 12$
$d = 6$
Substituting $d = 6$ in equation $(1)$:
$a + 2(6) = 16$
$a + 12 = 16$
$a = 4$
Thus,the $AP$ is $a, a+d, a+2d, a+3d, \ldots$
$4, 4+6, 4+12, 4+18, \ldots$
$4, 10, 16, 22, \ldots$

(C) The given $A.P.$ is $3, 8, 13, \ldots, 253.$
Here,the first term $a = 3$ and the common difference $d = 8 - 3 = 5.$
The last term $l = 253.$
To find the $n^{th}$ term from the end,we can reverse the $A.P.$ or use the formula: $n^{th}$ term from the end $= l - (n - 1)d.$
Here,$l = 253, n = 20, d = 5.$
$20^{th}$ term from the end $= 253 - (20 - 1) \times 5.$
$= 253 - (19 \times 5).$
$= 253 - 95.$
$= 158.$
Thus,the $20^{th}$ term from the last term is $158.$

(D) We know that the $n^{th}$ term of an $AP$ is given by $a_n = a + (n - 1)d$.
For the $4^{th}$ and $8^{th}$ terms:
$a_4 = a + 3d$
$a_8 = a + 7d$
Given $a_4 + a_8 = 24$,we have $(a + 3d) + (a + 7d) = 24$,which simplifies to $2a + 10d = 24$ or $a + 5d = 12$ $...(1)$
For the $6^{th}$ and $10^{th}$ terms:
$a_6 = a + 5d$
$a_{10} = a + 9d$
Given $a_6 + a_{10} = 44$,we have $(a + 5d) + (a + 9d) = 44$,which simplifies to $2a + 14d = 44$ or $a + 7d = 22$ $...(2)$
Subtracting equation $(1)$ from $(2)$:
$(a + 7d) - (a + 5d) = 22 - 12$
$2d = 10$
$d = 5$
Substituting $d = 5$ into equation $(1)$:
$a + 5(5) = 12$
$a + 25 = 12$
$a = -13$
The first three terms are:
$a_1 = a = -13$
$a_2 = a + d = -13 + 5 = -8$
$a_3 = a + 2d = -13 + 10 = -3$
Thus,the first three terms of the $AP$ are $-13, -8, -3$.

(A) It can be observed that the incomes that Subba Rao obtained in various years are in an $A.P.$ as every year,his salary is increased by ₹ $200$.
Therefore,the salaries of each year starting from $1995$ are $5000, 5200, 5400, \dots$
Here,the first term $a = 5000$ and the common difference $d = 200$.
Let after $n^{\text{th}}$ year,his salary be ₹ $7000$.
Using the formula for the $n^{\text{th}}$ term of an $A.P.$: $a_n = a + (n - 1)d$
Substituting the values: $7000 = 5000 + (n - 1)200$
$2000 = (n - 1)200$
$n - 1 = 10$
$n = 11$
Thus,in the $11^{\text{th}}$ year,his salary will be ₹ $7000$.

(B) Given that the savings follow an Arithmetic Progression $(AP)$ where:
First term $(a)$ = ₹ $5$
Common difference $(d)$ = ₹ $1.75$
$n$-th term $(a_n)$ = ₹ $20.75$
We use the formula for the $n$-th term of an $AP$: $a_n = a + (n - 1)d$
Substituting the given values:
$20.75 = 5 + (n - 1) \times 1.75$
Subtract $5$ from both sides:
$15.75 = (n - 1) \times 1.75$
Divide both sides by $1.75$:
$n - 1 = \frac{15.75}{1.75}$
$n - 1 = 9$
$n = 9 + 1 = 10$
Thus,in the $10$-th week,her savings will be ₹ $20.75$.

(C) Here,the first term $a = 8$ and the common difference $d = 3 - 8 = -5$. The number of terms $n = 22$.
We use the formula for the sum of the first $n$ terms of an $AP$:
$S_n = \frac{n}{2} [2a + (n - 1)d]$
Substituting the values:
$S_{22} = \frac{22}{2} [2(8) + (22 - 1)(-5)]$
$S_{22} = 11 [16 + 21(-5)]$
$S_{22} = 11 [16 - 105]$
$S_{22} = 11 [-89]$
$S_{22} = -979$
Thus,the sum of the first $22$ terms of the $AP$ is $-979$.

(D) Given: Sum of first $14$ terms $S_{14} = 1050$,first term $a = 10$,and number of terms $n = 14$.
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: $1050 = \frac{14}{2}[2(10) + (14 - 1)d]$.
$1050 = 7[20 + 13d]$.
$150 = 20 + 13d$.
$130 = 13d$,which gives $d = 10$.
The $n^{th}$ term of an $AP$ is given by $a_n = a + (n - 1)d$.
For the $20^{th}$ term: $a_{20} = 10 + (20 - 1) \times 10$.
$a_{20} = 10 + 19 \times 10 = 10 + 190 = 200$.
Thus,the $20^{th}$ term is $200$.

(A) Given the $AP$ is $24, 21, 18, \ldots$ where the first term $a = 24$ and the common difference $d = 21 - 24 = -3$.
We are given the sum of $n$ terms $S_n = 78$.
The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values,we get $78 = \frac{n}{2}[2(24) + (n - 1)(-3)]$.
$78 = \frac{n}{2}[48 - 3n + 3] = \frac{n}{2}[51 - 3n]$.
$156 = 51n - 3n^2$,which simplifies to $3n^2 - 51n + 156 = 0$.
Dividing by $3$,we get $n^2 - 17n + 52 = 0$.
Factoring the quadratic equation,we get $(n - 4)(n - 13) = 0$.
Thus,$n = 4$ or $n = 13$.
Both values are valid because the sum of the terms from $5$ to $13$ is zero.

(N/A) $(i)$ Let $S = 1 + 2 + 3 + \ldots + 1000$.
Using the formula $S_n = \frac{n}{2}(a + l)$ for the sum of the first $n$ terms of an Arithmetic Progression $(AP)$,where $a = 1$ and $l = 1000$:
$S_{1000} = \frac{1000}{2}(1 + 1000) = 500 \times 1001 = 500500$.
So,the sum of the first $1000$ positive integers is $500500$.
$(ii)$ Let $S_n = 1 + 2 + 3 + \ldots + n$.
Here,the first term $a = 1$ and the last term $l = n$.
Therefore,$S_n = \frac{n(1 + n)}{2}$ or $S_n = \frac{n(n + 1)}{2}$.
So,the sum of the first $n$ positive integers is given by $S_n = \frac{n(n + 1)}{2}$.

(C) The $n^{th}$ term is given by $a_{n} = 3 + 2n$.
To find the terms of the sequence,substitute $n = 1, 2, 3, \dots$:
$a_{1} = 3 + 2(1) = 5$
$a_{2} = 3 + 2(2) = 7$
$a_{3} = 3 + 2(3) = 9$
The sequence is $5, 7, 9, 11, \dots$
Since the difference between consecutive terms is constant $(7 - 5 = 2, 9 - 7 = 2)$,this is an Arithmetic Progression $(AP)$ with first term $a = 5$ and common difference $d = 2$.
The sum of the first $n$ terms of an $AP$ is given by $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 24$:
$S_{24} = \frac{24}{2}[2(5) + (24 - 1)2]$
$S_{24} = 12[10 + (23 \times 2)]$
$S_{24} = 12[10 + 46]$
$S_{24} = 12 \times 56 = 672$.
Thus,the sum of the first $24$ terms is $672$.

(D) Since the production increases uniformly by a fixed number every year,the number of $TV$ sets manufactured in the $1^{st}, 2^{nd}, 3^{rd}, \dots$ years forms an Arithmetic Progression $(AP)$.
Let $a$ be the production in the $1^{st}$ year and $d$ be the uniform increase in production every year.
The production in the $n^{th}$ year is given by $a_n = a + (n-1)d$.
Given,$a_3 = 600$ and $a_7 = 700$.
Thus,$a + 2d = 600$ (Equation $1$) and $a + 6d = 700$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 6d) - (a + 2d) = 700 - 600$.
$4d = 100$,which gives $d = 25$.
Substituting $d = 25$ in Equation $1$: $a + 2(25) = 600$.
$a + 50 = 600$,so $a = 550$.
Therefore,the production in the $1^{st}$ year is $550$ sets.

(A) Let the production in the $n^{th}$ year be represented by an arithmetic progression $a_n = a + (n-1)d$,where $a$ is the production in the first year and $d$ is the constant annual increase.
Given: $a_3 = 600$ and $a_7 = 700$.
Using the formula $a_n = a + (n-1)d$:
$a + 2d = 600$ (Equation $1$)
$a + 6d = 700$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$:
$(a + 6d) - (a + 2d) = 700 - 600$
$4d = 100$
$d = 25$
Substituting $d = 25$ into Equation $1$:
$a + 2(25) = 600$
$a + 50 = 600$
$a = 550$
Now,find the production in the $10^{th}$ year $(a_{10})$:
$a_{10} = a + 9d$
$a_{10} = 550 + 9(25)$
$a_{10} = 550 + 225 = 775$
Thus,the production in the $10^{th}$ year is $775$ sets.

(B) Let the production in the $n^{th}$ year be $a_n$. Since the production increases uniformly,it forms an Arithmetic Progression $(AP)$.
Given: $a_3 = 600$ and $a_7 = 700$.
The formula for the $n^{th}$ term is $a_n = a + (n-1)d$.
So,$a + 2d = 600$ (Equation $1$) and $a + 6d = 700$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 6d) - (a + 2d) = 700 - 600$,which gives $4d = 100$,so $d = 25$.
Substituting $d = 25$ in Equation $1$: $a + 2(25) = 600$,so $a + 50 = 600$,which gives $a = 550$.
The total production in the first $7$ years is the sum of the first $7$ terms,$S_7$.
The sum formula is $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_7 = \frac{7}{2}[2(550) + (7-1)(25)] = \frac{7}{2}[1100 + 150] = \frac{7}{2}[1250] = 7 \times 625 = 4375$.
Thus,the total production of $TV$ sets in the first $7$ years is $4375$.

(C) The given $A.P.$ is $2, 7, 12, \ldots$ up to $10$ terms.
For this $A.P.$:
First term $a = 2$.
Common difference $d = a_2 - a_1 = 7 - 2 = 5$.
Number of terms $n = 10$.
We use the sum formula for an $A.P.$:
$S_n = \frac{n}{2} [2a + (n - 1)d]$
Substituting the values:
$S_{10} = \frac{10}{2} [2(2) + (10 - 1)5]$
$S_{10} = 5 [4 + 9 \times 5]$
$S_{10} = 5 [4 + 45]$
$S_{10} = 5 \times 49 = 245$
Thus,the sum of the first $10$ terms is $245$.

(D) The given $AP$ is $-37, -33, -29, \ldots$ up to $12$ terms.
For this $AP$:
First term $a = -37$
Common difference $d = a_{2} - a_{1} = (-33) - (-37) = -33 + 37 = 4$
Number of terms $n = 12$
We use the formula for the sum of $n$ terms of an $AP$:
$S_{n} = \frac{n}{2} [2a + (n - 1)d]$
Substituting the values:
$S_{12} = \frac{12}{2} [2(-37) + (12 - 1)4]$
$S_{12} = 6 [-74 + 11 \times 4]$
$S_{12} = 6 [-74 + 44]$
$S_{12} = 6 (-30)$
$S_{12} = -180$
Thus,the sum of the first $12$ terms is $-180$.

(A) For the given $A.P.$: $0.6, 1.7, 2.8, \ldots,$ to $100$ terms.
Here,the first term $a = 0.6$.
The common difference $d = a_2 - a_1 = 1.7 - 0.6 = 1.1$.
The number of terms $n = 100$.
The sum of $n$ terms of an $A.P.$ is given by the formula:
$S_n = \frac{n}{2} [2a + (n - 1)d]$
Substituting the values:
$S_{100} = \frac{100}{2} [2(0.6) + (100 - 1) \times 1.1]$
$S_{100} = 50 [1.2 + 99 \times 1.1]$
$S_{100} = 50 [1.2 + 108.9]$
$S_{100} = 50 [110.1]$
$S_{100} = 5505$
Thus,the sum of the first $100$ terms is $5505$.

(D) For the given $A.P.$,the first term $a = \frac{1}{15}$ and the number of terms $n = 11$.
The common difference $d$ is calculated as:
$d = a_2 - a_1 = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$.
The sum of $n$ terms of an $A.P.$ is given by the formula:
$S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values $n = 11$,$a = \frac{1}{15}$,and $d = \frac{1}{60}$:
$S_{11} = \frac{11}{2} [2(\frac{1}{15}) + (11 - 1)(\frac{1}{60})]$.
$S_{11} = \frac{11}{2} [\frac{2}{15} + 10(\frac{1}{60})]$.
$S_{11} = \frac{11}{2} [\frac{2}{15} + \frac{1}{6}]$.
Taking the $LCM$ of $15$ and $6$,which is $30$:
$S_{11} = \frac{11}{2} [\frac{4 + 5}{30}] = \frac{11}{2} [\frac{9}{30}] = \frac{11}{2} [\frac{3}{10}] = \frac{33}{20}$.

The given series is $7 + 10 \frac{1}{2} + 14 + \ldots + 84$.
This is an Arithmetic Progression $(A.P.)$ where:
First term $(a)$ = $7$
Last term $(l)$ = $84$
Common difference $(d)$ = $10 \frac{1}{2} - 7 = \frac{21}{2} - 7 = \frac{7}{2}$.
Let $84$ be the $n^{\text{th}}$ term of this $A.P.$
Using the formula $l = a + (n - 1)d$:
$84 = 7 + (n - 1) \frac{7}{2}$
$77 = (n - 1) \frac{7}{2}$
$11 = (n - 1) \frac{1}{2}$
$22 = n - 1$
$n = 23$.
Now,find the sum using the formula $S_n = \frac{n}{2}(a + l)$:
$S_{23} = \frac{23}{2}(7 + 84)$
$S_{23} = \frac{23 \times 91}{2}$
$S_{23} = \frac{2093}{2}$
$S_{23} = 1046 \frac{1}{2}$.

(D) The given series is $34 + 32 + 30 + \ldots + 10$.
This is an Arithmetic Progression $(A.P.)$ where the first term $a = 34$,the common difference $d = 32 - 34 = -2$,and the last term $l = 10$.
Let $n$ be the number of terms. The formula for the $n^{\text{th}}$ term is $l = a + (n - 1)d$.
Substituting the values: $10 = 34 + (n - 1)(-2)$.
$10 - 34 = (n - 1)(-2) \implies -24 = (n - 1)(-2)$.
Dividing by $-2$: $12 = n - 1 \implies n = 13$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.
$S_{13} = \frac{13}{2}(34 + 10) = \frac{13}{2}(44) = 13 \times 22 = 286$.

(A) The given series is $-5 + (-8) + (-11) + \ldots + (-230)$.
This is an Arithmetic Progression $(A.P.)$ where the first term $a = -5$ and the last term $l = -230$.
The common difference $d = a_2 - a_1 = (-8) - (-5) = -8 + 5 = -3$.
Let $-230$ be the $n^{\text{th}}$ term of this $A.P.$
Using the formula $l = a + (n - 1)d$:
$-230 = -5 + (n - 1)(-3)$
$-230 + 5 = (n - 1)(-3)$
$-225 = (n - 1)(-3)$
$n - 1 = \frac{-225}{-3} = 75$
$n = 76$.
Now,the sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l)$:
$S_{76} = \frac{76}{2}(-5 + (-230))$
$S_{76} = 38(-235)$
$S_{76} = -8930$.

(B) Given that,$a = 5, d = 3, a_{n} = 50$.
Using the formula for the $n^{th}$ term of an $AP$:
$a_{n} = a + (n - 1)d$
$50 = 5 + (n - 1)3$
$45 = (n - 1)3$
$15 = n - 1$
$n = 16$.
Now,using the sum formula $S_{n} = \frac{n}{2}[a + a_{n}]$:
$S_{16} = \frac{16}{2}[5 + 50]$
$S_{16} = 8 \times 55$
$S_{16} = 440$.
Thus,$n = 16$ and $S_{n} = 440$.

(C) Given that,$a=7$ and $a_{13}=35$.
We know the formula for the $n^{th}$ term of an $AP$ is $a_{n}=a+(n-1)d$.
Substituting the values for $n=13$:
$a_{13}=a+(13-1)d$
$35=7+12d$
$35-7=12d$
$28=12d$
$d=\frac{28}{12}=\frac{7}{3}$.
Now,the formula for the sum of $n$ terms of an $AP$ is $S_{n}=\frac{n}{2}[a+a_{n}]$.
For $n=13$:
$S_{13}=\frac{13}{2}[a+a_{13}]$
$S_{13}=\frac{13}{2}[7+35]$
$S_{13}=\frac{13}{2}[42]$
$S_{13}=13 \times 21$
$S_{13}=273$.

(D) Given that,$a_{12}=37$ and $d=3$.
Using the formula for the $n^{th}$ term of an $AP$,$a_n = a + (n-1)d$.
For $n=12$,$a_{12} = a + (12-1)3$.
$37 = a + 11(3)$.
$37 = a + 33$.
$a = 37 - 33 = 4$.
Now,to find the sum of the first $12$ terms,we use the formula $S_n = \frac{n}{2}[a + a_n]$.
$S_{12} = \frac{12}{2}[4 + 37]$.
$S_{12} = 6(41)$.
$S_{12} = 246$.

(N/A) Given that,$a_{3} = 15$ and $S_{10} = 125$.
As the $n^{th}$ term of an $AP$ is $a_{n} = a + (n - 1)d$,
$a_{3} = a + 2d = 15$ $...(i)$
The sum of $n$ terms is $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
$S_{10} = \frac{10}{2}[2a + (10 - 1)d] = 125$
$5(2a + 9d) = 125$
$2a + 9d = 25$ $...(ii)$
Multiplying equation $(i)$ by $2$,we get:
$2a + 4d = 30$ $...(iii)$
Subtracting equation $(iii)$ from $(ii)$:
$(2a + 9d) - (2a + 4d) = 25 - 30$
$5d = -5$
$d = -1$
Substituting $d = -1$ in equation $(i)$:
$a + 2(-1) = 15$
$a - 2 = 15$
$a = 17$
Now,$a_{10} = a + (10 - 1)d$
$a_{10} = 17 + 9(-1)$
$a_{10} = 17 - 9 = 8$.
Thus,$d = -1$ and $a_{10} = 8$.

(N/A) Given that,$d = 5$ and $S_{9} = 75$.
The formula for the sum of $n$ terms of an $AP$ is $S_{n} = \frac{n}{2}[2a + (n-1)d]$.
Substituting the given values for $n=9$:
$75 = \frac{9}{2}[2a + (9-1)5]$
$75 = \frac{9}{2}[2a + 40]$
$75 = 9(a + 20)$
Dividing both sides by $3$:
$25 = 3(a + 20)$
$25 = 3a + 60$
$3a = 25 - 60$
$3a = -35$
$a = -\frac{35}{3}$
Now,to find the $9^{th}$ term $(a_{9})$,we use the formula $a_{n} = a + (n-1)d$:
$a_{9} = a + (9-1)d$
$a_{9} = -\frac{35}{3} + 8(5)$
$a_{9} = -\frac{35}{3} + 40$
$a_{9} = \frac{-35 + 120}{3}$
$a_{9} = \frac{85}{3}$

(A) Given that,$a=2, d=8, S_{n}=90$.
Using the formula for the sum of $n$ terms of an $AP$:
$S_{n} = \frac{n}{2}[2a + (n-1)d]$
$90 = \frac{n}{2}[2(2) + (n-1)8]$
$90 = \frac{n}{2}[4 + 8n - 8]$
$90 = \frac{n}{2}[8n - 4]$
$90 = n(4n - 2)$
$4n^{2} - 2n - 90 = 0$
Dividing by $2$:
$2n^{2} - n - 45 = 0$
$2n^{2} - 10n + 9n - 45 = 0$
$2n(n - 5) + 9(n - 5) = 0$
$(n - 5)(2n + 9) = 0$
Since $n$ must be a positive integer,$n = 5$.
Now,find $a_{n}$:
$a_{n} = a + (n-1)d$
$a_{5} = 2 + (5-1)8$
$a_{5} = 2 + 32 = 34$.
Thus,$n = 5$ and $a_{n} = 34$.

(A) Given that,$a=8, a_{n}=62, S_{n}=210$.
The formula for the sum of an $AP$ is $S_{n} = \frac{n}{2} [a + a_{n}]$.
Substituting the given values: $210 = \frac{n}{2} [8 + 62]$.
$210 = \frac{n}{2} (70)$.
$210 = 35n$.
$n = \frac{210}{35} = 6$.
Now,use the formula for the $n^{th}$ term: $a_{n} = a + (n - 1)d$.
$62 = 8 + (6 - 1)d$.
$62 - 8 = 5d$.
$54 = 5d$.
$d = \frac{54}{5}$.

(A) Given that,$a_{n}=4, d=2, S_{n}=-14$.
Using the formula for the $n^{th}$ term of an $AP$:
$a_{n} = a + (n - 1)d$
$4 = a + (n - 1)2$
$4 = a + 2n - 2$
$a = 6 - 2n$ --- $(i)$
Using the formula for the sum of $n$ terms of an $AP$:
$S_{n} = \frac{n}{2}[a + a_{n}]$
$-14 = \frac{n}{2}[a + 4]$
$-28 = n(a + 4)$
Substitute $a = 6 - 2n$ from $(i)$ into the equation:
$-28 = n(6 - 2n + 4)$
$-28 = n(10 - 2n)$
$-28 = 10n - 2n^{2}$
$2n^{2} - 10n - 28 = 0$
$n^{2} - 5n - 14 = 0$
$(n - 7)(n + 2) = 0$
Since $n$ must be a positive integer,$n = 7$.
Substitute $n = 7$ into equation $(i)$:
$a = 6 - 2(7)$
$a = 6 - 14$
$a = -8$
Thus,$n = 7$ and $a = -8$.

(B) Given that,$a=3, n=8, S_n=192$.
The formula for the sum of $n$ terms of an $AP$ is $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the given values into the formula:
$192 = \frac{8}{2}[2(3) + (8-1)d]$
$192 = 4[6 + 7d]$
Divide both sides by $4$:
$48 = 6 + 7d$
Subtract $6$ from both sides:
$42 = 7d$
Divide by $7$:
$d = 6$.

(C) Given that,the last term $l = 28$,the sum of terms $S_n = 144$,and the total number of terms $n = 9$.
The formula for the sum of an $AP$ when the first and last terms are known is:
$S_n = \frac{n}{2}(a + l)$
Substituting the given values into the formula:
$144 = \frac{9}{2}(a + 28)$
Multiply both sides by $\frac{2}{9}$:
$144 \times \frac{2}{9} = a + 28$
$16 \times 2 = a + 28$
$32 = a + 28$
$a = 32 - 28$
$a = 4$

(B) Let there be $n$ terms of this $AP$.
For this $AP$,the first term $a = 9$ and the common difference $d = 17 - 9 = 8$.
The sum of $n$ terms is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the given values: $636 = \frac{n}{2}[2(9) + (n - 1)8]$.
$636 = \frac{n}{2}[18 + 8n - 8]$.
$636 = \frac{n}{2}[10 + 8n]$.
$636 = n(5 + 4n)$.
$4n^2 + 5n - 636 = 0$.
Solving the quadratic equation using the factorization method: $4n^2 + 53n - 48n - 636 = 0$.
$n(4n + 53) - 12(4n + 53) = 0$.
$(4n + 53)(n - 12) = 0$.
This gives $n = -\frac{53}{4}$ or $n = 12$.
Since the number of terms $n$ must be a positive integer,we discard the negative fractional value.
Therefore,$n = 12$.

(A) Given that,
First term $a = 5$
Last term $l = 45$
Sum of terms $S_{n} = 400$
Using the formula for the sum of an $AP$: $S_{n} = \frac{n}{2}(a + l)$
Substituting the values: $400 = \frac{n}{2}(5 + 45)$
$400 = \frac{n}{2}(50)$
$400 = 25n$
$n = \frac{400}{25} = 16$
Now,using the formula for the $n^{th}$ term: $l = a + (n - 1)d$
Substituting the values: $45 = 5 + (16 - 1)d$
$45 - 5 = 15d$
$40 = 15d$
$d = \frac{40}{15} = \frac{8}{3}$
Thus,the number of terms is $16$ and the common difference is $\frac{8}{3}$.

(B) Given that,
First term $a = 17$
Last term $l = 350$
Common difference $d = 9$
Let there be $n$ terms in the $AP$.
The formula for the $n^{th}$ term is $l = a + (n - 1)d$.
Substituting the values: $350 = 17 + (n - 1)9$
$333 = (n - 1)9$
$n - 1 = 37$
$n = 38$
Now,the sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.
$S_{38} = \frac{38}{2}(17 + 350)$
$S_{38} = 19 \times 367 = 6973$.
Thus,there are $38$ terms and their sum is $6973$.

(C) Given: Common difference $d = 7$,number of terms $n = 22$,and the $22^{nd}$ term $a_{22} = 149$.
We know the formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n - 1)d$.
Substituting the values: $149 = a + (22 - 1) \times 7$.
$149 = a + 21 \times 7$.
$149 = a + 147$.
$a = 149 - 147 = 2$.
Now,the formula for the sum of the first $n$ terms is $S_n = \frac{n}{2}(a + a_n)$.
Substituting the values: $S_{22} = \frac{22}{2}(2 + 149)$.
$S_{22} = 11 \times 151$.
$S_{22} = 1661$.

(D) Given that,the second term $a_{2} = 14$ and the third term $a_{3} = 18$.
The common difference $d$ is given by $d = a_{3} - a_{2} = 18 - 14 = 4$.
Since $a_{2} = a + d$,we have $14 = a + 4$,which gives the first term $a = 10$.
The sum of the first $n$ terms of an $AP$ is given by $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 51$,$S_{51} = \frac{51}{2}[2(10) + (51 - 1)4]$.
$S_{51} = \frac{51}{2}[20 + (50)(4)]$.
$S_{51} = \frac{51}{2}[20 + 200] = \frac{51}{2}(220)$.
$S_{51} = 51 \times 110 = 5610$.

Given that,
$S_{7} = 49$
$S_{17} = 289$
Using the formula for the sum of the first $n$ terms of an $AP$: $S_{n} = \frac{n}{2}[2a + (n - 1)d]$
For $n = 7$:
$S_{7} = \frac{7}{2}[2a + (7 - 1)d] = 49$
$\frac{7}{2}(2a + 6d) = 49$
$7(a + 3d) = 49$
$a + 3d = 7$ $...(i)$
For $n = 17$:
$S_{17} = \frac{17}{2}[2a + (17 - 1)d] = 289$
$\frac{17}{2}(2a + 16d) = 289$
$17(a + 8d) = 289$
$a + 8d = 17$ $...(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 8d) - (a + 3d) = 17 - 7$
$5d = 10$
$d = 2$
Substituting $d = 2$ in equation $(i)$:
$a + 3(2) = 7$
$a + 6 = 7$
$a = 1$
Now,calculating the sum of the first $n$ terms:
$S_{n} = \frac{n}{2}[2a + (n - 1)d]$
$S_{n} = \frac{n}{2}[2(1) + (n - 1)(2)]$
$S_{n} = \frac{n}{2}[2 + 2n - 2]$
$S_{n} = \frac{n}{2}(2n)$
$S_{n} = n^{2}$

(B) Given the $n^{th}$ term of the sequence is $a_{n}=3+4 n$.
Calculating the first few terms:
$a_{1}=3+4(1)=7$
$a_{2}=3+4(2)=11$
$a_{3}=3+4(3)=15$
$a_{4}=3+4(4)=19$
Checking the difference between consecutive terms:
$a_{2}-a_{1}=11-7=4$
$a_{3}-a_{2}=15-11=4$
$a_{4}-a_{3}=19-15=4$
Since the difference $a_{k+1}-a_{k}=4$ is constant for all $k$,the sequence forms an Arithmetic Progression $(AP)$ with the first term $a=7$ and common difference $d=4$.
The sum of the first $n$ terms of an $AP$ is given by $S_{n}=\frac{n}{2}[2 a+(n-1) d]$.
For $n=15$:
$S_{15}=\frac{15}{2}[2(7)+(15-1) 4]$
$S_{15}=\frac{15}{2}[14+56]$
$S_{15}=\frac{15}{2}(70)$
$S_{15}=15 \times 35 = 525$.

(C) Given $a_{n}=9-5n$.
Calculating the first few terms:
$a_{1}=9-5(1)=4$
$a_{2}=9-5(2)=-1$
$a_{3}=9-5(3)=-6$
Checking the common difference:
$a_{2}-a_{1}=-1-4=-5$
$a_{3}-a_{2}=-6-(-1)=-5$
Since the difference $a_{k+1}-a_{k}$ is constant $(-5)$,the sequence forms an $AP$ with first term $a=4$ and common difference $d=-5$.
The sum of the first $n$ terms is given by $S_{n}=\frac{n}{2}[2a+(n-1)d]$.
For $n=15$:
$S_{15}=\frac{15}{2}[2(4)+(15-1)(-5)]$
$S_{15}=\frac{15}{2}[8+14(-5)]$
$S_{15}=\frac{15}{2}[8-70]$
$S_{15}=\frac{15}{2}(-62)$
$S_{15}=15(-31)=-465$.

(N/A) Given that,the sum of the first $n$ terms is $S_n = 4n - n^2$.
$1$. The first term $(a_1)$ is equal to $S_1$:
$a_1 = S_1 = 4(1) - (1)^2 = 4 - 1 = 3$.
$2$. The sum of the first two terms is $S_2$:
$S_2 = 4(2) - (2)^2 = 8 - 4 = 4$.
$3$. The second term $(a_2)$ is $S_2 - S_1$:
$a_2 = 4 - 3 = 1$.
$4$. The common difference $(d)$ is $a_2 - a_1 = 1 - 3 = -2$.
$5$. The $n^{th}$ term $(a_n)$ is given by $a + (n - 1)d$:
$a_n = 3 + (n - 1)(-2) = 3 - 2n + 2 = 5 - 2n$.
$6$. Finding specific terms:
$a_3 = 5 - 2(3) = 5 - 6 = -1$.
$a_{10} = 5 - 2(10) = 5 - 20 = -15$.
Thus,the first term is $3$,the sum of the first two terms is $4$,the second term is $1$,the $3^{rd}$ term is $-1$,the $10^{th}$ term is $-15$,and the $n^{th}$ term is $5 - 2n$.