Mix Examples - Arithmetic Progressions Questions in English

(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: $(A_{1}) \rightarrow (B_{4}), (A_{2}) \rightarrow (B_{5}), (A_{3}) \rightarrow (B_{1}), (A_{4}) \rightarrow (B_{2})$.

(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 $200$.

(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 $a = -8$.

(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 $6973$.

(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,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 $10$.

(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) 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 $n$ 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.

(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) 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 $3, 7, 11, 15, \dots$

(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 $a = 5$,the common difference $d = 13 - 5 = 8$,and $n = 20$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
$\therefore S_{20} = \frac{20}{2}[2(5) + (20 - 1)8]$
$\quad = 10[10 + 152]$
$\quad = 10[162] = 1620$.
Thus,the sum of the first $20$ terms of the $A.P.$ is $1620$.
Now,let the sum of $n$ terms be $S_n = 6440$.
$\therefore 6440 = \frac{n}{2}[2(5) + (n - 1)8]$
$\therefore 12880 = n[10 + 8n - 8]$
$\therefore 12880 = 8n^2 + 2n$
$\therefore 8n^2 + 2n - 12880 = 0$
Dividing by $2$,we get $4n^2 + n - 6440 = 0$.
Solving the quadratic equation: $4n^2 + 161n - 160n - 6440 = 0$
$\therefore n(4n + 161) - 40(4n + 161) = 0$
$\therefore (4n + 161)(n - 40) = 0$
Since $n$ must be a positive integer,$n = 40$ (as $n = -\frac{161}{4}$ is not possible).
Thus,the sum of $40$ terms of the $A.P.$ is $6440$.

(A) Let the four numbers in $A.P.$ be $(a-3d, a-d, a+d, a+3d).$
According to the first condition,the sum of the numbers is $36$:
$(a-3d) + (a-d) + (a+d) + (a+3d) = 36$
$4a = 36$
$a = 9$
According to the second condition,the product of the means exceeds the product of the extremes by $32$:
$(a-d)(a+d) = (a-3d)(a+3d) + 32$
$a^2 - d^2 = a^2 - 9d^2 + 32$
$8d^2 = 32$
$d^2 = 4$
$d = \pm 2$
Case $1$: If $a = 9$ and $d = 2$,the numbers are $(9-6, 9-2, 9+2, 9+6) = (3, 7, 11, 15).$
Case $2$: If $a = 9$ and $d = -2$,the numbers are $(9+6, 9+2, 9-2, 9-6) = (15, 11, 7, 3).$
Thus,the required four numbers are $3, 7, 11, 15$ or $15, 11, 7, 3.$

(N/A) Let the required five numbers in $A.P.$ be $a-2d, a-d, a, a+d, a+2d$.
By the first condition,
$(a-2d) + (a-d) + a + (a+d) + (a+2d) = 30$
$5a = 30$
$a = 6$
By the second condition,
$(a-2d)^2 + (a-d)^2 + a^2 + (a+d)^2 + (a+2d)^2 = 220$
$(a^2 - 4ad + 4d^2) + (a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) + (a^2 + 4ad + 4d^2) = 220$
$5a^2 + 10d^2 = 220$
$a^2 + 2d^2 = 44$
Substituting $a = 6$:
$6^2 + 2d^2 = 44$
$36 + 2d^2 = 44$
$2d^2 = 8$
$d^2 = 4$
$d = \pm 2$
If $a = 6$ and $d = 2$,the numbers are $6-4, 6-2, 6, 6+2, 6+4$,which are $2, 4, 6, 8, 10$.
If $a = 6$ and $d = -2$,the numbers are $6+4, 6+2, 6, 6-2, 6-4$,which are $10, 8, 6, 4, 2$.
Thus,the five numbers are $2, 4, 6, 8, 10$ or $10, 8, 6, 4, 2$.

(A) The given $A.P.$ is $85, 78, 71, \ldots$ where the first term $a = 85$ and the common difference $d = 78 - 85 = -7$.
For the $n^{th}$ term to be the first negative term,we set $a_{n} < 0$.
The formula for the $n^{th}$ term is $a_{n} = a + (n - 1)d$.
So,$85 + (n - 1)(-7) < 0$.
$85 - 7n + 7 < 0$.
$92 - 7n < 0$.
$7n > 92$.
$n > 13.14$.
Since $n$ must be an integer,the first negative term is the $14^{th}$ term $(n = 14)$.
To find $S_{14}$,we use the sum formula $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
$S_{14} = \frac{14}{2}[2(85) + (14 - 1)(-7)]$.
$S_{14} = 7[170 + 13(-7)]$.
$S_{14} = 7[170 - 91]$.
$S_{14} = 7[79] = 553$.

(A) Let the first term 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 $a_{12} = -13$,so $a + 11d = -13$ --- (Equation $1$).
The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given $S_4 = 24$,so $\frac{4}{2}[2a + 3d] = 24$,which simplifies to $2(2a + 3d) = 24$,or $2a + 3d = 12$ --- (Equation $2$).
Multiply Equation $1$ by $2$: $2a + 22d = -26$ --- (Equation $3$).
Subtract Equation $2$ from Equation $3$: $(2a + 22d) - (2a + 3d) = -26 - 12$,which gives $19d = -38$,so $d = -2$.
Substitute $d = -2$ into Equation $2$: $2a + 3(-2) = 12$,so $2a - 6 = 12$,$2a = 18$,$a = 9$.
Now,find the sum of the first $10$ terms $S_{10} = \frac{10}{2}[2(9) + (10-1)(-2)]$.
$S_{10} = 5[18 + 9(-2)] = 5[18 - 18] = 5(0) = 0$.

(N/A) Given: First term $a = 17$,last term $l = a_n = 350$,and common difference $d = 9$.
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.
Substituting the values: $350 = 17 + (n - 1)9$.
$350 - 17 = (n - 1)9$.
$333 = (n - 1)9$.
$n - 1 = 333 / 9 = 37$.
$n = 37 + 1 = 38$.
Now,to find the sum $S_n$ using the formula $S_n = (n/2)(a + l)$:
$S_{38} = (38 / 2)(17 + 350)$.
$S_{38} = 19 \times 367$.
$S_{38} = 6973$.
Thus,the number of terms is $38$ and the sum is $6973$.

(N/A) Since the production of the factory increases equally every year,the annual production forms an Arithmetic Progression $(A.P.)$.
For this $A.P.$,the $3^{rd}$ term $T_{3} = 600$ and the $7^{th}$ term $T_{7} = 700$.
The common difference $d$ is given by $d = \frac{T_{m} - T_{n}}{m - n}$.
Substituting $m = 7$ and $n = 3$,we get:
$d = \frac{700 - 600}{7 - 3} = \frac{100}{4} = 25$.
Using the formula $T_{n} = a + (n - 1)d$ for $T_{3}$:
$600 = a + (3 - 1)(25) \implies 600 = a + 50 \implies a = 550$.
Thus,the production in the $1^{st}$ year is $550 \, TV$.
For the $10^{th}$ year:
$T_{10} = a + 9d = 550 + 9(25) = 550 + 225 = 775 \, TV$.
For the total production in $7$ years $(S_{7})$:
$S_{n} = \frac{n}{2} [2a + (n - 1)d]$
$S_{7} = \frac{7}{2} [2(550) + (7 - 1)(25)] = \frac{7}{2} [1100 + 150] = \frac{7}{2} [1250] = 7 \times 625 = 4375 \, TV$.

(B) For the $A.P.$ $40, 36, 32, \ldots$,the first term $a = 40$ and common difference $d = 36 - 40 = -4$.
To find which term is $0$,we use the formula $a_n = a + (n - 1)d$.
Setting $a_n = 0$,we get $0 = 40 + (n - 1)(-4)$.
$4(n - 1) = 40 \implies n - 1 = 10 \implies n = 11$.
Thus,the $11^{th}$ term is $0$.
To find the sum of $n$ terms equal to $0$,we use $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Setting $S_n = 0$,we get $0 = \frac{n}{2}[2(40) + (n - 1)(-4)]$.
$0 = 80 - 4n + 4 \implies 4n = 84 \implies n = 21$.
Thus,the sum of $21$ terms is $0$.

(A) For the first $A.P.$: $a_1 = 63$,$d_1 = 65 - 63 = 2$. The $n^{th}$ term is $a_n = a_1 + (n - 1)d_1 = 63 + (n - 1)2 = 63 + 2n - 2 = 61 + 2n$.
For the second $A.P.$: $a_2 = 3$,$d_2 = 10 - 3 = 7$. The $n^{th}$ term is $a_n = a_2 + (n - 1)d_2 = 3 + (n - 1)7 = 3 + 7n - 7 = 7n - 4$.
Equating the two $n^{th}$ terms: $61 + 2n = 7n - 4$.
$61 + 4 = 7n - 2n$.
$65 = 5n$.
$n = 13$.
Thus,the $13^{th}$ term of both sequences is equal.

(A) Given: First term $a = 5$,common difference $d = 3$,and $n^{th}$ term $T_n = 50$.
Step $1$: Find $n$ using the formula $T_n = a + (n - 1)d$.
$50 = 5 + (n - 1)3$
$45 = (n - 1)3$
$15 = n - 1$
$n = 16$.
Step $2$: Find the sum of $n$ terms $S_n$ using the formula $S_n = \frac{n}{2}(a + T_n)$.
$S_{16} = \frac{16}{2}(5 + 50)$
$S_{16} = 8 \times 55$
$S_{16} = 440$.
Thus,$n = 16$ and $S_n = 440$.

(A) Given: $T_{n} = a + (n - 1)d = 4$ and $S_{n} = \frac{n}{2}[a + T_{n}] = -14$.
Substituting $T_{n} = 4$ in the sum formula: $\frac{n}{2}[a + 4] = -14 \implies n(a + 4) = -28$.
From the first equation: $a = 4 - (n - 1)d = 4 - 2n + 2 = 6 - 2n$.
Substitute $a$ in the sum equation: $n(6 - 2n + 4) = -28 \implies n(10 - 2n) = -28$.
$10n - 2n^{2} = -28 \implies 2n^{2} - 10n - 28 = 0 \implies n^{2} - 5n - 14 = 0$.
Factoring the quadratic: $(n - 7)(n + 2) = 0$.
Since $n$ must be a positive integer,$n = 7$.
Now,find $a$: $a = 6 - 2(7) = 6 - 14 = -8$.
Thus,$n = 7$ and $a = -8$.

(A) Let the three numbers in $A.P.$ be $(a - d)$,$a$,and $(a + d)$.
According to the problem,the sum of these numbers is $21$:
$(a - d) + a + (a + d) = 21$
$3a = 21$
$a = 7$
Now,the product of the first and the third number exceeds the second number by $6$:
$(a - d)(a + d) = a + 6$
$a^2 - d^2 = a + 6$
Substitute $a = 7$ into the equation:
$7^2 - d^2 = 7 + 6$
$49 - d^2 = 13$
$d^2 = 36$
$d = \pm 6$
Case $1$: If $d = 6$,the numbers are $(7 - 6), 7, (7 + 6)$,which are $1, 7, 13$.
Case $2$: If $d = -6$,the numbers are $(7 - (-6)), 7, (7 + (-6))$,which are $13, 7, 1$.
Thus,the numbers are $1, 7, 13$ or $13, 7, 1$.

(A) Given: First term $a = 10$,$n^{th}$ term $a_n = 40$,and sum of $n$ terms $S_n = 150$.
We know the formula for the sum of an $A.P.$ is $S_n = \frac{n}{2}(a + a_n)$.
Substituting the given values: $150 = \frac{n}{2}(10 + 40)$.
$150 = \frac{n}{2}(50)$.
$150 = 25n$.
$n = \frac{150}{25} = 6$.
Now,we use the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.
Substituting the values: $40 = 10 + (6 - 1)d$.
$40 - 10 = 5d$.
$30 = 5d$.
$d = \frac{30}{5} = 6$.
Thus,$n = 6$ and $d = 6$.

(D) Let the present ages of the four persons be $x_1, x_2, x_3,$ and $x_4$.
Five years ago,their ages were $(x_1 - 5), (x_2 - 5), (x_3 - 5),$ and $(x_4 - 5)$.
According to the problem,the sum of their ages five years ago was $60$ years:
$(x_1 - 5) + (x_2 - 5) + (x_3 - 5) + (x_4 - 5) = 60$
$(x_1 + x_2 + x_3 + x_4) - 20 = 60$
$x_1 + x_2 + x_3 + x_4 = 60 + 20$
$x_1 + x_2 + x_3 + x_4 = 80$
Thus,the sum of their present ages is $80$ years.

(D) Let the present ages of the five friends be $a_1, a_2, a_3, a_4,$ and $a_5$.
According to the problem,$y$ years ago,the sum of their ages was $x$.
So,$(a_1 - y) + (a_2 - y) + (a_3 - y) + (a_4 - y) + (a_5 - y) = x$.
This simplifies to $(a_1 + a_2 + a_3 + a_4 + a_5) - 5y = x$.
The sum of their present ages is $(a_1 + a_2 + a_3 + a_4 + a_5) = x + 5y$.