Arithmetic progression Questions in English

(C) The sum of interior angles of a polygon with $n$ sides is $(n-2) \times 180^{\circ}$.
Let the angles be in an $A.P.$ with first term $a$ and common difference $d = 6^{\circ}$.
The sum of the $A.P.$ is $\frac{n}{2}[2a + (n-1)d] = (n-2) \times 180^{\circ}$.
Given the largest angle $a + (n-1)d = 219^{\circ}$,so $a = 219^{\circ} - 6(n-1) = 225^{\circ} - 6n$.
Substituting $a$ into the sum formula:
$\frac{n}{2}[2(225 - 6n) + (n-1)6] = (n-2)180$
$n[225 - 6n + 3n - 3] = (n-2)180$
$n[222 - 3n] = 180n - 360$
$222n - 3n^2 = 180n - 360$
$3n^2 - 42n - 360 = 0$
Dividing by $3$: $n^2 - 14n - 120 = 0$
$(n - 20)(n + 6) = 0$
Since $n$ must be positive,$n = 20$.

(C) Let the $A.P.$ be $a, a+d, a+2d, \dots$ where $a$ and $d$ are positive integers.
Given $S_3 = a + (a+d) + (a+2d) = 3a + 3d = 54$,which simplifies to $a+d = 18$.
Thus,$a = 18-d$.
Since $a$ is a positive integer,$18-d > 0 \Rightarrow d < 18$.
Also,$S_{20} = \frac{20}{2} [2a + 19d] = 10[2(18-d) + 19d] = 10[36 - 2d + 19d] = 10[36 + 17d]$.
Given $1600 < 10(36 + 17d) < 1800$,dividing by $10$ gives $160 < 36 + 17d < 180$.
Subtracting $36$ gives $124 < 17d < 144$.
Dividing by $17$ gives $7.29 < d < 8.47$.
Since $d$ must be an integer,$d = 8$.
Then $a = 18 - 8 = 10$.
The $11^{\text{th}}$ term is $a_{11} = a + 10d = 10 + 10(8) = 10 + 80 = 90$.

(D) Let the given sum be $S = a_1 + (a_5 + a_{10} + \ldots + a_{2020}) + a_{2024} = 2233$.
In an $A.P.$,the sum of terms equidistant from the beginning and end is constant,i.e.,$a_k + a_{n-k+1} = a_1 + a_n$.
Here,$n = 2024$. The terms in the bracket are $a_{5k}$ for $k=1$ to $404$.
Note that $a_5 + a_{2020} = a_1 + a_{2024}$,$a_{10} + a_{2015} = a_1 + a_{2024}$,and so on.
There are $404$ terms in the sequence $5, 10, \ldots, 2020$.
Pairing these terms,we have $202$ pairs,each equal to $(a_1 + a_{2024})$.
Including the outer terms $a_1$ and $a_{2024}$,the total sum is $203(a_1 + a_{2024}) = 2233$.
Thus,$(a_1 + a_{2024}) = \frac{2233}{203} = 11$.
The sum of the $A.P.$ is $S_{2024} = \frac{2024}{2}(a_1 + a_{2024}) = 1012 \times 11 = 11132$.

(A) Let the first term be $a_1 = a$ and the common difference be $d$.
The sum of the first $12$ odd-indexed terms is given by $\sum_{k=1}^{12} a_{2k-1} = a_1 + a_3 + \ldots + a_{23} = -\frac{72}{5} a$.
This is an $A.P.$ with $12$ terms,first term $a$,and common difference $2d$.
The sum is $\frac{12}{2} [2a + (12-1)(2d)] = 6(2a + 22d) = 12a + 132d$.
Equating this to the given value: $12a + 132d = -\frac{72}{5} a$.
Multiplying by $5$: $60a + 660d = -72a$,which simplifies to $132a = -660d$,so $a = -5d$.
We are given $\sum_{k=1}^{n} a_k = 0$,which is $\frac{n}{2} [2a + (n-1)d] = 0$.
Since $n \neq 0$,we have $2a + (n-1)d = 0$.
Substituting $a = -5d$: $2(-5d) + (n-1)d = 0$.
$-10d + nd - d = 0 \Rightarrow (n-11)d = 0$.
Since $a_1 \neq 0$,$d \neq 0$,therefore $n - 11 = 0$,which gives $n = 11$.

(A) Let the number of terms be $n = 2k$. The terms are $a_1, a_2, \ldots, a_{2k}$.
Sum of even terms: $a_2 + a_4 + \ldots + a_{2k} = 30$.
Sum of odd terms: $a_1 + a_3 + \ldots + a_{2k-1} = 24$.
Subtracting the two sums: $(a_2 - a_1) + (a_4 - a_3) + \ldots + (a_{2k} - a_{2k-1}) = 30 - 24 = 6$.
Since each difference is the common difference $d$,we have $k \times d = 6$,so $n \times d = 2k \times d = 12$.
The last term exceeds the first by $\frac{21}{2}$,so $a_n - a_1 = (n-1)d = \frac{21}{2}$.
Substituting $nd = 12$: $12 - d = \frac{21}{2} \Rightarrow d = 12 - 10.5 = 1.5 = \frac{3}{2}$.
Since $nd = 12$,$n \times \frac{3}{2} = 12 \Rightarrow n = 8$.
Using the sum of odd terms: $\frac{k}{2}[2a_1 + (k-1)d] = 24$ where $k=4$ and $d=1.5$.
$2[2a_1 + 3(1.5)] = 24$ $\Rightarrow 2a_1 + 4.5 = 12$ $\Rightarrow 2a_1 = 7.5$ $\Rightarrow a_1 = 3.75 = \frac{15}{4}$.
Wait,re-evaluating $a_1$: $2[2a_1 + 4.5] = 24$ $\Rightarrow 2a_1 + 4.5 = 12$ $\Rightarrow 2a_1 = 7.5$ $\Rightarrow a_1 = 3.75$.
Actually,using $a_1 + a_3 + a_5 + a_7 = 4a_1 + (0+2+4+6)d = 4a_1 + 12d = 24$.
$4a_1 + 12(1.5) = 24$ $\Rightarrow 4a_1 + 18 = 24$ $\Rightarrow 4a_1 = 6$ $\Rightarrow a_1 = 1.5$.
The sequence is $1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12$.
The integer terms are $3, 6, 9, 12$. There are $4$ such terms.

(D) Let the elements of set $A$ be $(a-d, a, a+d)$. The sum is $3a = 36$,so $a = 12$. The product is $p = a(a^2 - d^2) = 12(144 - d^2)$.
Let the elements of set $B$ be $(b-D, b, b+D)$. The sum is $3b = 36$,so $b = 12$. The product is $q = b(b^2 - D^2) = 12(144 - D^2)$.
Given $\frac{p+q}{p-q} = \frac{19}{5}$. By componendo and dividendo,$\frac{p}{q} = \frac{19+5}{19-5} = \frac{24}{14} = \frac{12}{7}$.
Substituting $p$ and $q$: $\frac{12(144-d^2)}{12(144-D^2)} = \frac{12}{7}$.
Since $D = d+3$,$D^2 = (d+3)^2 = d^2 + 6d + 9$.
$\frac{144-d^2}{144-(d^2+6d+9)} = \frac{12}{7} \implies \frac{144-d^2}{135-d^2-6d} = \frac{12}{7}$.
$7(144-d^2) = 12(135-d^2-6d) \implies 1008 - 7d^2 = 1620 - 12d^2 - 72d$.
$5d^2 + 72d - 612 = 0$. Solving for $d$ using the quadratic formula: $d = \frac{-72 \pm \sqrt{5184 - 4(5)(-612)}}{10} = \frac{-72 \pm \sqrt{5184 + 12240}}{10} = \frac{-72 \pm \sqrt{17424}}{10} = \frac{-72 \pm 132}{10}$.
Since $d > 0$,$d = \frac{60}{10} = 6$. Then $D = 6+3 = 9$.
$p - q = 12(144 - d^2) - 12(144 - D^2) = 12(D^2 - d^2) = 12(81 - 36) = 12(45) = 540$.

(C) Given $a_6 = a + 5d = 7$ $(i)$
Given $S_7 = \frac{7}{2}(2a + 6d) = 7 \Rightarrow a + 3d = 1$ $(ii)$
Subtracting $(ii)$ from $(i)$: $(a + 5d) - (a + 3d) = 7 - 1$ $\Rightarrow 2d = 6$ $\Rightarrow d = 3$.
Substituting $d = 3$ in $(ii)$: $a + 3(3) = 1 \Rightarrow a = -8$.
Given $S_n = \frac{n}{2}[2a + (n-1)d] = 700$.
Substituting $a = -8$ and $d = 3$: $\frac{n}{2}[2(-8) + (n-1)3] = 700$.
$\frac{n}{2}[-16 + 3n - 3] = 700$ $\Rightarrow n(3n - 19) = 1400$ $\Rightarrow 3n^2 - 19n - 1400 = 0$.
Solving the quadratic equation: $(3n + 56)(n - 25) = 0$.
Since $n$ must be a positive integer,$n = 25$.
Thus,$a_n = a_{25} = a + 24d = -8 + 24(3) = -8 + 72 = 64$.

(A) Let the first term be $a$ and the common difference be $d$ for the given $AP$.
The $n^{th}$ term of an $AP$ is given by $a_n = a + (n-1)d$.
Given that $9 \times a_9 = 13 \times a_{13}$.
Substituting the formula for the terms:
$9[a + (9-1)d] = 13[a + (13-1)d]$
$9[a + 8d] = 13[a + 12d]$
$9a + 72d = 13a + 156d$
Rearranging the terms:
$13a - 9a + 156d - 72d = 0$
$4a + 84d = 0$
Dividing by $4$:
$a + 21d = 0$
Since the $22^{nd}$ term $a_{22} = a + (22-1)d = a + 21d$,
Therefore,$a_{22} = 0$.

(A) Given the sum $\sum_{r=1}^n(2r+1) = 440$.
Expanding the summation,we get the series $3 + 5 + 7 + \ldots + (2n+1) = 440$.
This is an arithmetic progression with first term $a = 3$,common difference $d = 2$,and number of terms $n$.
The sum of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values: $\frac{n}{2}[2(3) + (n-1)(2)] = 440$.
$\Rightarrow \frac{n}{2}[6 + 2n - 2] = 440$.
$\Rightarrow \frac{n}{2}[2n + 4] = 440$.
$\Rightarrow n(n + 2) = 440$.
$\Rightarrow n^2 + 2n - 440 = 0$.
Solving the quadratic equation: $(n + 22)(n - 20) = 0$.
Since $n$ must be positive,$n = 20$.

(D) Given equation is $x^{3}+a x^{2}+b x+c=0$.
Let the roots be $(\alpha, \beta, \gamma)$. Since they are in $AP$,we have $2 \beta = \alpha + \gamma$.
From the sum of roots,$\alpha + \beta + \gamma = -a$.
Substituting $\alpha + \gamma = 2 \beta$,we get $3 \beta = -a$,so $\beta = -\frac{a}{3}$.
Since $\beta$ is a root,it satisfies the equation:
$(-\frac{a}{3})^{3} + a(-\frac{a}{3})^{2} + b(-\frac{a}{3}) + c = 0$.
$-\frac{a^{3}}{27} + \frac{a^{3}}{9} - \frac{ab}{3} + c = 0$.
Multiplying by $27$,we get $-a^{3} + 3a^{3} - 9ab + 27c = 0$.
$2a^{3} - 9ab + 27c = 0$.
Therefore,$2a^{3} - 9ab = -27c$.

(A) Given,number of terms,$n = 51$.
Since $n$ is odd,the middle term is the $\left(\frac{n+1}{2}\right)$-th term.
$\text{Middle term} = \left(\frac{51+1}{2}\right) = 26\text{-th term}$.
Thus,$T_{26} = a + 25d = 300$.
The sum of the first $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$,where $l$ is the last term.
Here,$l = T_{51} = a + 50d$.
$S_{51} = \frac{51}{2}(a + a + 50d) = \frac{51}{2}(2a + 50d) = 51(a + 25d)$.
Substituting $a + 25d = 300$,we get:
$S_{51} = 51 \times 300 = 15300$.

(B) Given terms are in $AP$: $p(\frac{q+r}{qr}), q(\frac{p+r}{pr}), r(\frac{p+q}{pq})$.
Adding $1$ to each term,the sequence remains in $AP$:
$\frac{pq+pr+qr}{qr}, \frac{qp+qr+pr}{pr}, \frac{rp+rq+pq}{pq}$ are in $AP$.
Let $S = pq+pr+qr$. Then $\frac{S}{qr}, \frac{S}{pr}, \frac{S}{pq}$ are in $AP$.
Dividing each term by $S$ (assuming $S \neq 0$),we get $\frac{1}{qr}, \frac{1}{pr}, \frac{1}{pq}$ are in $AP$.
Multiplying each term by $pqr$,we get $p, q, r$ are in $AP$.

(A) The given series is $3+5+7+\ldots$ up to $n$ terms.
This is an Arithmetic Progression $(A.P.)$ with the first term $a = 3$ and common difference $d = 5 - 3 = 2$.
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 $a = 3$ and $d = 2$:
$S_n = \frac{n}{2}[2(3) + (n-1)2]$
$S_n = \frac{n}{2}[6 + 2n - 2]$
$S_n = \frac{n}{2}[2n + 4]$
$S_n = n(n + 2)$

(C) Given the sum of $n$ terms of an $AP$ is $S_{n} = n^{2} + n$.
We know that the first term $a_{1} = S_{1} = 1^{2} + 1 = 2$.
The sum of the first two terms is $S_{2} = 2^{2} + 2 = 4 + 2 = 6$.
The second term $a_{2} = S_{2} - S_{1} = 6 - 2 = 4$.
The common difference $d = a_{2} - a_{1} = 4 - 2 = 2$.

(B) The sequence of multiples of $5$ from $10$ to $95$ forms an arithmetic progression where the first term $a = 10$,the last term $l = 95$,and the common difference $d = 5$.
Using the formula for the $n^{th}$ term of an arithmetic progression: $l = a + (n - 1)d$.
Substituting the values: $95 = 10 + (n - 1)5$.
$85 = (n - 1)5$.
$n - 1 = 17$.
$n = 18$.
Thus,there are $18$ multiples of $5$ in the given range.

(B) Let the roots of $6x^3-11x^2+6x-1=0$ be $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Since they are in harmonic progression,$a, b, c$ are in arithmetic progression.
Substituting $x = \frac{1}{y}$ in the given equation,we get $6(\frac{1}{y})^3 - 11(\frac{1}{y})^2 + 6(\frac{1}{y}) - 1 = 0$,which simplifies to $y^3 - 6y^2 + 11y - 6 = 0$.
Now,consider the equation $x^3-6x^2+11x-6=0$.
Since $x=1$ satisfies the equation $(1-6+11-6=0)$,$(x-1)$ is a factor.
Dividing by $(x-1)$,we get $(x-1)(x^2-5x+6)=0$,which factors to $(x-1)(x-2)(x-3)=0$.
The roots are $1, 2, 3$.
Since $2-1 = 1$ and $3-2 = 1$,the roots are in Arithmetic Progression.

(A) Let the roots be $a-d, a, a+d$.
Since the roots are in $AP$,their sum is given by the coefficient of $x^2$:
$(a-d) + a + (a+d) = p$
$3a = p \implies a = \frac{p}{3}$.
Since $a$ is a root of the equation $x^3-p x^2+q x-r=0$,it must satisfy the equation:
$(\frac{p}{3})^3 - p(\frac{p}{3})^2 + q(\frac{p}{3}) - r = 0$
$\frac{p^3}{27} - \frac{p^3}{9} + \frac{pq}{3} - r = 0$
Multiply the entire equation by $27$:
$p^3 - 3p^3 + 9pq - 27r = 0$
$-2p^3 + 9pq - 27r = 0$
$2p^3 - 9pq + 27r = 0$.
Thus,the correct option is $A$.

(A) Let the roots of the cubic equation be in arithmetic progression as $\alpha-r, \alpha, \alpha+r$.
Sum of the roots $= \alpha-r+\alpha+\alpha+r = 3\alpha$.
From the given equation $x^3-b x^2+c x-d=0$,the sum of the roots is $b$.
Therefore,$3\alpha = b \Rightarrow \alpha = \frac{b}{3}$.
Since $\alpha$ is a root of the equation,it must satisfy $x^3-b x^2+c x-d=0$.
Substituting $x = \frac{b}{3}$:
$(\frac{b}{3})^3 - b(\frac{b}{3})^2 + c(\frac{b}{3}) - d = 0$
$\frac{b^3}{27} - \frac{b^3}{9} + \frac{bc}{3} - d = 0$
Multiplying the entire equation by $27$:
$b^3 - 3b^3 + 9bc - 27d = 0$
$-2b^3 + 9bc - 27d = 0$
$9bc = 2b^3 + 27d$.

(A) Let the roots of the cubic equation $x^3+3px^2+3qx-8=0$ be $a-d$,$a$,and $a+d$ since they are in an arithmetic progression.
According to Vieta's formulas:
Sum of roots: $(a-d) + a + (a+d) = -3p \implies 3a = -3p \implies a = -p$.
Since $a$ is a root,it must satisfy the equation:
$(-p)^3 + 3p(-p)^2 + 3q(-p) - 8 = 0$.
$-p^3 + 3p^3 - 3pq - 8 = 0$.
$2p^3 - 3pq = 8$.

(A) Let $S_5$ be the sum of integers divisible by $5$ in the range $[1, 100]$. These are $5, 10, \dots, 100$. This is an arithmetic progression with $a = 5$,$l = 100$,and $n = \frac{100}{5} = 20$. The sum $S_5 = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050$.
Let $S_{13}$ be the sum of integers divisible by $13$ in the range $[1, 100]$. These are $13, 26, 39, 52, 65, 78, 91$. Here $n = 7$. The sum $S_{13} = \frac{7}{2}(13 + 91) = \frac{7}{2}(104) = 7 \times 52 = 364$.
Let $S_{65}$ be the sum of integers divisible by both $5$ and $13$ (i.e.,divisible by $65$). The only such integer is $65$. So $S_{65} = 65$.
By the Principle of Inclusion-Exclusion,the required sum is $S = S_5 + S_{13} - S_{65} = 1050 + 364 - 65 = 1349$.

(B) Let the roots of the cubic equation be $A-d, A, A+d$.
Since the sum of the roots is $-a$,we have $(A-d) + A + (A+d) = -a$,which gives $3A = -a$,so $A = -\frac{a}{3}$.
Since $A$ is a root,it must satisfy the equation: $A^3 + aA^2 + bA + c = 0$.
Substituting $A = -\frac{a}{3}$:
$(-\frac{a}{3})^3 + a(-\frac{a}{3})^2 + b(-\frac{a}{3}) + c = 0$
$-\frac{a^3}{27} + \frac{a^3}{9} - \frac{ab}{3} + c = 0$
Multiplying by $27$:
$-a^3 + 3a^3 - 9ab + 27c = 0$
$2a^3 - 9ab + 27c = 0$
$9ab = 2a^3 + 27c$.

(A) Given the cubic equation $4x^3 - 12x^2 + 11x + m = 0$.
Let the roots be $A-d, A, A+d$.
Sum of roots $= (A-d) + A + (A+d) = -(-12)/4 = 3$.
$3A = 3 \Rightarrow A = 1$.
Since $A=1$ is a root,it must satisfy the equation: $4(1)^3 - 12(1)^2 + 11(1) + m = 0$.
$4 - 12 + 11 + m = 0$.
$3 + m = 0 \Rightarrow m = -3$.

(A) The given series is $(2+3) + (5+6) + (8+9) + \ldots$ up to $n$ pairs.
This can be written as $5 + 11 + 17 + \ldots$ up to $n$ terms.
This is an Arithmetic Progression with first term $a = 5$ and common difference $d = 6$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_n = \frac{n}{2}[2(5) + (n-1)6]$
$S_n = \frac{n}{2}[10 + 6n - 6]$
$S_n = \frac{n}{2}[6n + 4]$
$S_n = n(3n + 2) = 3n^2 + 2n$.

(B) Integers divisible by both $2$ and $3$ are divisible by their least common multiple,which is $6$.
The integers from $1$ to $50$ divisible by $6$ are $6, 12, 18, \dots, 48$.
This forms an Arithmetic Progression $(AP)$ where the first term $a = 6$,the last term $a_n = 48$,and the common difference $d = 6$.
Using the formula $a_n = a + (n - 1)d$:
$48 = 6 + (n - 1)6$
$42 = (n - 1)6$
$n - 1 = 7 \Rightarrow n = 8$.
The sum $S_n$ of an $AP$ is given by $S_n = \frac{n}{2}(a + a_n)$:
$S_8 = \frac{8}{2}(6 + 48) = 4(54) = 216$.
Since $216 = 6^3$,the correct option is $B$.

(D) Given the function $f(x) = \sum_{k=1}^n (x - a_k)^2$.
Expanding the summation,we get $f(x) = \sum_{k=1}^n (x^2 - 2xa_k + a_k^2)$.
This simplifies to $f(x) = nx^2 - 2x \sum_{k=1}^n a_k + \sum_{k=1}^n a_k^2$.
Since $f(x)$ is a quadratic expression of the form $ax^2 + bx + c$,it attains its minimum value at $x = -\frac{b}{2a}$.
Here,$a = n$ and $b = -2 \sum_{k=1}^n a_k$.
Thus,the minimum occurs at $x = -\frac{-2 \sum_{k=1}^n a_k}{2n} = \frac{\sum_{k=1}^n a_k}{n}$.
By definition,the arithmetic mean $A = \frac{\sum_{k=1}^n a_k}{n}$.
Therefore,$x = A$.

(C) The given series is $1+4+7+\cdots+(3n-2)$.
This is an arithmetic progression where the first term $a = 1$ and the common difference $d = 4-1 = 3$.
The number of terms is $n$.
The sum of the first $n$ terms of an arithmetic progression is given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values $a=1$ and $d=3$ into the formula:
$S_n = \frac{n}{2}[2(1) + (n-1)3]$
$S_n = \frac{n}{2}[2 + 3n - 3]$
$S_n = \frac{n(3n-1)}{2}$.

(A) Let the roots of $x^3+a x^2+b x+c=0$ be $\alpha-1, \alpha, \alpha+1$ as the roots are in $AP$ with common difference $1$.
Sum of roots $= (\alpha-1) + \alpha + (\alpha+1) = 3\alpha = -a \Rightarrow \alpha = -\frac{a}{3} \quad \dots(i)$
Sum of product of roots taken two at a time $= (\alpha-1)\alpha + \alpha(\alpha+1) + (\alpha-1)(\alpha+1) = b$
$\Rightarrow \alpha^2 - \alpha + \alpha^2 + \alpha + \alpha^2 - 1 = b$ $\Rightarrow 3\alpha^2 - 1 = b \quad \dots(ii)$
Product of roots $= (\alpha-1)\alpha(\alpha+1) = \alpha(\alpha^2-1) = -c \quad \dots(iii)$
Substitute $\alpha = -\frac{a}{3}$ into $(ii)$: $3(-\frac{a}{3})^2 - 1 = b$ $\Rightarrow \frac{a^2}{3} - 1 = b$ $\Rightarrow a^2 = 3(b+1)$.
Substitute $\alpha = -\frac{a}{3}$ into $(iii)$: $(-\frac{a}{3})((-\frac{a}{3})^2 - 1) = -c$
$\Rightarrow \frac{a}{3}(\frac{a^2}{9} - 1) = c$ $\Rightarrow \frac{a}{3}(\frac{3(b+1)}{9} - 1) = c$
$\Rightarrow \frac{a}{3}(\frac{b+1}{3} - 1) = c$ $\Rightarrow \frac{a}{3}(\frac{b-2}{3}) = c$ $\Rightarrow 9c = a(b-2)$.

(A) Let the roots of the equation $4x^3 - 12x^2 + 11x + k = 0$ be $\alpha - d, \alpha, \alpha + d$.
Sum of the roots $= (\alpha - d) + \alpha + (\alpha + d) = 3\alpha$.
From the equation,the sum of the roots is $-\frac{-12}{4} = 3$.
So,$3\alpha = 3 \Rightarrow \alpha = 1$.
Since $\alpha = 1$ is a root,it must satisfy the equation $4(1)^3 - 12(1)^2 + 11(1) + k = 0$.
$4 - 12 + 11 + k = 0$.
$3 + k = 0 \Rightarrow k = -3$.

(B) Let the roots of the cubic equation $32x^3 - 48x^2 + 22x - 3 = 0$ be $a-d$,$a$,and $a+d$.
From the sum of the roots,$(a-d) + a + (a+d) = -(\frac{-48}{32}) = \frac{48}{32} = \frac{3}{2}$.
Thus,$3a = \frac{3}{2}$,which implies $a = \frac{1}{2}$.
Since $a = \frac{1}{2}$ is a root,it must satisfy the equation: $32(\frac{1}{2})^3 - 48(\frac{1}{2})^2 + 22(\frac{1}{2}) - 3 = 32(\frac{1}{8}) - 48(\frac{1}{4}) + 11 - 3 = 4 - 12 + 11 - 3 = 0$.
Now,use the product of the roots taken two at a time: $(a-d)a + a(a+d) + (a-d)(a+d) = \frac{22}{32} = \frac{11}{16}$.
Substituting $a = \frac{1}{2}$: $(\frac{1}{2}-d)(\frac{1}{2}) + \frac{1}{2}(\frac{1}{2}+d) + (\frac{1}{4}-d^2) = \frac{11}{16}$.
$\frac{1}{4} - \frac{d}{2} + \frac{1}{4} + \frac{d}{2} + \frac{1}{4} - d^2 = \frac{11}{16}$.
$\frac{3}{4} - d^2 = \frac{11}{16}$.
$d^2 = \frac{3}{4} - \frac{11}{16} = \frac{12-11}{16} = \frac{1}{16}$.
Therefore,the square of the common difference is $\frac{1}{16}$.

(A) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+px+10=0$ and are in arithmetic progression $(AP)$.
Let the roots be $\beta-d, \beta, \beta+d$.
From the sum of roots,$(\beta-d) + \beta + (\beta+d) = 6$,which gives $3\beta = 6$,so $\beta = 2$.
Since $\beta = 2$ is a root,it must satisfy the equation: $2^3 - 6(2^2) + p(2) + 10 = 0$.
$8 - 24 + 2p + 10 = 0$ $\Rightarrow 2p - 6 = 0$ $\Rightarrow p = 3$.
The product of the roots is $\alpha\beta\gamma = -10$. Since $\beta = 2$,we have $\alpha\gamma = -5$.
Also,$\alpha+\gamma = 6 - 2 = 4$.
We use the identity $\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha))$.
Alternatively,$\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)((\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha))$.
Here,$\alpha+\beta+\gamma = 6$,$\alpha\beta+\beta\gamma+\gamma\alpha = p = 3$,and $\alpha\beta\gamma = -10$.
$\alpha^3+\beta^3+\gamma^3 - 3(-10) = 6(6^2 - 3(3))$.
$\alpha^3+\beta^3+\gamma^3 + 30 = 6(36 - 9) = 6(27) = 162$.
$\alpha^3+\beta^3+\gamma^3 = 162 - 30 = 132$.

(B) The given sequence is an arithmetic progression with first term $a = 148$ and common difference $d = -2$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values,$S_n = \frac{n}{2}[2(148) + (n-1)(-2)] = \frac{n}{2}[296 - 2n + 2] = \frac{n}{2}[298 - 2n] = n(149 - n)$.
The average of the first $n$ terms is $\frac{S_n}{n} = 149 - n$.
Given that the average is $125$,we have $149 - n = 125$.
Therefore,$n = 149 - 125 = 24$.

(B) Let the arithmetic progression be $a, a+d, a+2d, \dots, a+(n-1)d$.
Given the first term $a = 11$.
The sum of the first four terms is $a + (a+d) + (a+2d) + (a+3d) = 4a + 6d = 56$.
Substituting $a = 11$: $4(11) + 6d = 56$ $\Rightarrow 44 + 6d = 56$ $\Rightarrow 6d = 12$ $\Rightarrow d = 2$.
The sum of the last four terms is $t_{n-3} + t_{n-2} + t_{n-1} + t_n = 112$.
In an arithmetic progression,the sum of terms equidistant from the beginning and end is constant: $t_1 + t_n = t_2 + t_{n-1} = t_3 + t_{n-2} = t_4 + t_{n-3} = k$.
Thus,$(t_1 + t_n) + (t_2 + t_{n-1}) + (t_3 + t_{n-2}) + (t_4 + t_{n-3}) = 56 + 112 = 168$.
Since each pair equals $k$,we have $4k = 168 \Rightarrow k = 42$.
Therefore,$t_1 + t_n = 42$.
Substituting $t_1 = 11$: $11 + t_n = 42 \Rightarrow t_n = 31$.
Using the formula $t_n = a + (n-1)d$: $31 = 11 + (n-1)2$.
$20 = (n-1)2$ $\Rightarrow n-1 = 10$ $\Rightarrow n = 11$.

(C) Let the terms be $a_1, a_1+r, a_1+2r, \ldots, a_1+(n-1)r$.
The mean of their squares is $\frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)^2$.
The square of their mean is $\left(\frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)\right)^2$.
The difference is the variance of the $A$.$P$.,which is given by $\sigma^2 = \frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)^2 - \left(\frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)\right)^2$.
Using the formula for the variance of the first $n$ terms of an $A$.$P$.,$\sigma^2 = \frac{r^2(n^2-1)}{12}$.
Thus,the difference is $\frac{r^2(n^2-1)}{12}$.

(B) Given that $1, \log _9(3^{1-x}+2), \log _3(4 \cdot 3^x-1)$ are in $A.P.$
Since $2b = a + c$ for an $A.P.$,we have:
$2 \log _9(3^{1-x}+2) = 1 + \log _3(4 \cdot 3^x-1)$
Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we get $\log_9(y) = \frac{1}{2} \log_3(y)$:
$2 \cdot \frac{1}{2} \log _3(3^{1-x}+2) = \log _3 3 + \log _3(4 \cdot 3^x-1)$
$\log _3(3^{1-x}+2) = \log _3(3(4 \cdot 3^x-1))$
$3^{1-x}+2 = 12 \cdot 3^x - 3$
Let $3^x = t$. Then $\frac{3}{t} + 2 = 12t - 3$
$3 + 2t = 12t^2 - 3t$
$12t^2 - 5t - 3 = 0$
$(4t - 3)(3t + 1) = 0$
Since $t = 3^x > 0$,we have $t = \frac{3}{4}$.
$3^x = \frac{3}{4} \Rightarrow x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4$.

(A) Given the recurrence relation $h(p) = \frac{1}{2}\{h(p+1) + h(p-1)\}$,we can rewrite it as $2h(p) = h(p+1) + h(p-1)$,which implies $h(p+1) - h(p) = h(p) - h(p-1)$.
This indicates that the sequence $h(0), h(1), \ldots, h(100)$ forms an Arithmetic Progression ($A$.$P$.).
Let the common difference be $d$. Then $h(n) = h(0) + nd$.
Using $h(100) = 20$ and $h(0) = 5$,we have $20 = 5 + 100d$.
$100d = 15 \Rightarrow d = \frac{15}{100} = 0.15$.
Thus,$h(1) = h(0) + d = 5 + 0.15 = 5.15$.

(D) Let $d_1$ be the common difference for the first set $(a, 2b)$. The $(n+2)^{th}$ term is $2b = a + (n+1)d_1$,so $d_1 = \frac{2b-a}{n+1}$.
The $m^{th}$ arithmetic mean is $A_m = a + m d_1 = a + m \left( \frac{2b-a}{n+1} \right)$.
Let $d_2$ be the common difference for the second set $(2a, b)$. The $(n+2)^{th}$ term is $b = 2a + (n+1)d_2$,so $d_2 = \frac{b-2a}{n+1}$.
The $m^{th}$ arithmetic mean is $A'_m = 2a + m d_2 = 2a + m \left( \frac{b-2a}{n+1} \right)$.
Equating the two means: $a + m \left( \frac{2b-a}{n+1} \right) = 2a + m \left( \frac{b-2a}{n+1} \right)$.
Multiplying by $(n+1)$: $a(n+1) + m(2b-a) = 2a(n+1) + m(b-2a)$.
$an + a + 2bm - am = 2an + 2a + bm - 2am$.
Rearranging terms: $2bm - bm - am + 2am = 2an - an + 2a - a$.
$bm + am = an + a$.
$m(b+a) = a(n+1)$.
Thus,$\frac{a}{b} = \frac{m}{n+1-m}$.

(D) Let the six terms of the $AP$ be $a-5d, a-3d, a-d, a+d, a+3d, a+5d$ with common difference $2d$.
Sum of the terms $= (a-5d) + (a-3d) + (a-d) + (a+d) + (a+3d) + (a+5d) = 6a$.
Given $6a = 3$,so $a = \frac{1}{2}$.
Given $T_1 = 4T_3$,where $T_1 = a-5d$ and $T_3 = a-d$.
$a-5d = 4(a-d)$ $\Rightarrow a-5d = 4a-4d$ $\Rightarrow -3a = d$.
Substituting $a = \frac{1}{2}$,we get $d = -3 \times \frac{1}{2} = -\frac{3}{2}$.
The fifth term is $T_5 = a+3d$.
$T_5 = \frac{1}{2} + 3(-\frac{3}{2}) = \frac{1}{2} - \frac{9}{2} = -\frac{8}{2} = -4$.

(B) Let the terms of the sequence be $T_1, T_2, T_3, \ldots$ where $T_1 = \log a$,$T_2 = \log \frac{a^2}{b}$,and $T_3 = \log \frac{a^3}{b^2}$.
Using the logarithmic property $\log \frac{x}{y} = \log x - \log y$ and $\log x^n = n \log x$:
$T_1 = \log a$
$T_2 = \log a^2 - \log b = 2 \log a - \log b$
$T_3 = \log a^3 - \log b^2 = 3 \log a - 2 \log b$
Now,check the common difference $d = T_2 - T_1 = (2 \log a - \log b) - \log a = \log a - \log b$.
Check $T_3 - T_2 = (3 \log a - 2 \log b) - (2 \log a - \log b) = \log a - \log b$.
Since $T_2 - T_1 = T_3 - T_2$,the sequence is an $A$.$P$. with common difference $d = \log a - \log b$.

(B) Let the three positive real numbers in $A$.$P$. be $(b-d)$,$b$,and $(b+d)$,where $d$ is the common difference.
Given that their product is $4$,we have $(b-d)b(b+d) = 4$.
This simplifies to $b(b^2 - d^2) = 4$,which implies $b^3 - bd^2 = 4$.
Rearranging the terms,we get $b^3 = 4 + bd^2$.
Since $b$ and $d^2$ are positive real numbers,$bd^2 \geq 0$.
Therefore,$b^3 = 4 + bd^2 \geq 4$.
Taking the cube root on both sides,we get $b \geq 4^{1/3} = (2^2)^{1/3} = 2^{2/3}$.
Thus,the minimum possible value of $b$ is $2^{2/3}$.

(B) Given the sum of $n$ terms $S_n = 3n^2 + 5n$.
The first term $a = t_1 = S_1 = 3(1)^2 + 5(1) = 8$.
The sum of two terms $S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22$.
The second term $t_2 = S_2 - S_1 = 22 - 8 = 14$.
The common difference $d = t_2 - t_1 = 14 - 8 = 6$.
The $m$th term is given by $t_m = a + (m - 1)d = 164$.
Substituting the values: $8 + (m - 1)6 = 164$.
$6(m - 1) = 156$.
$m - 1 = 26$.
$m = 27$.

(B) By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality,for positive real numbers $x_1, x_2, \ldots, x_n$:
$\frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n}$
Let $x_1 = \frac{a_1}{a_2}, x_2 = \frac{a_2}{a_3}, \ldots, x_n = \frac{a_n}{a_1}$.
Then,the product $x_1 x_2 \ldots x_n = \frac{a_1}{a_2} \times \frac{a_2}{a_3} \times \ldots \times \frac{a_n}{a_1} = 1$.
Substituting this into the inequality:
$\frac{\frac{a_1}{a_2} + \frac{a_2}{a_3} + \ldots + \frac{a_n}{a_1}}{n} \geq \sqrt[n]{1} = 1$
Therefore,$\frac{a_1}{a_2} + \frac{a_2}{a_3} + \ldots + \frac{a_n}{a_1} \geq n$.
The minimum value is $n$,which occurs when $a_1 = a_2 = \ldots = a_n$.

(B) Given $a+b+c=21$ and $a, b, c$ are in $AP$,we have $2b = a+c$.
Substituting $a+c = 2b$ into the sum equation: $2b + b = 21$ $\Rightarrow 3b = 21$ $\Rightarrow b = 7$.
Since $a, b, c$ are in $AP$,let the common difference be $d$. Then $a = 7-d$,$b = 7$,and $c = 7+d$.
Since $a, b, c \in \mathbb{N}$,we must have $a \geq 1$,so $7-d \geq 1 \Rightarrow d \leq 6$.
Also,the condition $a \leq b \leq c$ implies $7-d \leq 7 \leq 7+d$,which means $d \geq 0$.
Possible values for $d$ are $0, 1, 2, 3, 4, 5, 6$.
For each $d$,we get a triplet $(7-d, 7, 7+d)$:
If $d=0: (7, 7, 7)$
If $d=1: (6, 7, 8)$
If $d=2: (5, 7, 9)$
If $d=3: (4, 7, 10)$
If $d=4: (3, 7, 11)$
If $d=5: (2, 7, 12)$
If $d=6: (1, 7, 13)$
There are $7$ such triplets.

(A) Given $a, 4, \alpha, b$ are in $A$.$P$. Let the common difference be $d$. Then $4 = a + d$,$\alpha = 4 + d$,and $b = 4 + 2d$.
Since $a = 4 - d$,we have $a, 4, \alpha, b$ as $4-d, 4, 4+d, 4+2d$.
The arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{5}{16}$,so $\frac{1}{2}(\frac{1}{4-d} + \frac{1}{4+2d}) = \frac{5}{16}$.
$\frac{4+2d+4-d}{(4-d)(4+2d)} = \frac{5}{8} \Rightarrow \frac{8+d}{16+4d-2d^2} = \frac{5}{8}$.
$64 + 8d = 80 + 20d - 10d^2$ $\Rightarrow 10d^2 - 12d - 16 = 0$ $\Rightarrow 5d^2 - 6d - 8 = 0$.
$(5d+4)(d-2) = 0$. Since $a > 2$,$4-d > 2 \Rightarrow d < 2$. Thus $d = -4/5$ is not possible for standard integer progression,but checking $d=2$ gives $a=2$,which contradicts $a>2$. Re-evaluating: if $d=2$,$a=2$. If $d=-4/5$,$a=24/5=4.8 > 2$.
Using $d = -4/5$: $a = 4.8, \alpha = 3.2, b = 2.4$.
Equation: $3.2x^2 - 4.8x + 2(3.2 - 4.8) = 0$ $\Rightarrow 3.2x^2 - 4.8x - 3.2 = 0$ $\Rightarrow x^2 - 1.5x - 1 = 0$.
Roots are $x = \frac{1.5 \pm \sqrt{2.25 + 4}}{2} = \frac{1.5 \pm 2.5}{2} = 2, -0.5$.
One root is $2 \in (0, 4)$ and another is $-0.5 \in (-2, 0)$.

(C) Let the common differences of the two $A.P.$s be $d_{1}$ and $d_{2}$ respectively.
Given that $d_{1} = d_{2} + 13$.
For the $A.P.$ $b_{n}$,we have $b_{31} = b_{1} + 30d_{2} = -277$ (Equation $1$) and $b_{43} = b_{1} + 42d_{2} = -385$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(b_{1} + 42d_{2}) - (b_{1} + 30d_{2}) = -385 - (-277)$
$12d_{2} = -108$
$d_{2} = -9$.
Thus,$d_{1} = -9 + 13 = 4$.
For the $A.P.$ $a_{m}$,we have $a_{78} = a_{1} + 77d_{1} = 327$.
Substituting $d_{1} = 4$:
$a_{1} + 77(4) = 327$
$a_{1} + 308 = 327$
$a_{1} = 327 - 308 = 19$.

(C) Given the sum of an $A$.$P$. is $S_n = \frac{n}{2}(a_1 + a_n) = \frac{525}{2}$ and common difference $d = a_2 - a_1 = -\frac{3}{4}$.
Substituting $a_n = \frac{1}{4} a_1$ into the sum formula:
$\frac{n}{2}(a_1 + \frac{a_1}{4}) = \frac{525}{2} \implies \frac{n}{2}(\frac{5a_1}{4}) = \frac{525}{2} \implies \frac{5a_1 n}{8} = \frac{525}{2} \implies a_1 n = 420$.
Using the formula $a_n = a_1 + (n-1)d$:
$\frac{1}{4} a_1 = a_1 + (n-1)(-\frac{3}{4}) \implies -\frac{3}{4} a_1 = -\frac{3}{4}(n-1) \implies a_1 = n-1$.
Substituting $a_1 = n-1$ into $a_1 n = 420$:
$(n-1)n = 420 \implies n^2 - n - 420 = 0 \implies (n-21)(n+20) = 0$.
Since $n > 0$,we have $n = 21$ and $a_1 = 21 - 1 = 20$.
Now,calculate $\sum_{i=1}^{17} a_i = \frac{17}{2}[2a_1 + (17-1)d]$:
$= \frac{17}{2}[2(20) + 16(-\frac{3}{4})] = \frac{17}{2}[40 - 12] = \frac{17}{2}[28] = 17 \times 14 = 238$.

(A) Let the four terms be $a-3d, a-d, a+d, a+3d$ where the common difference is $l=2d$.
Since the sum is $48$,we have $(a-3d)+(a-d)+(a+d)+(a+3d)=48$,which simplifies to $4a=48$,so $a=12$.
The product of the terms plus $l^4$ is given by $(a^2-9d^2)(a^2-d^2)+l^4=361$.
Since $l=2d$,$l^4=16d^4$. Substituting $a=12$:
$(144-9d^2)(144-d^2)+16d^4=361$
$20736 - 144d^2 - 1296d^2 + 9d^4 + 16d^4 = 361$
$25d^4 - 1440d^2 + 20375 = 0$
Dividing by $5$: $5d^4 - 288d^2 + 4075 = 0$.
Using the quadratic formula for $d^2$: $d^2 = \frac{288 \pm \sqrt{288^2 - 4(5)(4075)}}{10} = \frac{288 \pm \sqrt{82944 - 81500}}{10} = \frac{288 \pm \sqrt{1444}}{10} = \frac{288 \pm 38}{10}$.
$d^2 = \frac{326}{10} = 32.6$ (not an integer $d$) or $d^2 = \frac{250}{10} = 25$.
Thus $d=5$ (since $d$ must be an integer for $l=2d$ to be an integer).
The terms are $12-15, 12-5, 12+5, 12+15$,which are $-3, 7, 17, 27$.
The largest term is $27$.

(C) Given the sum of $n$ terms $S_n = \alpha n^2 + \beta n$.
We know that $a_n = S_n - S_{n-1}$.
$a_n = (\alpha n^2 + \beta n) - (\alpha(n-1)^2 + \beta(n-1))$
$a_n = \alpha n^2 + \beta n - (\alpha(n^2 - 2n + 1) + \beta n - \beta)$
$a_n = 2\alpha n - \alpha + \beta$.
Given $a_{10} = 59$,so $2\alpha(10) - \alpha + \beta = 59 \Rightarrow 19\alpha + \beta = 59$ (Equation $1$).
Given $a_6 = 7a_1$,so $2\alpha(6) - \alpha + \beta = 7(2\alpha(1) - \alpha + \beta)$.
$11\alpha + \beta = 7(\alpha + \beta) \Rightarrow 11\alpha + \beta = 7\alpha + 7\beta$.
$4\alpha = 6\beta$ $\Rightarrow 2\alpha = 3\beta$ $\Rightarrow \beta = \frac{2}{3}\alpha$ (Equation $2$).
Substitute Equation $2$ into Equation $1$:
$19\alpha + \frac{2}{3}\alpha = 59$ $\Rightarrow \frac{57\alpha + 2\alpha}{3} = 59$ $\Rightarrow \frac{59\alpha}{3} = 59$ $\Rightarrow \alpha = 3$.
Then $\beta = \frac{2}{3}(3) = 2$.
Therefore,$\alpha + \beta = 3 + 2 = 5$.

(D) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}(2a + (n-1)d)$.
Given $S_4 = 6$,we have $\frac{4}{2}(2a + 3d) = 6 \Rightarrow 2a + 3d = 3$ .... $(1)$
Given $S_6 = 4$,we have $\frac{6}{2}(2a + 5d) = 4$ $\Rightarrow 3(2a + 5d) = 4$ $\Rightarrow 2a + 5d = \frac{4}{3}$ .... $(2)$
Subtracting $(1)$ from $(2)$:
$(2a + 5d) - (2a + 3d) = \frac{4}{3} - 3$
$2d = \frac{4-9}{3} = -\frac{5}{3} \Rightarrow d = -\frac{5}{6}$
Substituting $d$ in $(1)$:
$2a + 3(-\frac{5}{6}) = 3$ $\Rightarrow 2a - \frac{5}{2} = 3$ $\Rightarrow 2a = 3 + \frac{5}{2} = \frac{11}{2}$ $\Rightarrow a = \frac{11}{4}$
Now,$S_{12} = \frac{12}{2}(2a + 11d) = 6(2(\frac{11}{4}) + 11(-\frac{5}{6}))$
$S_{12} = 6(\frac{11}{2} - \frac{55}{6}) = 6(\frac{33-55}{6}) = 33 - 55 = -22$