Mix Examples - Arithmetic Progressions Questions in English

(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 $6440$.

(10N-8) Given the sum of $n$ terms: $S_{n} = 5n^{2} - 3n$.
We know that the $n^{th}$ term $T_{n}$ is given by $T_{n} = S_{n} - S_{n-1}$ for $n > 1$.
First,find $S_{n-1}$:
$S_{n-1} = 5(n-1)^{2} - 3(n-1)$
$S_{n-1} = 5(n^{2} - 2n + 1) - 3n + 3$
$S_{n-1} = 5n^{2} - 10n + 5 - 3n + 3$
$S_{n-1} = 5n^{2} - 13n + 8$.
Now,calculate $T_{n}$:
$T_{n} = (5n^{2} - 3n) - (5n^{2} - 13n + 8)$
$T_{n} = 5n^{2} - 3n - 5n^{2} + 13n - 8$
$T_{n} = 10n - 8$.
For $n = 1$,$T_{1} = S_{1} = 5(1)^{2} - 3(1) = 2$.
Using the formula $T_{n} = 10n - 8$,for $n = 1$,$T_{1} = 10(1) - 8 = 2$.
Since both match,the $n^{th}$ term is $T_{n} = 10n - 8$.

(A) The multiples of $3$ lying between $250$ and $1000$ form an arithmetic progression ($A$.$P$.).
The first multiple of $3$ after $250$ is $252$,and the last multiple of $3$ before $1000$ is $999$.
So,the $A$.$P$. is $252, 255, 258, ..., 999$.
For this $A$.$P$.,the first term $a = 252$,the common difference $d = 3$,and the last term $l = 999$.
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.
$999 = 252 + (n - 1)3$
$999 - 252 = (n - 1)3$
$747 = (n - 1)3$
$n - 1 = 249$
$n = 250$.
Now,the sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.
$S_{250} = \frac{250}{2}(252 + 999)$
$S_{250} = 125 \times 1251$
$S_{250} = 1,56,375$.
Thus,the sum of the multiples of $3$ lying between $250$ and $1000$ is $1,56,375$.

(A) Let the three numbers in $A.P.$ be $(a-d)$,$a$,and $(a+d)$.
According to the first condition:
$(a-d) + a + (a+d) = 48$
$3a = 48$
$a = 16$
According to the second condition:
$(a-d)^2 + a^2 + (a+d)^2 = 800$
$(a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 800$
$3a^2 + 2d^2 = 800$
Substituting $a = 16$:
$3(16)^2 + 2d^2 = 800$
$3(256) + 2d^2 = 800$
$768 + 2d^2 = 800$
$2d^2 = 32$
$d^2 = 16$
$d = \pm 4$
Case $1$: If $a = 16$ and $d = 4$,the numbers are $(16-4), 16, (16+4)$,which are $12, 16, 20$.
Case $2$: If $a = 16$ and $d = -4$,the numbers are $(16-(-4)), 16, (16+(-4))$,which are $20, 16, 12$.
Thus,the required numbers are $12, 16, 20$ or $20, 16, 12$.

(C) For the given $A.P.$,the first term $a = 5$.
The common difference $d = 8 - 5 = 3$ and the sum of $n$ terms $S_n = 670$.
The formula for the sum of $n$ terms is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $670 = \frac{n}{2} [2(5) + (n - 1)3]$.
$1340 = n [10 + 3n - 3]$.
$1340 = n [3n + 7]$.
$3n^2 + 7n - 1340 = 0$.
Factoring the quadratic equation: $3n^2 + 67n - 60n - 1340 = 0$.
$n(3n + 67) - 20(3n + 67) = 0$.
$(3n + 67)(n - 20) = 0$.
Thus,$n = -\frac{67}{3}$ or $n = 20$.
Since the number of terms $n$ must be a positive integer $(n \in N)$,$n = 20$ is the only valid solution.

(A) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 4n + 7$.
To find the first term $a$,substitute $n = 1$:
$a = T_{1} = 4(1) + 7 = 11$.
To find the second term,substitute $n = 2$:
$T_{2} = 4(2) + 7 = 15$.
The common difference $d$ is given by $d = T_{2} - T_{1} = 15 - 11 = 4$.
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 $a = 11$ and $d = 4$:
$S_{n} = \frac{n}{2}[2(11) + (n - 1)4]$
$S_{n} = \frac{n}{2}[22 + 4n - 4]$
$S_{n} = \frac{n}{2}[4n + 18]$
$S_{n} = n[2n + 9]$
$S_{n} = 2n^{2} + 9n$.
Thus,the sum of $n$ terms is $2n^{2} + 9n$.

(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 $10, 8, 6, 4, 2$.

(C) The given $A.P.$ is $15, 20, 25, \ldots$
Here,the first term $a = 15$ and the common difference $d = 20 - 15 = 5$.
The number of terms $n = 30$.
The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $S_{30} = \frac{30}{2} [2(15) + (30 - 1)5]$.
$S_{30} = 15 [30 + (29 \times 5)]$.
$S_{30} = 15 [30 + 145]$.
$S_{30} = 15 \times 175$.
$S_{30} = 2625$.

(D) The given $A.P.$ is $30, 26, 22, \ldots$
Here,the first term $a = 30$ and the common difference $d = 26 - 30 = -4$.
The number of terms $n = 25$.
The formula for the sum of the first $n$ terms of an $A.P.$ is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $S_{25} = \frac{25}{2} [2(30) + (25 - 1)(-4)]$.
$S_{25} = \frac{25}{2} [60 + (24)(-4)]$.
$S_{25} = \frac{25}{2} [60 - 96]$.
$S_{25} = \frac{25}{2} [-36]$.
$S_{25} = 25 \times (-18) = -450$.
Thus,the sum of the first $25$ terms is $-450$.

(A) The formula for the sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Given values are $a = 11$,$d = 7$,and $n = 40$.
Substituting these values into the formula:
$S_{40} = \frac{40}{2} [2(11) + (40 - 1)7]$
$S_{40} = 20 [22 + (39 \times 7)]$
$S_{40} = 20 [22 + 273]$
$S_{40} = 20 [295]$
$S_{40} = 5900$.
Therefore,the sum of the first $40$ terms is $5900$.

(B) The two-digit natural numbers divisible by $6$ are $12, 18, 24, \dots, 96$.
This forms an arithmetic progression where the first term $a = 12$,the common difference $d = 6$,and the last term $l = 96$.
To find the number of terms $n$,we use the formula $a_n = a + (n - 1)d$.
$96 = 12 + (n - 1)6$
$84 = (n - 1)6$
$14 = n - 1$
$n = 15$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.
$S_{15} = \frac{15}{2}(12 + 96)$
$S_{15} = \frac{15}{2}(108)$
$S_{15} = 15 \times 54 = 810$.

(C) The given $A.P.$ is $8, 15, 22, \ldots$ where the first term $a = 8$ and the common difference $d = 15 - 8 = 7$.
Let the number of terms be $n$. The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Given $S_n = 1490$,we have $1490 = \frac{n}{2} [2(8) + (n - 1)7]$.
$2980 = n [16 + 7n - 7] = n [7n + 9]$.
$7n^2 + 9n - 2980 = 0$.
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $n = \frac{-9 \pm \sqrt{81 - 4(7)(-2980)}}{14} = \frac{-9 \pm \sqrt{81 + 83440}}{14} = \frac{-9 \pm \sqrt{83521}}{14}$.
Since $\sqrt{83521} = 289$,$n = \frac{-9 + 289}{14} = \frac{280}{14} = 20$.
Thus,$20$ terms are required.

(D) The given series is an Arithmetic Progression $(AP)$ where the first term $a = 3$ and the common difference $d = 7 - 3 = 4$.
The last term $l$ (or $a_n$) is $119$.
The formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n - 1)d$.
Substituting the values: $119 = 3 + (n - 1)4$.
$116 = (n - 1)4$.
$n - 1 = 29$,so $n = 30$.
The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$.
$S_{30} = \frac{30}{2}(3 + 119) = 15 \times 122 = 1830$.

(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 $a_5 = 20$,we have $a + 4d = 20$ (Equation $1$).
Given $a_{10} = 35$,we have $a + 9d = 35$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 9d) - (a + 4d) = 35 - 20$,which gives $5d = 15$,so $d = 3$.
Substituting $d = 3$ into Equation $1$: $a + 4(3) = 20$,so $a + 12 = 20$,which gives $a = 8$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.
For $n = 20$: $S_{20} = \frac{20}{2} [2(8) + (20-1)3]$.
$S_{20} = 10 [16 + 19 \times 3] = 10 [16 + 57] = 10 [73] = 730$.

(B) Given $A.P.$ is $31, 36, 41, \ldots$
Here,the first term $a = 31$ and the common difference $d = 36 - 31 = 5$.
Let the sum of $n$ terms be $S_n = 535$.
The formula for the sum of $n$ terms of an $A.P.$ is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $535 = \frac{n}{2} [2(31) + (n - 1)5]$.
$1070 = n [62 + 5n - 5]$.
$1070 = n [57 + 5n]$.
$5n^2 + 57n - 1070 = 0$.
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{-57 \pm \sqrt{57^2 - 4(5)(-1070)}}{2(5)}$.
$n = \frac{-57 \pm \sqrt{3249 + 21400}}{10}$.
$n = \frac{-57 \pm \sqrt{24649}}{10}$.
$n = \frac{-57 \pm 157}{10}$.
Since $n$ must be positive,$n = \frac{100}{10} = 10$.
Thus,the sum of $10$ terms is $535$.

(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 $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
$S_{14} = \frac{14}{2}[2(85) + (14 - 1)(-7)]$.
$S_{14} = 7[170 + 13(-7)]$.
$S_{14} = 7[170 - 91]$.
$S_{14} = 7[79] = 553$.

(D) 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 $a_5 = 30$,so $a + 4d = 30$ (Equation $1$).
Given $a_{12} = 65$,so $a + 11d = 65$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 11d) - (a + 4d) = 65 - 30$.
$7d = 35$,which gives $d = 5$.
Substituting $d = 5$ into Equation $1$: $a + 4(5) = 30 \implies a + 20 = 30 \implies a = 10$.
The sum of the first $n$ terms is $S_n = \frac{n}{2} [2a + (n-1)d]$.
For $n = 30$: $S_{30} = \frac{30}{2} [2(10) + (30-1)5]$.
$S_{30} = 15 [20 + 29 \times 5] = 15 [20 + 145] = 15 \times 165 = 2475$.

(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 $S_{10} = \frac{10}{2}[2(9) + (10-1)(-2)]$.
$S_{10} = 5[18 + 9(-2)] = 5[18 - 18] = 5(0) = 0$.

(A) The $n^{th}$ term of an $A.P.$ is given by the formula $T_{n} = S_{n} - S_{n-1}$ for $n > 1$.
Given $S_{n} = 7n^{2} - 3n$.
Then $S_{n-1} = 7(n-1)^{2} - 3(n-1) = 7(n^{2} - 2n + 1) - 3n + 3 = 7n^{2} - 14n + 7 - 3n + 3 = 7n^{2} - 17n + 10$.
Now,$T_{n} = (7n^{2} - 3n) - (7n^{2} - 17n + 10) = 7n^{2} - 3n - 7n^{2} + 17n - 10 = 14n - 10$.
For $n = 1$,$T_{1} = S_{1} = 7(1)^{2} - 3(1) = 7 - 3 = 4$.
Checking the formula $14n - 10$ for $n = 1$,we get $14(1) - 10 = 4$,which matches $T_{1}$.
Thus,the $n^{th}$ term is $T_{n} = 14n - 10$ for all $n \geq 1$.

(N/A) The $n^{th}$ term of an $A.P.$ is given by the formula $a_{n} = S_{n} - S_{n-1}$ for $n > 1$.
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,$a_{n} = (3n^{2} + 5n) - (3n^{2} - n - 2) = 3n^{2} + 5n - 3n^{2} + n + 2 = 6n + 2$.
For $n = 1$,$a_{1} = S_{1} = 3(1)^{2} + 5(1) = 8$.
Thus,the $n^{th}$ term is $a_{n} = 6n + 2$ for $n \geq 1$.

(A) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 8n + 3$.
To find the sum of the first $n$ terms $(S_{n})$,we use the formula $S_{n} = \frac{n}{2} [a + l]$,where $a$ is the first term and $l$ is the $n^{th}$ term.
First term $(a)$ = $T_{1} = 8(1) + 3 = 11$.
$n^{th}$ term $(l)$ = $T_{n} = 8n + 3$.
Substituting these values into the sum formula:
$S_{n} = \frac{n}{2} [11 + (8n + 3)]$
$S_{n} = \frac{n}{2} [8n + 14]$
$S_{n} = n [4n + 7]$
$S_{n} = 4n^{2} + 7n$.

(A) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 10 - 6n$.
To find the first term $a$,substitute $n = 1$:
$a = T_{1} = 10 - 6(1) = 4$.
To find the second term $T_{2}$,substitute $n = 2$:
$T_{2} = 10 - 6(2) = 10 - 12 = -2$.
The common difference $d = T_{2} - T_{1} = -2 - 4 = -6$.
The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_{n} = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values of $a$ and $d$:
$S_{n} = \frac{n}{2} [2(4) + (n - 1)(-6)]$
$S_{n} = \frac{n}{2} [8 - 6n + 6]$
$S_{n} = \frac{n}{2} [14 - 6n]$
$S_{n} = n(7 - 3n) = 7n - 3n^{2}$.
Thus,$S_{n} = -3n^{2} + 7n$.

(B) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
Given,the $2^{nd}$ term $a_2 = a + d = 2$ --- $(1)$.
The $7^{th}$ term $a_7 = a + 6d = 22$ --- $(2)$.
Subtracting equation $(1)$ from equation $(2)$:
$(a + 6d) - (a + d) = 22 - 2$
$5d = 20$
$d = 4$.
Substituting $d = 4$ in equation $(1)$:
$a + 4 = 2$
$a = -2$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.
For $n = 30$:
$S_{30} = \frac{30}{2} [2(-2) + (30 - 1)4]$
$S_{30} = 15 [-4 + 29 \times 4]$
$S_{30} = 15 [-4 + 116]$
$S_{30} = 15 \times 112 = 1680$.
Thus,the sum of the first $30$ terms is $1680$.

(C) The $n^{th}$ term of an $A.P.$ is given by the formula $a_{n} = S_{n} - S_{n-1}$.
Given $S_{n} = \frac{3n^{2}}{2} + \frac{5n}{2}$.
To find the $25^{th}$ term $(a_{25})$,we use $a_{25} = S_{25} - S_{24}$.
First,calculate $S_{25} = \frac{3(25)^{2}}{2} + \frac{5(25)}{2} = \frac{3(625) + 125}{2} = \frac{1875 + 125}{2} = \frac{2000}{2} = 1000$.
Next,calculate $S_{24} = \frac{3(24)^{2}}{2} + \frac{5(24)}{2} = \frac{3(576) + 120}{2} = \frac{1728 + 120}{2} = \frac{1848}{2} = 924$.
Therefore,$a_{25} = 1000 - 924 = 76$.

(N/A) Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Similarly,$S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d] = 2an + 2n^2d - nd$.
And $S_{3n} = \frac{3n}{2}[2a + (3n-1)d] = 3an + \frac{3n(3n-1)d}{2}$.
Now,consider the expression $3(S_{2n} - S_n)$:
$S_{2n} - S_n = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$
$= \frac{n}{2} [2(2a + 2nd - d) - (2a + nd - d)]$
$= \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d]$
$= \frac{n}{2} [2a + 3nd - d] = \frac{n}{2} [2a + (3n-1)d]$.
Multiplying this by $3$,we get $3(S_{2n} - S_n) = 3 \times \frac{n}{2} [2a + (3n-1)d] = \frac{3n}{2} [2a + (3n-1)d] = S_{3n}$.
Hence,$S_{3n} = 3(S_{2n} - S_n)$ is proved.

(D) Let the first terms of the two $A.P.s$ be $a_1$ and $a_2$,and their common differences be $d_1$ and $d_2$ respectively.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Given,$\frac{S_{n1}}{S_{n2}} = \frac{\frac{n}{2} [2a_1 + (n - 1)d_1]}{\frac{n}{2} [2a_2 + (n - 1)d_2]} = \frac{7n + 1}{4n + 27}$.
This simplifies to $\frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{7n + 1}{4n + 27}$.
We need to find the ratio of the $m^{th}$ terms: $\frac{a_m}{a'_m} = \frac{a_1 + (m - 1)d_1}{a_2 + (m - 1)d_2}$.
Multiply the numerator and denominator of the required ratio by $2$: $\frac{2a_1 + 2(m - 1)d_1}{2a_2 + 2(m - 1)d_2}$.
Comparing this with the given sum ratio,we set $n - 1 = 2(m - 1)$,which gives $n = 2m - 1$.
Substitute $n = 2m - 1$ into the ratio $\frac{7n + 1}{4n + 27}$:
Ratio $= \frac{7(2m - 1) + 1}{4(2m - 1) + 27} = \frac{14m - 7 + 1}{8m - 4 + 27} = \frac{14m - 6}{8m + 23}$.
Thus,the ratio of the $m^{th}$ terms is $(14m - 6) : (8m + 23)$.

(A) Let the four prizes be in an Arithmetic Progression $(AP)$.
Let the first prize be $a$.
Since each successive prize reduces by $Rs. 20$,the common difference $d = -20$.
The number of prizes $n = 4$.
The sum of the prizes $S_n = 280$.
The formula for the sum of $n$ terms of an $AP$ is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $280 = \frac{4}{2} [2a + (4 - 1)(-20)]$.
$280 = 2 [2a + 3(-20)]$.
$140 = 2a - 60$.
$2a = 200$,so $a = 100$.
The four prizes are $a, a+d, a+2d, a+3d$.
Substituting $a = 100$ and $d = -20$: $100, 100-20, 100-40, 100-60$.
Thus,the prize amounts are $Rs. 100, Rs. 80, Rs. 60, Rs. 40$.

(C) The given $A.P.$ is $50, 46, 42, \ldots$
Here,the first term $a = 50$ and the common difference $d = 46 - 50 = -4$.
The number of terms $n = 20$.
The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $S_{20} = \frac{20}{2} [2(50) + (20 - 1)(-4)]$.
$S_{20} = 10 [100 + 19(-4)]$.
$S_{20} = 10 [100 - 76]$.
$S_{20} = 10 [24] = 240$.
Therefore,the sum of the first $20$ terms is $240$.

(D) The given $A.P.$ is $3, \frac{9}{2}, 6, \frac{15}{2}, \ldots$
Here,the first term $a = 3$.
The common difference $d = \frac{9}{2} - 3 = \frac{9-6}{2} = \frac{3}{2}$.
The number of terms $n = 25$.
The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2} [2a + (n-1)d]$.
Substituting the values: $S_{25} = \frac{25}{2} [2(3) + (25-1)(\frac{3}{2})]$.
$S_{25} = \frac{25}{2} [6 + 24(\frac{3}{2})]$.
$S_{25} = \frac{25}{2} [6 + 12(3)]$.
$S_{25} = \frac{25}{2} [6 + 36] = \frac{25}{2} [42]$.
$S_{25} = 25 \times 21 = 525$.

(A) The given $A.P.$ is $5, 2, -1, \ldots$
Here,the first term $a = 5$.
The common difference $d = 2 - 5 = -3$.
The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values of $a$ and $d$:
$S_n = \frac{n}{2}[2(5) + (n - 1)(-3)]$
$S_n = \frac{n}{2}[10 - 3n + 3]$
$S_n = \frac{n}{2}[13 - 3n]$
$S_n = \frac{13n}{2} - \frac{3n^2}{2}$
Thus,$S_n = -\frac{3}{2}n^2 + \frac{13}{2}n$.

(N/A) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 5 - 6n$.
To find the first term $(a)$,substitute $n = 1$:
$a = T_{1} = 5 - 6(1) = -1$.
To find the second term $(T_{2})$,substitute $n = 2$:
$T_{2} = 5 - 6(2) = 5 - 12 = -7$.
The common difference $(d)$ is $T_{2} - T_{1} = -7 - (-1) = -6$.
The sum of the first $n$ terms $(S_{n})$ is given by the formula $S_{n} = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values of $a = -1$ and $d = -6$:
$S_{n} = \frac{n}{2} [2(-1) + (n - 1)(-6)]$
$S_{n} = \frac{n}{2} [-2 - 6n + 6]$
$S_{n} = \frac{n}{2} [4 - 6n]$
$S_{n} = n(2 - 3n) = -3n^{2} + 2n$.

(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 $6973$.

(D) Given: First term $a = 2$,last term $l = 50$,and sum $S_n = 442$.
The formula for the sum of an $A.P.$ is $S_n = \frac{n}{2}(a + l)$.
Substituting the values: $442 = \frac{n}{2}(2 + 50)$.
$442 = \frac{n}{2}(52) \implies 442 = 26n$.
$n = \frac{442}{26} = 17$.
Now,use the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.
$50 = 2 + (17 - 1)d$.
$50 - 2 = 16d$.
$48 = 16d$.
$d = \frac{48}{16} = 3$.
Therefore,the common difference $d$ is $3$.

(A) The multiples of $5$ between $84$ and $719$ form an Arithmetic Progression $(AP)$.
The first multiple of $5$ after $84$ is $a = 85$.
The last multiple of $5$ before $719$ is $l = 715$.
The common difference is $d = 5$.
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.
$715 = 85 + (n - 1)5$.
$630 = (n - 1)5$.
$n - 1 = 126$.
$n = 127$.
The sum of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$.
$S_{127} = \frac{127}{2}(85 + 715)$.
$S_{127} = \frac{127}{2}(800)$.
$S_{127} = 127 \times 400 = 50,800$.

(B) Let the first term be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
For the $5^{th}$ term: $a + 4d = 30$ --- $(1)$
For the $12^{th}$ term: $a + 11d = 100$ --- $(2)$
Subtracting equation $(1)$ from $(2)$:
$(a + 11d) - (a + 4d) = 100 - 30$
$7d = 70$
$d = 10$
Substituting $d = 10$ in equation $(1)$:
$a + 4(10) = 30$
$a + 40 = 30$
$a = -10$
The sum of the first $n$ terms is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
For $n = 20$:
$S_{20} = \frac{20}{2} [2(-10) + (20 - 1)10]$
$S_{20} = 10 [-20 + 190]$
$S_{20} = 10 [170]$
$S_{20} = 1700$.

(C) The given series is $7, 10.5, 14, \ldots, 84$.
This is an Arithmetic Progression $(AP)$ where the first term $a = 7$ and the common difference $d = 10.5 - 7 = 3.5 = \frac{7}{2}$.
Let the number of terms be $n$. The $n^{th}$ term is given by $a_n = a + (n - 1)d$.
Substituting the values: $84 = 7 + (n - 1) \times \frac{7}{2}$.
$84 - 7 = (n - 1) \times \frac{7}{2} \implies 77 = (n - 1) \times \frac{7}{2}$.
$(n - 1) = 77 \times \frac{2}{7} = 11 \times 2 = 22$.
$n = 22 + 1 = 23$.
The sum of $n$ terms of an $AP$ is $S_n = \frac{n}{2} (a + a_n)$.
$S_{23} = \frac{23}{2} (7 + 84) = \frac{23}{2} \times 91 = \frac{2093}{2} = 1046.5 = 1046 \frac{1}{2}$.

(D) Given the $n^{th}$ term of the $A.P.$ is $T_n = 4n + 3$.
To find the first term $(a)$: Substitute $n = 1$,$a = T_1 = 4(1) + 3 = 7$.
To find the $60^{th}$ term $(l)$: Substitute $n = 60$,$l = T_{60} = 4(60) + 3 = 240 + 3 = 243$.
The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$.
For $n = 60$,$S_{60} = \frac{60}{2}(7 + 243)$.
$S_{60} = 30(250) = 7500$.
Therefore,the sum of the first $60$ terms is $7500$.

(A) Let the digits at the hundreds place,tens place,and units place of the original number be $a-d$,$a$,and $a+d$ respectively.
$\therefore$ The original number $= 100(a-d) + 10(a) + (a+d) = 111a - 99d$.
Given that the sum of the digits is $15$:
$(a-d) + a + (a+d) = 15$
$3a = 15$
$a = 5$.
When the digits are reversed,the new number is:
$100(a+d) + 10(a) + (a-d) = 111a + 99d$.
Given that the new number exceeds the original number by $594$:
$(111a + 99d) - (111a - 99d) = 594$
$198d = 594$
$d = 3$.
Therefore,the digits are:
Hundreds place: $a-d = 5-3 = 2$
Tens place: $a = 5$
Units place: $a+d = 5+3 = 8$
The original number is $258$.

(B) Let the required roll number be $n$.
According to the problem,the sum of all roll numbers smaller than $n$ is equal to the sum of all roll numbers greater than $n$.
This can be written as: $1 + 2 + 3 + \ldots + (n - 1) = (n + 1) + (n + 2) + \ldots + 49$.
To solve this,we add the sum of numbers from $1$ to $n$ to both sides:
$1 + 2 + \ldots + (n - 1) + n = (n + 1) + (n + 2) + \ldots + 49 + n$.
The left side becomes the sum of the first $n$ natural numbers: $\frac{n(n + 1)}{2}$.
The right side becomes the sum of the first $49$ natural numbers: $\frac{49(50)}{2}$.
Thus,$2 \times \frac{n(n + 1)}{2} = \frac{49 \times 50}{2} + \frac{n(n + 1)}{2}$ is not the correct approach; rather,we use the property: $\sum_{i=1}^{n-1} i = \sum_{i=n+1}^{49} i$.
We know $\sum_{i=1}^{n-1} i = \frac{(n-1)n}{2}$.
The sum $\sum_{i=n+1}^{49} i = \sum_{i=1}^{49} i - \sum_{i=1}^{n} i = \frac{49 \times 50}{2} - \frac{n(n+1)}{2}$.
Equating the two: $\frac{n^2 - n}{2} = \frac{2450}{2} - \frac{n^2 + n}{2}$.
$n^2 - n = 2450 - n^2 - n$.
$2n^2 = 2450$.
$n^2 = 1225$.
$n = \sqrt{1225} = 35$.
Therefore,the required roll number is $35$.

(C) Let the first terms of the two $A.P.s$ be $a_1$ and $a_2$,and their common differences be $d_1$ and $d_2$ respectively.
The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given the ratio of the sums of $n$ terms:
$\frac{S_{n,1}}{S_{n,2}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{4n+3}{5n-7}$
$\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{4n+3}{5n-7}$
We need to find the ratio of the $15^{th}$ terms,which is $\frac{a_1 + 14d_1}{a_2 + 14d_2}$.
To get $14d$ in the expression,we set $\frac{n-1}{2} = 14$,which gives $n-1 = 28$,or $n = 29$.
Substituting $n = 29$ in the ratio:
$\frac{2a_1 + 28d_1}{2a_2 + 28d_2} = \frac{4(29)+3}{5(29)-7}$
$\frac{2(a_1 + 14d_1)}{2(a_2 + 14d_2)} = \frac{116+3}{145-7}$
$\frac{a_1 + 14d_1}{a_2 + 14d_2} = \frac{119}{138}$
Thus,the ratio of the $15^{th}$ terms is $\frac{119}{138}$.

(D) Since $2x$,$x+10$,and $3x+2$ are three consecutive terms of an $A.P.$,the common difference between consecutive terms must be equal.
Therefore,$(x+10) - 2x = (3x+2) - (x+10)$.
Simplifying both sides:
$-x + 10 = 2x - 8$.
Rearranging the terms to solve for $x$:
$10 + 8 = 2x + x$.
$18 = 3x$.
$x = \frac{18}{3} = 6$.
Thus,the value of $x$ is $6$.

(A) For the $A.P.$ $54, 51, 48, \dots$,the first term $a = 54$,the common difference $d = 51 - 54 = -3$,and the sum of $n$ terms $S_{n} = 513$.
Using the formula $S_{n} = \frac{n}{2}[2a + (n - 1)d]$:
$513 = \frac{n}{2}[2(54) + (n - 1)(-3)]$
$1026 = n[108 - 3n + 3]$
$1026 = n[111 - 3n]$
$1026 = 111n - 3n^{2}$
$3n^{2} - 111n + 1026 = 0$
Dividing by $3$:
$n^{2} - 37n + 342 = 0$
Factoring the quadratic equation:
$(n - 18)(n - 19) = 0$
Thus,$n = 18$ or $n = 19$.
Reason for two answers: The $19^{th}$ term of the $A.P.$ is $T_{19} = a + 18d = 54 + 18(-3) = 54 - 54 = 0$. Since the $19^{th}$ term is $0$,adding it to the sum of the first $18$ terms does not change the total sum. Therefore,both $S_{18}$ and $S_{19}$ equal $513$.

(B) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
$T_n = a + (n - 1)d$
Given $T_3 = 7$,so $a + 2d = 7$ --- $(1)$
Given $T_7 = 3(T_3) + 2$,so $a + 6d = 3(7) + 2 = 23$ --- $(2)$
Subtracting equation $(1)$ from $(2)$:
$(a + 6d) - (a + 2d) = 23 - 7$
$4d = 16$
$d = 4$
Substituting $d = 4$ in equation $(1)$:
$a + 2(4) = 7$
$a + 8 = 7$
$a = -1$
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 20$:
$S_{20} = \frac{20}{2}[2(-1) + (20 - 1)4]$
$S_{20} = 10[-2 + 19(4)]$
$S_{20} = 10[-2 + 76]$
$S_{20} = 10[74] = 740$.
Thus,the sum of the first $20$ terms is $740$.

(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 $(S_{7})$:
$S_{n} = \frac{n}{2} [2a + (n - 1)d]$
$S_{7} = \frac{7}{2} [2(550) + (7 - 1)(25)] = \frac{7}{2} [1100 + 150] = \frac{7}{2} [1250] = 7 \times 625 = 4375 \, TV$.

(D) The amounts paid in monthly instalments form an Arithmetic Progression ($A$.$P$.).
First term $a = 20$.
Common difference $d = 15$.
Total debt $S_n = 3250$.
The sum of $n$ terms of an $A$.$P$. is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values: $3250 = \frac{n}{2}[2(20) + (n-1)15]$.
$6500 = n[40 + 15n - 15]$.
$6500 = n[25 + 15n]$.
$15n^2 + 25n - 6500 = 0$.
Dividing by $5$: $3n^2 + 5n - 1300 = 0$.
Factoring the quadratic equation: $3n^2 + 65n - 60n - 1300 = 0$.
$n(3n + 65) - 20(3n + 65) = 0$.
$(n - 20)(3n + 65) = 0$.
Since $n$ must be a positive integer,$n = 20$.
Thus,it will take $20$ months to clear the debt.

(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 $0$.

(B) The given $A.P.$ is $53, 48, 43, \ldots$
Here,the first term $a = 53$ and the common difference $d = 48 - 53 = -5$.
Let the $n^{th}$ term be the first negative term,so $a_n < 0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $53 + (n - 1)(-5) < 0$.
$53 - 5n + 5 < 0$.
$58 - 5n < 0$.
$58 < 5n$.
$n > 11.6$.
Since $n$ must be an integer,the first integer greater than $11.6$ is $n = 12$.
Now,find the $12^{th}$ term: $a_{12} = 53 + (12 - 1)(-5) = 53 + 11(-5) = 53 - 55 = -2$.
Thus,the $12^{th}$ term is the first negative term,which is $-2$.

(C) The given $A.P.$ is $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots$
Here,the first term $a = \sqrt{2}$.
The common difference $d = 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2}$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
For the $10^{th}$ term $(n = 10)$:
$a_{10} = \sqrt{2} + (10 - 1)(2 \sqrt{2})$
$a_{10} = \sqrt{2} + 9(2 \sqrt{2})$
$a_{10} = \sqrt{2} + 18 \sqrt{2}$
$a_{10} = 19 \sqrt{2}$.

(D) The given $A.P.$ is $-40, -15, 10, \ldots$
Here,the first term $a = -40$.
The common difference $d = -15 - (-40) = -15 + 40 = 25$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
To find the $20^{th}$ term $(n = 20)$:
$a_{20} = -40 + (20 - 1) \times 25$
$a_{20} = -40 + 19 \times 25$
$a_{20} = -40 + 475$
$a_{20} = 435$.
Therefore,the $20^{th}$ term is $435$.