Mix Examples - Arithmetic Progressions Questions in English

(A) Given $A.P.$ is $3, 8, 13, \ldots, 248$.
Here,the first term $a = 3$ and the common difference $d = 8 - 3 = 5$.
Let the $n^{th}$ term be $a_n = 248$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $248 = 3 + (n - 1)5$.
$248 - 3 = (n - 1)5$.
$245 = (n - 1)5$.
$n - 1 = 245 / 5 = 49$.
$n = 49 + 1 = 50$.
Therefore,the $50^{th}$ term of the $A.P.$ is $248$.

(B) The $n^{th}$ term of an $A.P.$ is given by the formula $a_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.
Given,the $6^{th}$ term $a_6 = 19$,so $a + 5d = 19$ (Equation $1$).
Given,the $17^{th}$ term $a_{17} = 41$,so $a + 16d = 41$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(a + 16d) - (a + 5d) = 41 - 19$
$11d = 22$
$d = 2$
Substituting $d = 2$ into Equation $1$:
$a + 5(2) = 19$
$a + 10 = 19$
$a = 9$
Now,find the $40^{th}$ term $a_{40}$:
$a_{40} = a + (40 - 1)d$
$a_{40} = 9 + 39(2)$
$a_{40} = 9 + 78$
$a_{40} = 87$.

(N/A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
According to the problem,$10 \times a_{10} = 15 \times a_{15}$.
Substituting the formula for the terms: $10(a + 9d) = 15(a + 14d)$.
Divide both sides by $5$: $2(a + 9d) = 3(a + 14d)$.
Expand the brackets: $2a + 18d = 3a + 42d$.
Rearrange the terms: $2a - 3a = 42d - 18d$.
$-a = 24d$,which implies $a = -24d$.
Now,we need to find the $25^{th}$ term,$a_{25} = a + (25 - 1)d = a + 24d$.
Substitute $a = -24d$ into the expression: $a_{25} = -24d + 24d = 0$.
Thus,the $25^{th}$ term of the $A.P.$ is $0$.

(D) The given sequence is an $A.P.$ with the first term $a = 3$ and common difference $d = 5 - 3 = 2$.
Let there be $n$ terms in this $A.P.$ The last term $l = 201$.
Using the formula $l = a + (n - 1)d$:
$201 = 3 + (n - 1)2$
$198 = (n - 1)2$
$n - 1 = 99$
$n = 100$.
The $k^{th}$ term from the end is given by the formula $(n - k + 1)^{th}$ term from the beginning.
Here,$k = 12$ and $n = 100$.
So,the $12^{th}$ term from the end is the $(100 - 12 + 1)^{th} = 89^{th}$ term from the beginning.
$a_{89} = a + (89 - 1)d$
$a_{89} = 3 + 88 \times 2$
$a_{89} = 3 + 176 = 179$.

(A) The given $A.P.$ is $1, 4, 7, \ldots, 118$.
Here,the first term $a = 1$ and the common difference $d = 4 - 1 = 3$.
The last term $l = 118$.
To find the $n^{th}$ term from the end of an $A.P.$,we use the formula: $a_n (\text{from end}) = l - (n - 1)d$.
Here,$n = 15$,$l = 118$,and $d = 3$.
Substituting the values: $a_{15} = 118 - (15 - 1) \times 3$.
$a_{15} = 118 - (14 \times 3)$.
$a_{15} = 118 - 42$.
$a_{15} = 76$.
Therefore,the $15^{th}$ term from the end is $76$.

(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) Let the three numbers in $A.P.$ be $(a-d)$, $a$, and $(a+d)$.
According to the problem, the sum is $(a-d) + a + (a+d) = -3$.
$3a = -3$, which gives $a = -1$.
The product of the numbers is $(a-d) \cdot a \cdot (a+d) = 8$.
Substituting $a = -1$, we get $(-1-d)(-1)(-1+d) = 8$.
$-1(1-d^2) = 8$, which simplifies to $d^2 - 1 = 8$.
$d^2 = 9$, so $d = \pm 3$.
If $d = 3$, the numbers are $(-1-3), -1, (-1+3)$, which are $-4, -1, 2$.
If $d = -3$, the numbers are $(-1-(-3)), -1, (-1+(-3))$, which are $2, -1, -4$.
Thus, the numbers are $-4, -1, 2$ or $2, -1, -4$.

(2, 6, 10, 14) Let the four numbers in $A.P.$ be $(a-3d), (a-d), (a+d), (a+3d)$.
Given that the sum of the numbers is $32$:
$(a-3d) + (a-d) + (a+d) + (a+3d) = 32$
$4a = 32 \implies a = 8$.
The numbers are $(8-3d), (8-d), (8+d), (8+3d)$.
The product of extremes is $(8-3d)(8+3d) = 64 - 9d^2$.
The product of means is $(8-d)(8+d) = 64 - d^2$.
Given the ratio of the product of extremes to the product of means is $7:15$:
$\frac{64-9d^2}{64-d^2} = \frac{7}{15}$
$15(64-9d^2) = 7(64-d^2)$
$960 - 135d^2 = 448 - 7d^2$
$512 = 128d^2$
$d^2 = 4 \implies d = 2$ (since the numbers are in increasing order,$d > 0$).
Substituting $a=8$ and $d=2$:
$8-3(2) = 2$
$8-2 = 6$
$8+2 = 10$
$8+3(2) = 14$
The numbers are $2, 6, 10, 14$.

(A) For three terms $a, b, c$ to be in an $A.P.$,the condition is $2b = a + c$.
Here,$a = x+1$,$b = 3x$,and $c = 4x+2$.
Substituting these values into the condition:
$2(3x) = (x+1) + (4x+2)$
$6x = 5x + 3$
$6x - 5x = 3$
$x = 3$
Therefore,the value of $x$ is $3$.

(N/A) To prove that the given terms $(a-b)^{2}, (a^{2}+b^{2}),$ and $(a+b)^{2}$ form an $A.P.$,we need to show that the difference between consecutive terms is constant.
Let the terms be $T_1 = (a-b)^2$,$T_2 = (a^2+b^2)$,and $T_3 = (a+b)^2$.
First,calculate the difference between the second and first term $(d_1)$:
$d_1 = T_2 - T_1 = (a^2 + b^2) - (a - b)^2$
$d_1 = (a^2 + b^2) - (a^2 - 2ab + b^2)$
$d_1 = a^2 + b^2 - a^2 + 2ab - b^2 = 2ab$
Next,calculate the difference between the third and second term $(d_2)$:
$d_2 = T_3 - T_2 = (a + b)^2 - (a^2 + b^2)$
$d_2 = (a^2 + 2ab + b^2) - (a^2 + b^2)$
$d_2 = a^2 + 2ab + b^2 - a^2 - b^2 = 2ab$
Since $d_1 = d_2 = 2ab$,the common difference is constant.
Therefore,the given terms form an $A.P.$

(N/A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given:
$S_1 = \frac{n}{2}[2a + (n-1)d]$
$S_2 = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$
$S_3 = \frac{3n}{2}[2a + (3n-1)d]$
Now,consider the expression $3(S_2 - S_1)$:
$S_2 - S_1 = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$
$= \frac{n}{2} [2(2a + 2nd - d) - (2a + nd - d)]$
$= \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d]$
$= \frac{n}{2} [2a + 3nd - d]$
$= \frac{n}{2} [2a + (3n-1)d]$
Multiplying by $3$:
$3(S_2 - S_1) = 3 \times \frac{n}{2} [2a + (3n-1)d] = \frac{3n}{2} [2a + (3n-1)d] = S_3$.
Hence,$S_3 = 3(S_2 - S_1)$ is proved.

(A) 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 the ratio of sums: $\frac{S_{n1}}{S_{n2}} = \frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{5n - 3}{7n + 2}$.
This simplifies to $\frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{5n - 3}{7n + 2}$.
We need the ratio of the $m^{th}$ terms: $\frac{a_m1}{a_m2} = \frac{a_1 + (m - 1)d_1}{a_2 + (m - 1)d_2}$.
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 sum ratio expression,we set $n - 1 = 2(m - 1)$,which gives $n = 2m - 1$.
Substitute $n = 2m - 1$ into the ratio $\frac{5n - 3}{7n + 2}$:
Ratio $= \frac{5(2m - 1) - 3}{7(2m - 1) + 2} = \frac{10m - 5 - 3}{14m - 7 + 2} = \frac{10m - 8}{14m - 5}$.
Thus,the ratio of the $m^{th}$ terms is $(10m - 8) : (14m - 5)$.

(A) The given $A.P.$ is $6, 2, -2, \ldots$
Here,the first term $a = 6$ and the common difference $d = 2 - 6 = -4$.
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 $d$ into the formula:
$S_{n} = \frac{n}{2} [2(6) + (n - 1)(-4)]$
$S_{n} = \frac{n}{2} [12 - 4n + 4]$
$S_{n} = \frac{n}{2} [16 - 4n]$
$S_{n} = n(8 - 2n) = 8n - 2n^{2} = -2n^{2} + 8n$.

(B) Given $A.P.$ is $9, 17, 25, \ldots$
Here,the first term $a = 9$ and common difference $d = 17 - 9 = 8$.
Let the sum of $n$ terms be $S_n = 636$.
The formula for the sum of $n$ terms is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $636 = \frac{n}{2} [2(9) + (n - 1)8]$.
$636 = \frac{n}{2} [18 + 8n - 8]$.
$636 = \frac{n}{2} [10 + 8n]$.
$636 = n(5 + 4n)$.
$4n^2 + 5n - 636 = 0$.
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{-5 \pm \sqrt{25 - 4(4)(-636)}}{8} = \frac{-5 \pm \sqrt{25 + 10176}}{8} = \frac{-5 \pm \sqrt{10201}}{8} = \frac{-5 \pm 101}{8}$.
Since $n$ must be positive,$n = \frac{96}{8} = 12$.
Thus,the sum of $12$ terms is $636$.

(C) The given series is an Arithmetic Progression $(AP)$ where the first term $a = -5$ and the common difference $d = (-8) - (-5) = -3$.
The last term $l$ or $a_n = -230$.
Using the formula for the $n^{th}$ term of an $AP$: $a_n = a + (n - 1)d$.
$-230 = -5 + (n - 1)(-3)$.
$-230 + 5 = (n - 1)(-3)$.
$-225 = (n - 1)(-3)$.
$n - 1 = \frac{-225}{-3} = 75$.
$n = 76$.
Now,using the sum formula $S_n = \frac{n}{2}(a + l)$:
$S_{76} = \frac{76}{2}(-5 + (-230))$.
$S_{76} = 38 \times (-235)$.
$S_{76} = -8930$.

(D) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 7 - 3n$.
To find the first term $(a)$,substitute $n = 1$:
$a = T_{1} = 7 - 3(1) = 4$.
To find the $25^{th}$ term $(T_{25})$,substitute $n = 25$:
$T_{25} = 7 - 3(25) = 7 - 75 = -68$.
The sum of the first $n$ terms of an $A.P.$ is given by $S_{n} = \frac{n}{2}(a + T_{n})$.
For $n = 25$:
$S_{25} = \frac{25}{2}(4 + (-68))$
$S_{25} = \frac{25}{2}(-64)$
$S_{25} = 25 \times (-32) = -800$.
Thus,the sum of the first $25$ terms is $-800$.

(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 $a = -8$.

(C) The sum of the first $n$ terms of an $A.P.$ is given by the formula:
$S_n = \frac{n}{2} [2a + (n-1)d]$
Given values are $a = 3$,$n = 8$,and $S_n = 192$.
Substituting these values into the formula:
$192 = \frac{8}{2} [2(3) + (8-1)d]$
$192 = 4 [6 + 7d]$
Divide both sides by $4$:
$48 = 6 + 7d$
$48 - 6 = 7d$
$42 = 7d$
$d = \frac{42}{7} = 6$
Therefore,the common difference $d$ is $6$.

(D) The formula for the sum of $n$ terms of an $A.P.$ is given by $S_{n} = \frac{n}{2}(a + T_{n})$,where $a$ is the first term and $T_{n}$ is the $n^{th}$ term.
Given values are $S_{n} = 144$,$T_{n} = 28$,and $n = 9$.
Substituting these values into the formula:
$144 = \frac{9}{2}(a + 28)$
Multiply both sides by $2$ and divide by $9$:
$144 \times \frac{2}{9} = a + 28$
$16 \times 2 = a + 28$
$32 = a + 28$
$a = 32 - 28$
$a = 4$
Therefore,the first term $a$ is $4$.

(A) Let the savings of the first year be $a$.
Since the savings increase by $Rs. 100$ every year,this forms an Arithmetic Progression $(AP)$ where the common difference $d = 100$.
The number of years is $n = 10$,and the total savings $S_{10} = 16,500$.
The formula for the sum of $n$ terms of an $AP$ is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $16,500 = \frac{10}{2} [2a + (10 - 1) \times 100]$.
$16,500 = 5 [2a + 900]$.
$3,300 = 2a + 900$.
$2a = 3,300 - 900 = 2,400$.
$a = 1,200$.
Thus,the savings of the first year is $Rs. 1,200$.

(B) The radii of the $14$ discs form an arithmetic progression: $1, 2, 3, \dots, 14$.
The circumference of a disc with radius $r$ is given by $C = 2\pi r$.
The sum of the circumferences of all $14$ discs is $S = 2\pi(1) + 2\pi(2) + 2\pi(3) + \dots + 2\pi(14)$.
$S = 2\pi(1 + 2 + 3 + \dots + 14)$.
Using the sum formula for the first $n$ natural numbers, $S_n = \frac{n(n+1)}{2}$, where $n = 14$:
$S = 2\pi \times \frac{14(14+1)}{2} = 2\pi \times \frac{14 \times 15}{2} = 2\pi \times 105 = 210\pi$.
Taking $\pi \approx \frac{22}{7}$:
$S = 210 \times \frac{22}{7} = 30 \times 22 = 660\, cm$.

(C) Let the required number be $x.$ The sum of natural numbers smaller than $x$ is given by the sum of integers from $1$ to $x-1.$
This sum is $S_1 = \frac{(x-1)x}{2}.$
The sum of natural numbers greater than $x$ up to $288$ is the sum of integers from $x+1$ to $288.$
This sum is $S_2 = \sum_{i=1}^{288} i - \sum_{i=1}^{x} i = \frac{288 \times 289}{2} - \frac{x(x+1)}{2}.$
According to the problem,$S_1 = S_2.$
$\frac{x^2 - x}{2} = \frac{288 \times 289}{2} - \frac{x^2 + x}{2}.$
Multiplying by $2,$ we get $x^2 - x = 288 \times 289 - x^2 - x.$
$2x^2 = 288 \times 289.$
$x^2 = 144 \times 289.$
$x = \sqrt{144 \times 289} = 12 \times 17 = 204.$
Thus,the required number is $204.$

(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) Let the four numbers in $A.P.$ be $(a-3d), (a-d), (a+d), (a+3d)$.
Given that the sum is $50$:
$(a-3d) + (a-d) + (a+d) + (a+3d) = 50$
$4a = 50 \implies a = 12.5$.
The smallest number is $(a-3d)$ and the greatest number is $(a+3d)$.
Given that the greatest number is four times the smallest:
$(a+3d) = 4(a-3d)$
$12.5 + 3d = 4(12.5 - 3d)$
$12.5 + 3d = 50 - 12d$
$15d = 37.5 \implies d = 2.5$.
The numbers are:
$12.5 - 3(2.5) = 12.5 - 7.5 = 5$
$12.5 - 2.5 = 10$
$12.5 + 2.5 = 15$
$12.5 + 3(2.5) = 12.5 + 7.5 = 20$.
Thus,the numbers are $5, 10, 15, 20$ or in reverse order $20, 15, 10, 5$.

(B) The given $A.P.$ is $108, 103, 98, \ldots$
Here,the first term $a = 108$ and the common difference $d = 103 - 108 = -5$.
We want to find the first negative term,so we set the $n^{th}$ term $a_n < 0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $108 + (n - 1)(-5) < 0$.
$108 - 5n + 5 < 0$.
$113 - 5n < 0$.
$113 < 5n$.
$n > \frac{113}{5} = 22.6$.
Since $n$ must be an integer,the smallest integer greater than $22.6$ is $23$.
Therefore,the $23^{rd}$ term is the first negative term.

(C) The given $A.P.$ is $-1, -\frac{5}{6}, -\frac{2}{3}, \ldots, \frac{10}{3}$.
Here,the first term $a = -1$.
The common difference $d = -\frac{5}{6} - (-1) = -\frac{5}{6} + 1 = \frac{1}{6}$.
The last term $a_n = \frac{10}{3}$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $\frac{10}{3} = -1 + (n - 1)\frac{1}{6}$.
Adding $1$ to both sides: $\frac{10}{3} + 1 = (n - 1)\frac{1}{6}$.
$\frac{13}{3} = (n - 1)\frac{1}{6}$.
Multiplying both sides by $6$: $13 \times 2 = n - 1$.
$26 = n - 1$.
$n = 27$.
Thus,the number of terms is $27$.

(D) Given: Number of terms $n = 60$,first term $a = 7$,and last term $a_{60} = 125$.
Using the formula for the $n^{th}$ term of an $A.P.$,$a_n = a + (n - 1)d$.
For the $60^{th}$ term: $125 = 7 + (60 - 1)d$.
$125 - 7 = 59d$.
$118 = 59d$.
$d = 118 / 59 = 2$.
Now,find the $32^{nd}$ term $(a_{32})$:
$a_{32} = a + (32 - 1)d$.
$a_{32} = 7 + 31(2)$.
$a_{32} = 7 + 62 = 69$.
Therefore,the $32^{nd}$ term is $69$.

(A) The given $A.P.$ is $5.5, 11, 16.5, \ldots$
Here,the first term $a = 5.5$ and the common difference $d = 11 - 5.5 = 5.5$.
We need to find $n$ such that the $n^{th}$ term $a_n = 550$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $550 = 5.5 + (n - 1)5.5$.
$550 - 5.5 = (n - 1)5.5$.
$544.5 = (n - 1)5.5$.
$n - 1 = \frac{544.5}{5.5} = 99$.
$n = 99 + 1 = 100$.
Therefore,the $100^{th}$ term of the $A.P.$ is $550$.

(B) The given $A.P.$ is $5, 9, 13, \ldots, 101$.
Here,the first term $a = 5$ and the common difference $d = 9 - 5 = 4$.
The last term $l = 101$.
To find the $n^{th}$ term from the end,the formula is $l - (n - 1)d$.
Here,$n = 9$,$l = 101$,and $d = 4$.
Substituting these values: $101 - (9 - 1) \times 4 = 101 - 8 \times 4 = 101 - 32 = 69$.
Therefore,the $9^{th}$ term from the end is $69$.

(C) To find the $n^{th}$ term from the end of an $A.P.$,we can reverse the sequence or use the formula: $l - (n - 1)d$,where $l$ is the last term,$n$ is the position from the end,and $d$ is the common difference of the original sequence.
Given $A.P.$ is $40, 35, 30, \ldots, -200$.
Here,the first term $a = 40$,common difference $d = 35 - 40 = -5$,and the last term $l = -200$.
We need to find the $10^{th}$ term from the end.
Using the formula: $a_n (\text{from end}) = l - (n - 1)d$.
Substituting the values: $a_{10} = -200 - (10 - 1)(-5)$.
$a_{10} = -200 - (9)(-5)$.
$a_{10} = -200 + 45$.
$a_{10} = -155$.
Thus,the $10^{th}$ term from the end is $-155$.

(A) Let the three numbers in $A.P.$ be $(a-d)$,$a$,and $(a+d)$.
According to the problem,their sum is $(a-d) + a + (a+d) = 12$.
This simplifies to $3a = 12$,so $a = 4$.
The numbers are $(4-d)$,$4$,and $(4+d)$.
The sum of their cubes is $(4-d)^3 + 4^3 + (4+d)^3 = 288$.
Expanding the cubes: $(64 - 48d + 12d^2 - d^3) + 64 + (64 + 48d + 12d^2 + d^3) = 288$.
Combining terms: $192 + 24d^2 = 288$.
$24d^2 = 96$,which gives $d^2 = 4$,so $d = \pm 2$.
If $d = 2$,the numbers are $(4-2), 4, (4+2)$,which are $2, 4, 6$.
If $d = -2$,the numbers are $(4-(-2)), 4, (4+(-2))$,which are $6, 4, 2$.
Thus,the numbers are $2, 4, 6$ or $6, 4, 2$.

(B) Let the four angles of the quadrilateral be $a-3d, a-d, a+d, a+3d$.
Here,the common difference between consecutive terms is $2d = 10^{\circ}$,so $d = 5^{\circ}$.
The sum of the angles of a quadrilateral is $360^{\circ}$.
Therefore,$(a-3d) + (a-d) + (a+d) + (a+3d) = 360^{\circ}$.
$4a = 360^{\circ} \implies a = 90^{\circ}$.
The angles are:
$a-3d = 90^{\circ} - 3(5^{\circ}) = 90^{\circ} - 15^{\circ} = 75^{\circ}$.
$a-d = 90^{\circ} - 5^{\circ} = 85^{\circ}$.
$a+d = 90^{\circ} + 5^{\circ} = 95^{\circ}$.
$a+3d = 90^{\circ} + 3(5^{\circ}) = 90^{\circ} + 15^{\circ} = 105^{\circ}$.
Thus,the angles are $75^{\circ}, 85^{\circ}, 95^{\circ}, 105^{\circ}$.

(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 $d = 6$.

(C) The given Arithmetic Progression $(A.P.)$ is $-2, -4, -6, \ldots$
The common difference $d$ of an $A.P.$ is calculated by subtracting the first term from the second term: $d = a_2 - a_1$.
Here,$a_1 = -2$ and $a_2 = -4$.
Therefore,$d = -4 - (-2) = -4 + 2 = -2$.
Thus,the common difference $d$ is $-2$.

(D) The $n^{th}$ term of the $A.P.$ is given by $T_n = 3n - 1$.
To find the common difference $d$,we calculate the first two terms of the sequence.
For $n = 1$,$T_1 = 3(1) - 1 = 3 - 1 = 2$.
For $n = 2$,$T_2 = 3(2) - 1 = 6 - 1 = 5$.
The common difference $d$ is given by $d = T_2 - T_1$.
Substituting the values,$d = 5 - 2 = 3$.
Alternatively,for an $A.P.$ of the form $T_n = an + b$,the common difference $d$ is equal to the coefficient of $n$,which is $3$.

(A) The formula for the $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
Here,the first term $a = 5$,the common difference $d = 3$,and the term number $n = 15$.
Substituting these values into the formula:
$a_{15} = 5 + (15 - 1) \times 3$
$a_{15} = 5 + 14 \times 3$
$a_{15} = 5 + 42$
$a_{15} = 47$.
Therefore,the $15^{th}$ term of the $A.P.$ is $47$.

(B) The given $A.P.$ is $1, 11, 21, \ldots$
Here,the first term $a = 1$.
The common difference $d = 11 - 1 = 10$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
To find the $20^{th}$ term $(n = 20)$:
$a_{20} = 1 + (20 - 1) \times 10$
$a_{20} = 1 + 19 \times 10$
$a_{20} = 1 + 190$
$a_{20} = 191$.
Thus,the $20^{th}$ term is $191$.

(C) The $n^{th}$ term of the $A.P.$ is given by the formula $T_n = 5n - 2$.
To find the $12^{th}$ term,we substitute $n = 12$ into the given expression.
$T_{12} = 5(12) - 2$
$T_{12} = 60 - 2$
$T_{12} = 58$
Therefore,the $12^{th}$ term of the $A.P.$ is $58$.

(D) The $n^{th}$ term of an $A.P.$ is given by the formula $a_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.
Given,$a_7 = 34$,so $a + 6d = 34$ (Equation $1$).
Given,$a_{13} = 64$,so $a + 12d = 64$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(a + 12d) - (a + 6d) = 64 - 34$
$6d = 30$
$d = 5$.
Substituting $d = 5$ into Equation $1$:
$a + 6(5) = 34$
$a + 30 = 34$
$a = 4$.
Now,to find the $18^{th}$ term $(a_{18})$:
$a_{18} = a + 17d$
$a_{18} = 4 + 17(5)$
$a_{18} = 4 + 85$
$a_{18} = 89$.

(A) The $n^{th}$ term of an $A.P.$ can be calculated using the formula $T_n = S_n - S_{n-1}$,where $S_n$ is the sum of the first $n$ terms.
Given $S_n = 3n^2 + 5n$.
Then,$S_{n-1} = 3(n-1)^2 + 5(n-1)$.
$S_{n-1} = 3(n^2 - 2n + 1) + 5n - 5$.
$S_{n-1} = 3n^2 - 6n + 3 + 5n - 5$.
$S_{n-1} = 3n^2 - n - 2$.
Now,$T_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2)$.
$T_n = 3n^2 + 5n - 3n^2 + n + 2$.
$T_n = 6n + 2$.

(A) The sum of $n$ terms of an $A.P.$ is given by $S_{n} = 2n^{2} + 5n$.
To find the $n^{th}$ term $(T_{n})$,we use the formula $T_{n} = S_{n} - S_{n-1}$.
Given $S_{n} = 2n^{2} + 5n$.
Then,$S_{n-1} = 2(n-1)^{2} + 5(n-1)$.
$S_{n-1} = 2(n^{2} - 2n + 1) + 5n - 5$.
$S_{n-1} = 2n^{2} - 4n + 2 + 5n - 5$.
$S_{n-1} = 2n^{2} + n - 3$.
Now,$T_{n} = S_{n} - S_{n-1} = (2n^{2} + 5n) - (2n^{2} + n - 3)$.
$T_{n} = 2n^{2} + 5n - 2n^{2} - n + 3$.
$T_{n} = 4n + 3$.

(C) The sum of the first $n$ terms of an Arithmetic Progression $(AP)$ is given by the formula: $S_n = \frac{n}{2} [2a + (n-1)d]$.
Given values are $a = 2$,$d = 4$,and $n = 40$.
Substituting these values into the formula:
$S_{40} = \frac{40}{2} [2(2) + (40-1)4]$
$S_{40} = 20 [4 + (39 \times 4)]$
$S_{40} = 20 [4 + 156]$
$S_{40} = 20 [160]$
$S_{40} = 3200$.
Therefore,the correct option is $C$.

(D) The given $A.P.$ is $71, 68, 65, \ldots$.
Here,the first term $a = 71$ and the common difference $d = 68 - 71 = -3$.
We want to find the first negative term,so we set the $n^{th}$ term $a_n < 0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $71 + (n - 1)(-3) < 0$.
$71 - 3n + 3 < 0$.
$74 - 3n < 0$.
$74 < 3n$.
$n > \frac{74}{3} \approx 24.66$.
Since $n$ must be an integer,the smallest integer greater than $24.66$ is $25$.
Therefore,the $25^{th}$ term is the first negative term.

(A) The $n^{th}$ term of an $A.P.$ is given by the formula $T_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.
For $T_{25}$,we have $T_{25} = a + (25 - 1)d = a + 24d$.
For $T_{20}$,we have $T_{20} = a + (20 - 1)d = a + 19d$.
Now,calculating the difference: $T_{25} - T_{20} = (a + 24d) - (a + 19d)$.
$T_{25} - T_{20} = a + 24d - a - 19d = 5d$.
Therefore,the correct option is $A$.

(B) The first $n$ odd natural numbers form an Arithmetic Progression $(AP)$ where the first term $a = 1$ and the common difference $d = 2$.
The sequence is $1, 3, 5, \dots, (2n-1)$.
The sum of an $AP$ is given by the formula $S_n = \frac{n}{2} [2a + (n-1)d]$.
Substituting the values: $S_n = \frac{n}{2} [2(1) + (n-1)2]$.
$S_n = \frac{n}{2} [2 + 2n - 2]$.
$S_n = \frac{n}{2} [2n]$.
$S_n = n^2$.
Therefore,the sum of the first $n$ odd natural numbers is $n^2$.

(C) The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.
This forms an Arithmetic Progression $(AP)$ where the first term $a = 2$ and the common difference $d = 2$.
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: $S_n = \frac{n}{2}[2(2) + (n - 1)2]$.
$S_n = \frac{n}{2}[4 + 2n - 2]$.
$S_n = \frac{n}{2}[2n + 2]$.
$S_n = \frac{n}{2} \cdot 2(n + 1)$.
$S_n = n(n + 1)$.

(D) Given that for an $A.P.$,the first term $a = 10$ and the $10^{th}$ term $a_{10} = 100$.
We need to find the sum of the first $10$ terms,denoted by $S_{10}$.
The formula for the sum of the first $n$ terms of an $A.P.$ when the first and last terms are known is given by:
$S_n = \frac{n}{2} (a + a_n)$
Here,$n = 10$,$a = 10$,and $a_{10} = 100$.
Substituting these values into the formula:
$S_{10} = \frac{10}{2} (10 + 100)$
$S_{10} = 5 \times 110$
$S_{10} = 550$
Thus,the sum of the first $10$ terms is $550$.

(A) The given $A.P.$ is $1, 21, 41, \ldots$
Here,the first term $a = 1$.
The common difference $d = 21 - 1 = 20$.
The number of terms $n = 20$.
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: $S_{20} = \frac{20}{2} [2(1) + (20 - 1)20]$.
$S_{20} = 10 [2 + (19 \times 20)]$.
$S_{20} = 10 [2 + 380]$.
$S_{20} = 10 [382]$.
$S_{20} = 3820$.
Therefore,the sum of the first $20$ terms is $3820$.

(D) The given sequence is $1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots$.
In this sequence,each term is the sum of the two preceding terms:
$1 + 1 = 2$
$1 + 2 = 3$
$2 + 3 = 5$
$3 + 5 = 8$
$5 + 8 = 13$,and so on.
This pattern defines the Fibonacci sequence,where $F_n = F_{n-1} + F_{n-2}$ for $n > 2$ with $F_1 = 1$ and $F_2 = 1$.
Therefore,the correct option is $D$.