Mix Examples - Arithmetic Progressions Questions in English

(C) Let the first term be $a$, the common difference be $d$, and the number of terms be $n$ for the given $AP$.
Given $AP$ is $7, 10, 13, \dots$
Here, $a = 7$ and $d = 10 - 7 = 3$.
Let the $n^{th}$ term $T_n = 55$.
The formula for the $n^{th}$ term of an $AP$ is $T_n = a + (n - 1)d$.
Substituting the values, we get $55 = 7 + (n - 1) \times 3$.
$55 - 7 = (n - 1) \times 3$.
$48 = (n - 1) \times 3$.
$n - 1 = 48 / 3 = 16$.
$n = 16 + 1 = 17$.
Since $n$ is a positive integer, $55$ is the $17^{th}$ term of the given $AP$.

(D) Since the terms $k^{2}+4k+8, 2k^{2}+3k+6,$ and $3k^{2}+4k+4$ are in an $AP$,the difference between consecutive terms must be equal.
Let the terms be $a_1, a_2, a_3$.
Then,$a_2 - a_1 = a_3 - a_2$.
Substituting the values:
$(2k^{2}+3k+6) - (k^{2}+4k+8) = (3k^{2}+4k+4) - (2k^{2}+3k+6)$
Simplifying both sides:
$k^{2} - k - 2 = k^{2} + k - 2$
Subtracting $k^{2}$ and adding $2$ to both sides:
$-k = k$
$2k = 0$
$k = 0$

(A) Let the three parts of the number $207$ be $(a - d)$,$a$,and $(a + d)$,which are in $AP$.
According to the given condition:
Sum of these parts $= 207$
$(a - d) + a + (a + d) = 207$
$3a = 207$
$a = 69$
Given that the product of the two smaller parts is $4623$:
$a(a - d) = 4623$
$69(69 - d) = 4623$
$69 - d = \frac{4623}{69}$
$69 - d = 67$
$d = 69 - 67 = 2$
Therefore,the three parts are:
First part $= a - d = 69 - 2 = 67$
Second part $= a = 69$
Third part $= a + d = 69 + 2 = 71$
Thus,the required three parts are $67, 69, 71$.

(B) Let the angles of the triangle be $a-d, a, a+d$ in $AP$.
The sum of the angles of a triangle is $180^{\circ}$.
$(a-d) + a + (a+d) = 180^{\circ}$
$3a = 180^{\circ} \Rightarrow a = 60^{\circ}$.
The angles are $60^{\circ}-d, 60^{\circ}, 60^{\circ}+d$.
Given that the greatest angle is twice the least angle:
$60^{\circ}+d = 2(60^{\circ}-d)$
$60^{\circ}+d = 120^{\circ}-2d$
$3d = 60^{\circ} \Rightarrow d = 20^{\circ}$.
Substituting $d = 20^{\circ}$ into the angles:
Least angle: $60^{\circ}-20^{\circ} = 40^{\circ}$.
Middle angle: $60^{\circ}$.
Greatest angle: $60^{\circ}+20^{\circ} = 80^{\circ}$.
Thus,the angles are $80^{\circ}, 60^{\circ}, 40^{\circ}$.

(N/A) For the first $AP$: $9, 7, 5, \ldots$
First term $a_1 = 9$,common difference $d_1 = 7 - 9 = -2$.
The $n$th term is $T_n = a_1 + (n - 1)d_1 = 9 + (n - 1)(-2) = 9 - 2n + 2 = 11 - 2n$.
For the second $AP$: $24, 21, 18, \ldots$
First term $a_2 = 24$,common difference $d_2 = 21 - 24 = -3$.
The $n$th term is $T_n = a_2 + (n - 1)d_2 = 24 + (n - 1)(-3) = 24 - 3n + 3 = 27 - 3n$.
Given that the $n$th terms are equal:
$11 - 2n = 27 - 3n$
$3n - 2n = 27 - 11$
$n = 16$.
Substituting $n = 16$ into either expression for the $n$th term:
$T_{16} = 11 - 2(16) = 11 - 32 = -21$.
Thus,the value of $n$ is $16$ and the term is $-21$.

(D) Let the first term and common difference of an $AP$ be $a$ and $d,$ respectively.
According to the question,
$a_{3} + a_{8} = 7$ and $a_{7} + a_{14} = -3$
Using the formula $a_{n} = a + (n - 1)d$:
$(a + 2d) + (a + 7d) = 7 \Rightarrow 2a + 9d = 7$ ....$(i)$
$(a + 6d) + (a + 13d) = -3 \Rightarrow 2a + 19d = -3$ ....$(ii)$
Subtracting Eq. $(i)$ from Eq. $(ii)$:
$(2a + 19d) - (2a + 9d) = -3 - 7$
$10d = -10 \Rightarrow d = -1$
Substituting $d = -1$ in Eq. $(i)$:
$2a + 9(-1) = 7$
$2a - 9 = 7 \Rightarrow 2a = 16 \Rightarrow a = 8$
The $10^{\text{th}}$ term is $a_{10} = a + 9d = 8 + 9(-1) = 8 - 9 = -1$.

(A) Given $AP: -2, -4, -6, \ldots, -100$.
Here,the first term $(a) = -2$,the common difference $(d) = -4 - (-2) = -2$,and the last term $(l) = -100$.
We know that 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 and $d$ is the common difference.
Therefore,the $12^{\text{th}}$ term from the end is:
$a_{12} = -100 - (12 - 1)(-2)$
$a_{12} = -100 - (11)(-2)$
$a_{12} = -100 + 22$
$a_{12} = -78$.
Hence,the $12^{\text{th}}$ term from the end is $-78$.

(B) Given $A.P.$ is $53, 48, 43, \ldots$
The first term $(a) = 53$ and the common difference $(d) = 48 - 53 = -5$.
Let the $n^{th}$ term of the $A.P.$ be the first negative term.
Therefore,$T_n < 0$.
Using the formula for the $n^{th}$ term of an $A.P.$,$T_n = a + (n - 1)d$.
Substituting the values,we get:
$53 + (n - 1)(-5) < 0$
$53 - 5n + 5 < 0$
$58 - 5n < 0$
$5n > 58$
$n > 11.6$
Since $n$ must be a positive integer,the smallest integer greater than $11.6$ is $12$.
Thus,the $12^{th}$ term is the first negative term.
Verification: $T_{12} = 53 + (12 - 1)(-5) = 53 + 11(-5) = 53 - 55 = -2$,which is less than $0$.

(C) The numbers between $10$ and $300$ that leave a remainder of $3$ when divided by $4$ form an Arithmetic Progression $(AP)$.
The first number greater than $10$ that satisfies the condition is $11$ (since $11 = 4 \times 2 + 3$).
The last number less than $300$ that satisfies the condition is $299$ (since $299 = 4 \times 74 + 3$).
Thus,the sequence is $11, 15, 19, 23, \dots, 299$.
In this $AP$,the first term $a = 11$,the common difference $d = 4$,and the last term $a_n = 299$.
Using the formula for the $n^{th}$ term of an $AP$: $a_n = a + (n - 1)d$.
Substituting the values: $299 = 11 + (n - 1)4$.
$299 - 11 = (n - 1)4$.
$288 = (n - 1)4$.
$n - 1 = 288 / 4$.
$n - 1 = 72$.
$n = 73$.
Therefore,there are $73$ such numbers.

(D) Here,the first term $(a) = -\frac{4}{3}$,common difference $(d) = -1 - (-\frac{4}{3}) = -1 + \frac{4}{3} = \frac{1}{3}$,and the last term $(l) = 4 \frac{1}{3} = \frac{13}{3}$.
Since the $n^{th}$ term of an $AP$ is given by $l = a_n = a + (n - 1)d$,we have:
$\frac{13}{3} = -\frac{4}{3} + (n - 1) \frac{1}{3}$
Multiplying by $3$ on both sides:
$13 = -4 + (n - 1)$
$13 + 4 = n - 1$
$n - 1 = 17$
$n = 18$ (which is an even number).
Since $n$ is even,the two middle most terms are the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ terms,i.e.,the $9^{th}$ and $10^{th}$ terms.
$a_9 = a + 8d = -\frac{4}{3} + 8(\frac{1}{3}) = \frac{-4 + 8}{3} = \frac{4}{3}$.
$a_{10} = a + 9d = -\frac{4}{3} + 9(\frac{1}{3}) = \frac{-4 + 9}{3} = \frac{5}{3}$.
Sum of the two middle most terms $= a_9 + a_{10} = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3$.

(A) Let the first term,common difference,and the number of terms of an $AP$ be $a$,$d$,and $n$ respectively.
Given that,first term $(a) = -5$ and last term $(l) = 45$.
Sum of the terms of the $AP = 120 \Rightarrow S_n = 120$.
We know that,if the last term of an $AP$ is known,the sum of $n$ terms is given by:
$S_n = \frac{n}{2}(a + l)$
$120 = \frac{n}{2}(-5 + 45)$
$120 \times 2 = 40 \times n$
$240 = 40n \Rightarrow n = 6$.
Now,to find the common difference,we use the formula for the $n^{th}$ term:
$l = a + (n - 1)d$
$45 = -5 + (6 - 1)d$
$45 + 5 = 5d$
$50 = 5d \Rightarrow d = 10$.
Thus,the number of terms is $6$ and the common difference is $10$.

(B) The given series is an Arithmetic Progression $(AP)$ where the first term $a = 1$ and the common difference $d = -2 - 1 = -3$.
The last term is $l = a_n = -236$.
Using the formula for the $n^{th}$ term of an $AP$: $a_n = a + (n - 1)d$.
$-236 = 1 + (n - 1)(-3)$
$-237 = (n - 1)(-3)$
$n - 1 = 79$
$n = 80$.
Now,the sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$.
$S_{80} = \frac{80}{2}(1 + (-236))$
$S_{80} = 40 \times (-235)$
$S_{80} = -9400$.

(C) The given series is $4-\frac{1}{n}, 4-\frac{2}{n}, 4-\frac{3}{n}, \ldots$ up to $n$ terms.
This is an Arithmetic Progression $(AP)$ where the first term $a = 4 - \frac{1}{n}$.
The common difference $d = (4 - \frac{2}{n}) - (4 - \frac{1}{n}) = -\frac{2}{n} + \frac{1}{n} = -\frac{1}{n}$.
The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values of $a$ and $d$:
$S_n = \frac{n}{2} [2(4 - \frac{1}{n}) + (n-1)(-\frac{1}{n})]$
$S_n = \frac{n}{2} [8 - \frac{2}{n} - 1 + \frac{1}{n}]$
$S_n = \frac{n}{2} [7 - \frac{1}{n}]$
$S_n = \frac{n}{2} [\frac{7n-1}{n}]$
$S_n = \frac{7n-1}{2}$.

(D) The given series is an Arithmetic Progression $(AP)$ with first term $A = \frac{a-b}{a+b}$.
The common difference $D$ is calculated as:
$D = \frac{3a-2b}{a+b} - \frac{a-b}{a+b} = \frac{3a-2b-a+b}{a+b} = \frac{2a-b}{a+b}$.
The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2A + (n-1)D]$.
For $n = 11$:
$S_{11} = \frac{11}{2} \left[ 2 \left( \frac{a-b}{a+b} \right) + (11-1) \left( \frac{2a-b}{a+b} \right) \right]$
$S_{11} = \frac{11}{2} \left[ \frac{2a-2b}{a+b} + \frac{10(2a-b)}{a+b} \right]$
$S_{11} = \frac{11}{2} \left[ \frac{2a-2b + 20a - 10b}{a+b} \right]$
$S_{11} = \frac{11}{2} \left[ \frac{22a - 12b}{a+b} \right]$
$S_{11} = \frac{11 \cdot 2(11a - 6b)}{2(a+b)} = \frac{11(11a - 6b)}{a+b} = \frac{121a - 66b}{a+b}$.

(D) Given $AP$ is $-2, -7, -12, \ldots$
Let the $n^{th}$ term of the $AP$ be $T_n = -77$.
The first term $a = -2$ and the common difference $d = -7 - (-2) = -5$.
The formula for the $n^{th}$ term is $T_n = a + (n - 1)d$.
Substituting the values: $-77 = -2 + (n - 1)(-5)$.
$-77 + 2 = (n - 1)(-5) \Rightarrow -75 = (n - 1)(-5)$.
$n - 1 = \frac{-75}{-5} = 15 \Rightarrow n = 16$.
So,the $16^{th}$ term of the $AP$ is $-77$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}[a + l]$,where $l$ is the last term.
$S_{16} = \frac{16}{2}[-2 + (-77)] = 8[-79] = -632$.
Thus,the sum of this $AP$ up to the term $-77$ is $-632$.

(B) Given that the $n^{th}$ term of the series is $a_{n} = 3 - 4n$.
For $n = 1$,$a_{1} = 3 - 4(1) = 3 - 4 = -1$.
For $n = 2$,$a_{2} = 3 - 4(2) = 3 - 8 = -5$.
For $n = 3$,$a_{3} = 3 - 4(3) = 3 - 12 = -9$.
For $n = 4$,$a_{4} = 3 - 4(4) = 3 - 16 = -13$.
So,the series is $-1, -5, -9, -13, \ldots$
We observe that the common difference $d$ is:
$a_{2} - a_{1} = -5 - (-1) = -4$
$a_{3} - a_{2} = -9 - (-5) = -4$
$a_{4} - a_{3} = -13 - (-9) = -4$
Since the difference between consecutive terms is constant $(d = -4)$,the series forms an $AP$.
The sum of $n$ terms of an $AP$ is given by $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 20$,$a = -1$,and $d = -4$:
$S_{20} = \frac{20}{2}[2(-1) + (20 - 1)(-4)]$
$S_{20} = 10[-2 + (19)(-4)]$
$S_{20} = 10[-2 - 76]$
$S_{20} = 10 \times (-78) = -780$.
Thus,the sum of $20$ terms is $-780$.

(A) We know that the $n^{th}$ term of an $AP$ is given by $a_{n} = S_{n} - S_{n-1}$.
Given $S_{n} = n(4n + 1) = 4n^{2} + n$.
Now,$S_{n-1} = (n-1)(4(n-1) + 1) = (n-1)(4n - 4 + 1) = (n-1)(4n - 3) = 4n^{2} - 3n - 4n + 3 = 4n^{2} - 7n + 3$.
Therefore,$a_{n} = (4n^{2} + n) - (4n^{2} - 7n + 3) = 4n^{2} + n - 4n^{2} + 7n - 3 = 8n - 3$.
To find the $AP$,we substitute values for $n$:
For $n = 1$,$a_{1} = 8(1) - 3 = 5$.
For $n = 2$,$a_{2} = 8(2) - 3 = 16 - 3 = 13$.
For $n = 3$,$a_{3} = 8(3) - 3 = 24 - 3 = 21$.
Thus,the required $AP$ is $5, 13, 21, \ldots$

(D) The $n^{th}$ term of an $AP$ is given by $a_{n} = S_{n} - S_{n-1}$.
Given $S_{n} = 3n^{2} + 5n$.
Then $S_{n-1} = 3(n-1)^{2} + 5(n-1) = 3(n^{2} - 2n + 1) + 5n - 5 = 3n^{2} - 6n + 3 + 5n - 5 = 3n^{2} - n - 2$.
Now,$a_{n} = (3n^{2} + 5n) - (3n^{2} - n - 2) = 3n^{2} + 5n - 3n^{2} + n + 2 = 6n + 2$.
Since $a_{k} = 164$,we substitute $n = k$ in the expression for $a_{n}$:
$6k + 2 = 164$.
$6k = 164 - 2 = 162$.
$k = 162 / 6 = 27$.

The sum of the first $n$ terms of an $AP$ is given by the formula: $S_{n} = \frac{n}{2}[2a + (n - 1)d]$ ... $(i)$
Calculating $S_{8}$:
$S_{8} = \frac{8}{2}[2a + (8 - 1)d] = 4(2a + 7d) = 8a + 28d$
Calculating $S_{4}$:
$S_{4} = \frac{4}{2}[2a + (4 - 1)d] = 2(2a + 3d) = 4a + 6d$
Now,calculating the difference $(S_{8} - S_{4})$:
$S_{8} - S_{4} = (8a + 28d) - (4a + 6d) = 4a + 22d$ ... $(ii)$
Calculating $S_{12}$:
$S_{12} = \frac{12}{2}[2a + (12 - 1)d] = 6(2a + 11d) = 12a + 66d$
From equation $(ii)$,we can see that $3(S_{8} - S_{4}) = 3(4a + 22d) = 12a + 66d$.
Since $S_{12} = 12a + 66d$ and $3(S_{8} - S_{4}) = 12a + 66d$,it is proved that $S_{12} = 3(S_{8} - S_{4})$.

(B) Let the first term be $a$ and the common difference be $d$ for the $AP$.
The $n^{\text{th}}$ term of an $AP$ is given by $T_n = a + (n - 1)d$.
Given,$T_4 = -15$:
$a + 3d = -15$ --- $(i)$
Given,$T_9 = -30$:
$a + 8d = -30$ --- $(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 8d) - (a + 3d) = -30 - (-15)$
$5d = -15$
$d = -3$
Substituting $d = -3$ in equation $(i)$:
$a + 3(-3) = -15$
$a - 9 = -15$
$a = -6$
The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 17$:
$S_{17} = \frac{17}{2}[2(-6) + (17 - 1)(-3)]$
$S_{17} = \frac{17}{2}[-12 + 16(-3)]$
$S_{17} = \frac{17}{2}[-12 - 48]$
$S_{17} = \frac{17}{2}[-60]$
$S_{17} = 17 \times (-30) = -510$.
Thus,the sum of the first $17$ terms is $-510$.

(A) Let $a$ be the first term and $d$ be the common difference of the $AP$.
The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given $S_6 = 36$:
$\frac{6}{2}[2a + (6-1)d] = 36$
$3[2a + 5d] = 36$
$2a + 5d = 12$ ....$(i)$
Given $S_{16} = 256$:
$\frac{16}{2}[2a + (16-1)d] = 256$
$8[2a + 15d] = 256$
$2a + 15d = 32$ ....$(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(2a + 15d) - (2a + 5d) = 32 - 12$
$10d = 20$
$d = 2$
Substituting $d = 2$ into equation $(i)$:
$2a + 5(2) = 12$
$2a + 10 = 12$
$2a = 2$
$a = 1$
Now,find the sum of the first $10$ terms $(S_{10})$:
$S_{10} = \frac{10}{2}[2a + (10-1)d]$
$S_{10} = 5[2(1) + 9(2)]$
$S_{10} = 5[2 + 18]$
$S_{10} = 5 \times 20 = 100$.
Thus,the sum of the first $10$ terms is $100$.

(B) The total number of terms is $n = 11$,which is odd.
Therefore,the middle most term is the $\left(\frac{n+1}{2}\right)^{th}$ term,which is the $\left(\frac{11+1}{2}\right)^{th} = 6^{th}$ term.
Given that the $6^{th}$ term $a_6 = 30$.
Using the formula for the $n^{th}$ term of an $AP$,$a_n = a + (n-1)d$,we have $a + 5d = 30$ ... $(i)$.
The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
For $n = 11$,$S_{11} = \frac{11}{2}[2a + (11-1)d] = \frac{11}{2}[2a + 10d]$.
Factoring out $2$,we get $S_{11} = 11(a + 5d)$.
Substituting the value from equation $(i)$,$S_{11} = 11 \times 30 = 330$.

(C) To find the sum of the last ten terms,we can write the given $AP$ in reverse order.
The given $AP$ is $8, 10, 12, \ldots, 126$.
The reverse $AP$ is $126, 124, 122, \ldots, 10, 8$.
In this reversed $AP$,the first term $(a) = 126$ and the common difference $(d) = 124 - 126 = -2$.
We need to find the sum of the first $10$ terms of this reversed $AP$ using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 10$:
$S_{10} = \frac{10}{2} [2(126) + (10 - 1)(-2)]$
$S_{10} = 5 [252 + 9(-2)]$
$S_{10} = 5 [252 - 18]$
$S_{10} = 5 \times 234 = 1170$.
Thus,the sum of the last ten terms is $1170$.

(D) To find the sum of the first seven numbers that are multiples of both $2$ and $9$,we first find the least common multiple $(LCM)$ of $2$ and $9$.
Since $2$ and $9$ are coprime,their $LCM$ is $2 \times 9 = 18$.
The numbers that are multiples of both $2$ and $9$ are multiples of $18$,which form an arithmetic progression: $18, 36, 54, \ldots$
Here,the first term $a = 18$ and the common difference $d = 18$.
We need to find the sum of the first $n = 7$ terms using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values: $S_7 = \frac{7}{2}[2(18) + (7 - 1)18]$.
$S_7 = \frac{7}{2}[36 + 6 \times 18] = \frac{7}{2}[36 + 108] = \frac{7}{2}[144]$.
$S_7 = 7 \times 72 = 504$.

(A) Let $n$ be the number of terms required to make the sum $-55$.
Here,the first term $(a) = -15$ and the common difference $(d) = -13 - (-15) = 2$.
The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values: $-55 = \frac{n}{2}[2(-15) + (n-1)2]$.
$-55 = \frac{n}{2}[-30 + 2n - 2] = \frac{n}{2}[2n - 32] = n(n - 16)$.
$n^2 - 16n + 55 = 0$.
Factoring the quadratic equation: $n^2 - 11n - 5n + 55 = 0 \Rightarrow n(n - 11) - 5(n - 11) = 0$.
$(n - 5)(n - 11) = 0$,so $n = 5$ or $n = 11$.
The reason for the double answer is that the sum of the terms from the $6^{th}$ term to the $11^{th}$ term is zero. Specifically,the terms are $-15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5$. The sum of the $6^{th}$ to $11^{th}$ terms is $(-5) + (-3) + (-1) + 1 + 3 + 5 = 0$. Thus,adding these terms to the sum of the first $5$ terms does not change the total sum.

(B) Given that,for the first $AP$,the first term $a = 8$ and the common difference $d = 20$.
Let the number of terms be $n$.
The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_n = \frac{n}{2}[2(8) + (n-1)20] = \frac{n}{2}[16 + 20n - 20] = \frac{n}{2}[20n - 4] = n(10n - 2) = 10n^2 - 2n$ .....$(i)$
For the second $AP$,the first term $a' = -30$ and the common difference $d' = 8$.
The sum of the first $2n$ terms is $S'_{2n} = \frac{2n}{2}[2a' + (2n-1)d']$.
$S'_{2n} = n[2(-30) + (2n-1)8] = n[-60 + 16n - 8] = n[16n - 68] = 16n^2 - 68n$ .....$(ii)$
According to the problem,$S_n = S'_{2n}$.
$10n^2 - 2n = 16n^2 - 68n$
$6n^2 - 66n = 0$
$6n(n - 11) = 0$
Since $n$ must be a positive integer,$n = 11$.

(C) Let her pocket money be $Rs.\, x$.
She puts money into her piggy bank following the pattern: $1, 2, 3, \dots, 31$ (since January has $31$ days).
This forms an Arithmetic Progression $(AP)$ where the first term $a = 1$,the common difference $d = 1$,and the number of terms $n = 31$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
$S_{31} = \frac{31}{2}[2(1) + (31 - 1)(1)] = \frac{31}{2}[2 + 30] = \frac{31 \times 32}{2} = 31 \times 16 = 496$.
So,she put $Rs.\, 496$ into her piggy bank.
She also spent $Rs.\, 204$ and had $Rs.\, 100$ left.
The total pocket money $x$ is the sum of the money put in the piggy bank,the money spent,and the money remaining:
$x = 496 + 204 + 100 = 800$.
Therefore,her pocket money for the month was $Rs.\, 800$.

(D) Given that,Yasmeen's savings follow an arithmetic progression $(AP)$.
First term $(a) = 32$.
Common difference $(d) = 36 - 32 = 4$.
Total savings $(S_n) = 2000$.
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: $2000 = \frac{n}{2}[2(32) + (n - 1)4]$.
$2000 = \frac{n}{2}[64 + 4n - 4]$.
$2000 = \frac{n}{2}[60 + 4n]$.
$2000 = n(30 + 2n)$.
$2000 = 30n + 2n^2$.
Dividing by $2$: $n^2 + 15n - 1000 = 0$.
Factoring the quadratic equation: $n^2 + 40n - 25n - 1000 = 0$.
$n(n + 40) - 25(n + 40) = 0$.
$(n + 40)(n - 25) = 0$.
Since $n$ cannot be negative,$n = 25$.
Therefore,she will save $Rs.\, 2000$ in $25$ months.

(2, 6, 10, 14) Let the four consecutive numbers in $AP$ be $(a-3d), (a-d), (a+d), (a+3d)$.
Given that their sum is $32$:
$(a-3d) + (a-d) + (a+d) + (a+3d) = 32$
$4a = 32$
$a = 8$
Given the ratio of the product of the first and last terms to the product of the middle terms is $7:15$:
$\frac{(a-3d)(a+3d)}{(a-d)(a+d)} = \frac{7}{15}$
$\frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15}$
Substitute $a = 8$:
$\frac{64 - 9d^2}{64 - d^2} = \frac{7}{15}$
$15(64 - 9d^2) = 7(64 - d^2)$
$960 - 135d^2 = 448 - 7d^2$
$512 = 128d^2$
$d^2 = 4$
$d = \pm 2$
If $a = 8$ and $d = 2$,the numbers are $(8-6), (8-2), (8+2), (8+6)$,which are $2, 6, 10, 14$.
If $a = 8$ and $d = -2$,the numbers are $(8+6), (8+2), (8-2), (8-6)$,which are $14, 10, 6, 2$.

(B) The given series $1, 4, 7, 10, \ldots, x$ is an Arithmetic Progression $(AP)$ where the first term $a = 1$ and the common difference $d = 4 - 1 = 3$.
Let the number of terms be $n$. The $n^{th}$ term is given by $a_n = a + (n - 1)d$.
Thus,$x = 1 + (n - 1) \times 3 = 3n - 2$.
The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$,where $l$ is the last term $x$.
Given $S_n = 287$,we have $287 = \frac{n}{2}(1 + x)$.
Substituting $x = 3n - 2$ into the equation:
$287 = \frac{n}{2}(1 + 3n - 2)$
$574 = n(3n - 1)$
$3n^2 - n - 574 = 0$.
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{1 \pm \sqrt{(-1)^2 - 4(3)(-574)}}{2(3)} = \frac{1 \pm \sqrt{1 + 6888}}{6} = \frac{1 \pm \sqrt{6889}}{6} = \frac{1 \pm 83}{6}$.
Since $n$ must be a positive integer,$n = \frac{84}{6} = 14$.
Now,find $x$: $x = 3n - 2 = 3(14) - 2 = 42 - 2 = 40$.

(C) Let the first term be $a$ and the common difference be $d$ of an $AP$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
For the first five terms: $S_5 = \frac{5}{2}[2a + 4d] = 5(a + 2d) = 5a + 10d$.
For the first seven terms: $S_7 = \frac{7}{2}[2a + 6d] = 7(a + 3d) = 7a + 21d$.
Given $S_5 + S_7 = 167$,we have $(5a + 10d) + (7a + 21d) = 167$,which simplifies to $12a + 31d = 167$ (Eq. $1$).
Given the sum of the first ten terms $S_{10} = 235$,we have $\frac{10}{2}[2a + 9d] = 235$,which simplifies to $5(2a + 9d) = 235$,or $2a + 9d = 47$ (Eq. $2$).
Multiplying Eq. $2$ by $6$,we get $12a + 54d = 282$ (Eq. $3$).
Subtracting Eq. $1$ from Eq. $3$: $(12a + 54d) - (12a + 31d) = 282 - 167$,which gives $23d = 115$,so $d = 5$.
Substituting $d = 5$ into Eq. $2$: $2a + 9(5) = 47 \Rightarrow 2a + 45 = 47 \Rightarrow 2a = 2 \Rightarrow a = 1$.
The sum of the first twenty terms is $S_{20} = \frac{20}{2}[2a + (20-1)d] = 10[2(1) + 19(5)] = 10[2 + 95] = 10[97] = 970$.

(D) The integers between $1$ and $500$ that are multiples of both $2$ and $5$ must be multiples of the $LCM(2, 5) = 10$.
The sequence of these integers is $10, 20, 30, \dots, 490$.
This forms an Arithmetic Progression $(AP)$ where the first term $a = 10$, the common difference $d = 10$, and the last term $l = 490$.
To find the number of terms $(n)$, we use the formula $a_n = a + (n - 1)d$:
$490 = 10 + (n - 1)10$
$480 = (n - 1)10$
$n - 1 = 48$
$n = 49$.
The sum of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$:
$S_{49} = \frac{49}{2}(10 + 490)$
$S_{49} = \frac{49}{2} \times 500$
$S_{49} = 49 \times 250 = 12250$.

(A) Integers from $1$ to $500$ that are multiples of both $2$ and $5$ are multiples of $\text{lcm}(2, 5) = 10$.
These integers form an arithmetic progression: $10, 20, 30, \ldots, 500$.
Here,the first term $a = 10$,the common difference $d = 10$,and the last term $l = 500$.
Using the formula for the $n$-th term: $a_n = a + (n - 1)d = l$.
$10 + (n - 1)10 = 500$.
$(n - 1)10 = 490$.
$n - 1 = 49$,so $n = 50$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.
$S_{50} = \frac{50}{2}(10 + 500) = 25 \times 510 = 12750$.

(B) To find the sum of integers from $1$ to $500$ which are multiples of $2$ or $5$, we use the principle of inclusion-exclusion:
Sum $= (\text{Sum of multiples of } 2) + (\text{Sum of multiples of } 5) - (\text{Sum of multiples of } \text{LCM}(2, 5) = 10)$.
$1$. Multiples of $2$: $2, 4, 6, \dots, 500$. This is an $AP$ with $a=2, l=500, d=2$.
Number of terms $n_1$: $500 = 2 + (n_1 - 1)2 \Rightarrow 498 = 2(n_1 - 1) \Rightarrow n_1 - 1 = 249 \Rightarrow n_1 = 250$.
Sum $S_1 = \frac{250}{2}(2 + 500) = 125 \times 502 = 62750$.
$2$. Multiples of $5$: $5, 10, 15, \dots, 500$. This is an $AP$ with $a=5, l=500, d=5$.
Number of terms $n_2$: $500 = 5 + (n_2 - 1)5 \Rightarrow 495 = 5(n_2 - 1) \Rightarrow n_2 - 1 = 99 \Rightarrow n_2 = 100$.
Sum $S_2 = \frac{100}{2}(5 + 500) = 50 \times 505 = 25250$.
$3$. Multiples of $10$: $10, 20, 30, \dots, 500$. This is an $AP$ with $a=10, l=500, d=10$.
Number of terms $n_3$: $500 = 10 + (n_3 - 1)10 \Rightarrow 490 = 10(n_3 - 1) \Rightarrow n_3 - 1 = 49 \Rightarrow n_3 = 50$.
Sum $S_3 = \frac{50}{2}(10 + 500) = 25 \times 510 = 12750$.
Total Sum $= S_1 + S_2 - S_3 = 62750 + 25250 - 12750 = 88000 - 12750 = 75250$.

(C) Let $a$ be the first term and $d$ be the common difference of the $AP$.
The $n^{\text{th}}$ term of an $AP$ is given by $a_n = a + (n-1)d$.
According to the first condition,$a_8 = \frac{1}{2} a_2$.
$a + 7d = \frac{1}{2}(a + d)$
$2a + 14d = a + d$
$a + 13d = 0$ --- $(i)$
According to the second condition,$a_{11} = \frac{1}{3} a_4 + 1$.
$a + 10d = \frac{1}{3}(a + 3d) + 1$
$3a + 30d = a + 3d + 3$
$2a + 27d = 3$ --- $(ii)$
From equation $(i)$,$a = -13d$. Substituting this into equation $(ii)$:
$2(-13d) + 27d = 3$
$-26d + 27d = 3$
$d = 3$
Substituting $d = 3$ into equation $(i)$:
$a + 13(3) = 0$
$a = -39$
Now,find the $15^{\text{th}}$ term $(a_{15})$:
$a_{15} = a + 14d$
$a_{15} = -39 + 14(3)$
$a_{15} = -39 + 42 = 3$.

(A) Given,total number of terms $n = 37$.
The middle term is the $\left(\frac{37+1}{2}\right)$-th term,which is the $19$-th term.
So,the three middlemost terms are the $18$-th,$19$-th,and $20$-th terms.
According to the given condition,the sum of the three middlemost terms is $225$:
$a_{18} + a_{19} + a_{20} = 225$
$(a + 17d) + (a + 18d) + (a + 19d) = 225$
$3a + 54d = 225$
$a + 18d = 75$ .....$(i)$
The sum of the last three terms is $429$:
$a_{35} + a_{36} + a_{37} = 429$
$(a + 34d) + (a + 35d) + (a + 36d) = 429$
$3a + 105d = 429$
$a + 35d = 143$ .....$(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 35d) - (a + 18d) = 143 - 75$
$17d = 68$
$d = 4$
Substituting $d = 4$ into equation $(i)$:
$a + 18(4) = 75$
$a + 72 = 75$
$a = 3$
The required $AP$ is $a, a+d, a+2d, a+3d, \dots$
Substituting the values,we get: $3, 3+4, 3+2(4), 3+3(4), \dots$
Thus,the $AP$ is $3, 7, 11, 15, \dots$

(A) The integers between $100$ and $200$ that are divisible by $9$ form an arithmetic progression.
The first integer greater than $100$ divisible by $9$ is $108$ $(9 \times 12 = 108)$.
The last integer less than $200$ divisible by $9$ is $198$ $(9 \times 22 = 198)$.
Thus,the sequence is $108, 117, 126, \ldots, 198$.
Here,the first term $a = 108$,the common difference $d = 9$,and the last term $l = a_n = 198$.
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.
$198 = 108 + (n - 1)9$.
$198 - 108 = (n - 1)9$.
$90 = (n - 1)9$.
$n - 1 = 10 \implies n = 11$.
Now,find the sum $S_n$ using the formula $S_n = \frac{n}{2}(a + l)$.
$S_{11} = \frac{11}{2}(108 + 198)$.
$S_{11} = \frac{11}{2}(306)$.
$S_{11} = 11 \times 153 = 1683$.
Therefore,the sum of the integers between $100$ and $200$ that are divisible by $9$ is $1683$.

(B) The sum of integers between $100$ and $200$ not divisible by $9$ is calculated as: (Sum of all integers between $100$ and $200$) $-$ (Sum of integers between $100$ and $200$ divisible by $9$).
Step $1$: Sum of all integers between $100$ and $200$ (i.e., $101, 102, \dots, 199$).
Here, $a = 101$, $l = 199$, and $d = 1$.
Number of terms $n = 199 - 101 + 1 = 99$.
Sum $S_{99} = \frac{n}{2}(a + l) = \frac{99}{2}(101 + 199) = \frac{99}{2}(300) = 99 \times 150 = 14850$.
Step $2$: Sum of integers between $100$ and $200$ divisible by $9$.
The first multiple of $9$ after $100$ is $108$ $(9 \times 12)$ and the last is $198$ $(9 \times 22)$.
This forms an Arithmetic Progression: $108, 117, \dots, 198$.
Here, $a = 108$, $l = 198$, $d = 9$.
$198 = 108 + (n-1)9 \Rightarrow 90 = (n-1)9 \Rightarrow n-1 = 10 \Rightarrow n = 11$.
Sum $S_{11} = \frac{11}{2}(108 + 198) = \frac{11}{2}(306) = 11 \times 153 = 1683$.
Step $3$: Required sum $= 14850 - 1683 = 13167$.

(C) The sum of integers between $100$ and $200$ not divisible by $9$ is calculated as: (Sum of all integers between $100$ and $200$) $-$ (Sum of integers between $100$ and $200$ divisible by $9$).
Step $1$: Sum of all integers between $100$ and $200$.
The integers are $101, 102, \dots, 199$. This is an Arithmetic Progression $(AP)$ with $a = 101$, $l = 199$, and $d = 1$.
Number of terms $n$: $199 = 101 + (n - 1)1 \implies n - 1 = 98 \implies n = 99$.
Sum $S_{99} = \frac{99}{2}(101 + 199) = \frac{99}{2}(300) = 99 \times 150 = 14850$.
Step $2$: Sum of integers between $100$ and $200$ divisible by $9$.
The first integer divisible by $9$ after $100$ is $108$ $(9 \times 12)$ and the last is $198$ $(9 \times 22)$.
This is an $AP$ with $a = 108$, $l = 198$, $d = 9$.
Number of terms $n$: $198 = 108 + (n - 1)9 \implies 90 = (n - 1)9 \implies n - 1 = 10 \implies n = 11$.
Sum $S_{11} = \frac{11}{2}(108 + 198) = \frac{11}{2}(306) = 11 \times 153 = 1683$.
Step $3$: Required sum $= 14850 - 1683 = 13167$.

(A) Let $a$ and $d$ be the first term and common difference of an $AP$.
Given that,$a_{11} : a_{18} = 2 : 3$.
$\Rightarrow \frac{a + 10d}{a + 17d} = \frac{2}{3}$.
$\Rightarrow 3a + 30d = 2a + 34d$.
$\Rightarrow a = 4d$ ....$(i)$.
Now,$a_5 = a + 4d = 4d + 4d = 8d$.
$a_{21} = a + 20d = 4d + 20d = 24d$.
Therefore,$a_5 : a_{21} = 8d : 24d = 1 : 3$.
Now,the sum of the first five terms,$S_5 = \frac{5}{2}[2a + (5-1)d] = \frac{5}{2}[2(4d) + 4d] = \frac{5}{2}(12d) = 30d$.
And the sum of the first $21$ terms,$S_{21} = \frac{21}{2}[2a + (21-1)d] = \frac{21}{2}[2(4d) + 20d] = \frac{21}{2}(28d) = 294d$.
So,the ratio of the sum of the first five terms to the sum of the first $21$ terms is $S_5 : S_{21} = 30d : 294d = 5 : 49$.

(A) Given that the $AP$ is $a, b, \dots, c.$
Here,the first term $= a,$ and the common difference $d = b - a.$
The last term is $l = a_n = c.$
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d.$
Substituting the values: $c = a + (n - 1)(b - a).$
$(n - 1) = \frac{c - a}{b - a}.$
$n = \frac{c - a}{b - a} + 1 = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a} \dots (i).$
The sum of an $AP$ is given by $S_n = \frac{n}{2}(a + l).$
Substituting $n$ from $(i)$ and $l = c:$
$S_n = \frac{1}{2} \left( \frac{b + c - 2a}{b - a} \right) (a + c).$
Thus,$S_n = \frac{(a + c)(b + c - 2a)}{2(b - a)}.$ Hence proved.

(B) The given series is $-4, -1, 2, \ldots, x$,which forms an Arithmetic Progression $(AP)$.
Here,the first term $a = -4$ and the common difference $d = (-1) - (-4) = 3$.
Let the number of terms be $n$. The $n^{th}$ term is $x = a + (n - 1)d$.
$x = -4 + (n - 1)3 \Rightarrow x = -4 + 3n - 3 \Rightarrow x = 3n - 7 \Rightarrow n = \frac{x + 7}{3}$.
The sum of the $AP$ is given by $S_n = \frac{n}{2}(a + l)$,where $l = x$ is the last term.
$437 = \frac{n}{2}(-4 + x)$.
Substituting $n = \frac{x + 7}{3}$ into the equation:
$437 = \frac{x + 7}{2 \times 3}(-4 + x) = \frac{(x + 7)(x - 4)}{6}$.
$437 \times 6 = x^2 + 3x - 28$.
$2622 = x^2 + 3x - 28 \Rightarrow x^2 + 3x - 2650 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2650)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 10600}}{2} = \frac{-3 \pm \sqrt{10609}}{2}$.
$x = \frac{-3 \pm 103}{2}$.
Since $x$ must be positive for the series to sum to $437$,we take $x = \frac{100}{2} = 50$.

(A) The instalments form an Arithmetic Progression $(AP)$ where the first term $a = 1000$ and the common difference $d = 100$.
$1$. The $30^{\text{th}}$ instalment is given by the formula $a_n = a + (n - 1)d$.
For $n = 30$,$a_{30} = 1000 + (30 - 1) \times 100 = 1000 + 2900 = Rs. 3900$.
$2$. The total amount paid in $30$ instalments is given by the sum formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.
$S_{30} = \frac{30}{2} [2(1000) + (30 - 1) \times 100] = 15 [2000 + 2900] = 15 \times 4900 = Rs. 73500$.
$3$. The remaining loan amount after $30$ instalments is $Total Loan - S_{30} = 118000 - 73500 = Rs. 44500$.

(A) Given,total number of flags $= 27$. The flags are placed at intervals of $2 \,m$. The middle-most flag is the $14^{th}$ flag. Ruchi starts from the $14^{th}$ flag position.
For placing flags on the left side: There are $13$ flags to the left. To place the $1^{st}$ flag (at $2 \,m$ from center),she travels $2 \,m$ and returns $2 \,m$ (Total $4 \,m$). To place the $2^{nd}$ flag (at $4 \,m$ from center),she travels $4 \,m$ and returns $4 \,m$ (Total $8 \,m$). This continues up to the $13^{th}$ flag (at $26 \,m$ from center),where she travels $26 \,m$ and returns $26 \,m$ (Total $52 \,m$).
The total distance for the left side is $4 + 8 + 12 + \dots + 52$. This is an Arithmetic Progression $(AP)$ with $a = 4$,$d = 4$,and $n = 13$.
Sum $= \frac{n}{2} [2a + (n-1)d] = \frac{13}{2} [2(4) + (12)(4)] = \frac{13}{2} [8 + 48] = \frac{13}{2} [56] = 13 \times 28 = 364 \,m$.
Similarly,for the right side,the total distance is also $364 \,m$.
Total distance covered $= 364 + 364 = 728 \,m$.
The maximum distance travelled carrying a flag is the distance to the furthest flag from the center,which is $26 \,m$.

(C) For the given $A.P.$,the first term $a = 5$ and the common difference $d = 12$.
The formula for the $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.
To find the $25^{th}$ term,we substitute $n = 25$,$a = 5$,and $d = 12$ into the formula:
$T_{25} = 5 + (25 - 1) \times 12$
$T_{25} = 5 + 24 \times 12$
$T_{25} = 5 + 288$
$T_{25} = 293$
Thus,the $25^{th}$ term of the $A.P.$ is $293$.

(D) For the given $A.P.$,the $11^{th}$ term $T_{11} = 38$ and the $16^{th}$ term $T_{16} = 73$.
For an $A.P.$,the common difference $d$ is given by:
$d = \frac{T_{m} - T_{n}}{m - n}$
Substituting $m = 16$ and $n = 11$:
$d = \frac{T_{16} - T_{11}}{16 - 11} = \frac{73 - 38}{5} = \frac{35}{5} = 7$
Now,using the formula for the $n^{th}$ term $T_{n} = a + (n - 1)d$:
$T_{11} = a + 10d$
$38 = a + 10(7)$
$38 = a + 70$
$a = 38 - 70 = -32$
Finally,find the $31^{st}$ term $T_{31}$:
$T_{31} = a + 30d$
$T_{31} = -32 + 30(7)$
$T_{31} = -32 + 210 = 178$
Thus,the $31^{st}$ term of the $A.P.$ is $178$.

(A) For the given $A.P.$ $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots$,the first term $a = 20$ and the common difference $d = 19 \frac{1}{4} - 20 = -\frac{3}{4}$.
Suppose the $n^{th}$ term of the $A.P.$ is its first negative term,so $T_n < 0$.
Using the formula $T_n = a + (n - 1)d$,we have $20 + (n - 1)(-\frac{3}{4}) < 0$.
$20 < (n - 1)(\frac{3}{4})$.
$20 \times \frac{4}{3} < n - 1$.
$\frac{80}{3} < n - 1$.
$26.66 + 1 < n$.
$n > 27.66$.
Since $n$ must be a natural number,the smallest integer greater than $27.66$ is $28$.
Therefore,the $28^{th}$ term is the first negative term of the $A.P.$

(B) For the given $A.P.,$ the first term $a = 3$ and the common difference $d = 8 - 3 = 5$.
If $253$ is the $n^{th}$ term of the $A.P.,$ then
$253 = 3 + (n - 1)5$
$250 = 5(n - 1)$
$50 = n - 1$
$n = 51$
Thus,there are $51$ terms in the given finite $A.P.$
The $20^{th}$ term from the end is equivalent to the $(51 - 20 + 1)^{th} = 32^{nd}$ term from the beginning.
$T_{32} = a + (32 - 1)d$
$T_{32} = 3 + 31(5)$
$T_{32} = 3 + 155$
$T_{32} = 158$
Hence,the $20^{th}$ term from the end of the given $A.P.$ is $158.$

(C) For the given $A.P.$,the first term $a = 7$ and the $10^{th}$ term $T_{10} = 61$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_{n} = a + (n - 1)d$.
For the $10^{th}$ term: $T_{10} = a + 9d$.
Substituting the values: $61 = 7 + 9d$.
$54 = 9d$,which gives $d = 6$.
Now,to find the $25^{th}$ term: $T_{25} = a + 24d$.
$T_{25} = 7 + 24(6) = 7 + 144 = 151$.
Thus,the common difference is $6$ and the $25^{th}$ term is $151$.

(D) For the given $A.P.$,the common difference $d = 5$ and the $15^{th}$ term $T_{15} = 72$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_{n} = a + (n - 1)d$,where $a$ is the first term.
Substituting the values for the $15^{th}$ term:
$72 = a + (15 - 1) \times 5$
$72 = a + 14 \times 5$
$72 = a + 70$
$a = 72 - 70 = 2$
Now,to find the $50^{th}$ term $(T_{50})$:
$T_{50} = a + (50 - 1)d$
$T_{50} = 2 + 49 \times 5$
$T_{50} = 2 + 245$
$T_{50} = 247$
Thus,the first term of the $A.P.$ is $2$ and the $50^{th}$ term is $247$.