Progression and Sequence Questions in English

(B) In an $A.P.$, the sum of terms equidistant from the beginning and the end is constant.
Given the sum $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 147$.
There are $6$ terms in this sequence. We can pair them as $(a_1 + a_{16}) + (a_4 + a_{13}) + (a_7 + a_{10}) = 147$.
Since $a_1 + a_{16} = a_4 + a_{13} = a_7 + a_{10} = \lambda$, we have $3\lambda = 147$, which gives $\lambda = 49$.
Now, we need to find $S = a_1 + a_6 + a_{11} + a_{16}$.
Pairing these terms, we get $S = (a_1 + a_{16}) + (a_6 + a_{11})$.
Since the sum of equidistant terms is constant in an $A.P.$, $a_1 + a_{16} = a_6 + a_{11} = \lambda = 49$.
Therefore, $S = 49 + 49 = 98$.

(B) Given the equation $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} = K$.
Applying the property of Componendo and Dividendo,$\frac{(a+bx) + (a-bx)}{(a+bx) - (a-bx)} = \frac{2a}{2bx} = \frac{a}{bx}$.
Similarly,for the other terms,we get $\frac{a}{bx} = \frac{b}{cx} = \frac{c}{dx}$.
Since $x \ne 0$,we can cancel $x$ from the denominators: $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$.
This implies that the ratio of consecutive terms is constant,which is the definition of a Geometric Progression $(G.P.)$.
Therefore,$a, b, c, d$ are in $G.P.$

(C) Given the ratio of the sum of $n$ terms of two $A.P.$s is $\frac{s_n}{S_n} = \frac{3n - 13}{7n + 13}$.
Let $s_n = k(3n^2 - 13n)$ and $S_n = k(7n^2 + 13n)$ for some constant $k$.
We need to find the ratio $\frac{s_n}{S_{2n}}$.
Substitute $2n$ for $n$ in the expression for $S_n$:
$S_{2n} = k(7(2n)^2 + 13(2n)) = k(7(4n^2) + 26n) = k(28n^2 + 26n)$.
Now,the ratio is $\frac{s_n}{S_{2n}} = \frac{k(3n^2 - 13n)}{k(28n^2 + 26n)} = \frac{3n^2 - 13n}{28n^2 + 26n} = \frac{n(3n - 13)}{n(28n + 26)} = \frac{3n - 13}{28n + 26}$.

(A) Let $S = \sum_{n=0}^{\infty} \frac{(n+1)^2}{7^n} = \frac{1^2}{7^0} + \frac{2^2}{7^1} + \frac{3^2}{7^2} + \frac{4^2}{7^3} + \dots$
Multiply by $\frac{1}{7}$:
$\frac{S}{7} = \frac{1^2}{7^1} + \frac{2^2}{7^2} + \frac{3^2}{7^3} + \dots$
Subtracting the two equations:
$S - \frac{S}{7} = 1 + \frac{2^2 - 1^2}{7^1} + \frac{3^2 - 2^2}{7^2} + \frac{4^2 - 3^2}{7^3} + \dots$
$\frac{6S}{7} = 1 + \frac{3}{7} + \frac{5}{7^2} + \frac{7}{7^3} + \dots$
This is an arithmetico-geometric series. Let $T = 1 + \frac{3}{7} + \frac{5}{7^2} + \frac{7}{7^3} + \dots$
$\frac{T}{7} = \frac{1}{7} + \frac{3}{7^2} + \frac{5}{7^3} + \dots$
Subtracting these:
$T - \frac{T}{7} = 1 + \frac{2}{7} + \frac{2}{7^2} + \frac{2}{7^3} + \dots$
$\frac{6T}{7} = 1 + \frac{2/7}{1 - 1/7} = 1 + \frac{2/7}{6/7} = 1 + \frac{1}{3} = \frac{4}{3}$
$T = \frac{4}{3} \times \frac{7}{6} = \frac{14}{9}$
Since $\frac{6S}{7} = T$,we have $\frac{6S}{7} = \frac{14}{9} \Rightarrow S = \frac{14}{9} \times \frac{7}{6} = \frac{98}{54} = \frac{49}{27}$.

(B) Let $n = 2015$. The general term of the series is $T_k = k(n - k + 1)$ for $k = 1, 2, \dots, n$.
We need to find the sum $S = \sum_{k=1}^{n} k(n - k + 1) = \sum_{k=1}^{n} (nk - k^2 + k) = (n+1) \sum_{k=1}^{n} k - \sum_{k=1}^{n} k^2$.
Using the standard summation formulas $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$S = (n+1) \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$.
Factor out $\frac{n(n+1)}{6}$:
$S = \frac{n(n+1)}{6} [3(n+1) - (2n+1)] = \frac{n(n+1)}{6} [3n + 3 - 2n - 1] = \frac{n(n+1)(n+2)}{6}$.
Substituting $n = 2015$:
$S = \frac{2015 \times 2016 \times 2017}{6} = 2015 \times 336 \times 2017$.
Comparing this with the given options,the correct value is $336 \times 2015 \times 2017$.

(B) Let the three numbers in $G.P.$ be $3^a, 3^b, 3^c$ where $1 \le a < b < c \le 20$.
For these to be in $G.P.$, the condition is $(3^b)^2 = 3^a \cdot 3^c$, which implies $2b = a + c$.
This means $a + c$ must be an even number, which occurs if $a$ and $c$ are both even or both odd.
Case $1$: $a$ and $c$ are both odd. There are $10$ odd numbers in the set ${1, 2, \dots, 20}$. We choose $2$ distinct odd numbers for $a$ and $c$ in $^{10}C_2$ ways. Once $a$ and $c$ are chosen, $b$ is uniquely determined as $(a+c)/2$.
Number of ways $= ^{10}C_2 = 45$.
Case $2$: $a$ and $c$ are both even. There are $10$ even numbers in the set. We choose $2$ distinct even numbers for $a$ and $c$ in $^{10}C_2$ ways. Once $a$ and $c$ are chosen, $b$ is uniquely determined as $(a+c)/2$.
Number of ways $= ^{10}C_2 = 45$.
Total ways $= 45 + 45 = 90$.

(A) The given series is $S = \frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + ..........\infty$.
We can rewrite the terms as $S = \frac{1}{3} (\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..........\infty)$.
Multiplying by $2$,we get $2S = \frac{2}{3} (\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + ..........\infty)$.
Note that the denominators are triangular numbers $T_n = \frac{n(n+1)}{2}$. Thus,the series inside is $\sum_{n=2}^{\infty} \frac{2}{n(n+1)} = 2 \sum_{n=2}^{\infty} (\frac{1}{n} - \frac{1}{n+1})$.
This is a telescoping series: $2 [(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + ..........] = 2 (\frac{1}{2}) = 1$.
Therefore,$2S = \frac{2}{3} (1)$,which gives $S = \frac{1}{3}$.

(C) Given that the terms $\log _{5} 2, \log _{5}(2^{x}-3)$,and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$
For three terms $a, b, c$ to be in $A.P.$,the condition is $2b = a + c$.
Substituting the given values: $2 \log _{5}(2^{x}-3) = \log _{5} 2 + \log _{5}(\frac{17}{2}+2^{x-1})$.
Using the property $\log m + \log n = \log(mn)$,we get: $\log _{5}(2^{x}-3)^{2} = \log _{5}(2 \cdot (\frac{17}{2}+2^{x-1}))$.
Removing the logarithm: $(2^{x}-3)^{2} = 17 + 2 \cdot 2^{x-1}$.
Since $2 \cdot 2^{x-1} = 2^{x}$,the equation becomes: $(2^{x}-3)^{2} = 17 + 2^{x}$.
Let $2^{x} = y$. Then $(y-3)^{2} = 17 + y$.
$y^{2} - 6y + 9 = 17 + y \Rightarrow y^{2} - 7y - 8 = 0$.
Factoring the quadratic: $(y-8)(y+1) = 0$.
So,$y = 8$ or $y = -1$.
Since $y = 2^{x}$,$2^{x} = 8$ gives $x = 3$. $2^{x} = -1$ is not possible for real $x$.
Thus,$x = 3$.

(D) The number of terms in successive groups are $1, 2, 3, \dots, n$. Thus,the $n^{th}$ group contains $n$ terms in an $A$.$P$. with a common difference of $d = 2$.
First,we find the first term of the $n^{th}$ group. The first terms of the groups are $1, 3, 7, 13, \dots$. Let $a_n$ be the first term of the $n^{th}$ group. The differences between consecutive first terms are $2, 4, 6, \dots$,which form an $A$.$P$.
$a_n = a_1 + \sum_{k=1}^{n-1} (2k) = 1 + 2 \cdot \frac{(n-1)n}{2} = 1 + n^2 - n = n^2 - n + 1$.
The $n^{th}$ group is an $A$.$P$. with $n$ terms,first term $a = n^2 - n + 1$,and common difference $d = 2$.
The sum of the $n^{th}$ group is $S_n = \frac{n}{2} [2a + (n-1)d]$.
$S_n = \frac{n}{2} [2(n^2 - n + 1) + (n-1)2]$
$S_n = \frac{n}{2} [2n^2 - 2n + 2 + 2n - 2]$
$S_n = \frac{n}{2} [2n^2] = n^3$.

(B) By the Arithmetic Mean-Geometric Mean Inequality ($AM$-$GM$),we have $\frac{a+b+c+d}{4} \geq (abcd)^{1/4}$.
Raising both sides to the power of $4$,we get $\frac{(a+b+c+d)^4}{256} \geq abcd$,which implies $(a+b+c+d)^4 \leq 256 abcd$.
The given condition is $256 abcd \geq (a+b+c+d)^4$,which is the reverse of the standard $AM$-$GM$ inequality.
Equality holds if and only if $a = b = c = d$.
Given the equation $3a + b + 2c + 5d = 11$,and substituting $a = b = c = d = k$,we get $3k + k + 2k + 5k = 11$,which simplifies to $11k = 11$,so $k = 1$.
Thus,$a = b = c = d = 1$.
Substituting these values into the expression $a^3 + b + c^2 + 5d$,we get $1^3 + 1 + 1^2 + 5(1) = 1 + 1 + 1 + 5 = 8$.

(A) We are given the sum $S = \sum\limits_{r = 0}^{100} {(r^2 + 4r + 4)(r + 1)!}$.
Note that $r^2 + 4r + 4 = (r + 2)^2$.
So,$S = \sum\limits_{r = 0}^{100} {(r + 2)^2 (r + 1)!}$.
We can rewrite $(r + 2)^2$ as $(r + 2)(r + 2) = (r + 2)(r + 3 - 1) = (r + 2)(r + 3) - (r + 2)$.
Thus,$S = \sum\limits_{r = 0}^{100} {(r + 2)(r + 3)(r + 1)! - (r + 2)(r + 1)!}$.
Since $(r + 3)(r + 2)(r + 1)! = (r + 3)!$ and $(r + 2)(r + 1)! = (r + 2)!$,we have:
$S = \sum\limits_{r = 0}^{100} {(r + 3)! - (r + 2)!}$.
This is a telescoping sum:
$S = [(3! - 2!) + (4! - 3!) + (5! - 4!) + ... + (103! - 102!)]$.
All intermediate terms cancel out,leaving $S = 103! - 2! = 103! - 2$.

(C) Given ${x_r} = \cos(\pi/3^r) - i\sin(\pi/3^r)$.
We need to find the product $P = x_1 \cdot x_2 \cdot x_3 \cdots \infty$.
Using the property of complex numbers,$(\cos \theta_1 - i\sin \theta_1)(\cos \theta_2 - i\sin \theta_2) = \cos(\theta_1 + \theta_2) - i\sin(\theta_1 + \theta_2)$.
Thus,$P = \cos(\sum_{r=1}^{\infty} \frac{\pi}{3^r}) - i\sin(\sum_{r=1}^{\infty} \frac{\pi}{3^r})$.
The sum in the argument is an infinite geometric series: $S = \frac{\pi}{3} + \frac{\pi}{9} + \frac{\pi}{27} + \cdots$.
Here,the first term $a = \pi/3$ and the common ratio $r = 1/3$.
The sum of an infinite geometric series is $S = \frac{a}{1-r} = \frac{\pi/3}{1 - 1/3} = \frac{\pi/3}{2/3} = \frac{\pi}{2}$.
Therefore,$P = \cos(\pi/2) - i\sin(\pi/2) = 0 - i(1) = -i$.

(A) Let $S_n = x_n + y_n + z_n + w_n$. Since the sum of arithmetic progressions is also an arithmetic progression, let $S_n = A + (n-1)D$.
Given $S_4 = A + 3D = 8$ and $S_{10} = A + 9D = 20$.
Subtracting the first equation from the second: $(A + 9D) - (A + 3D) = 20 - 8 \Rightarrow 6D = 12 \Rightarrow D = 2$.
Substituting $D = 2$ into $A + 3D = 8$, we get $A + 6 = 8 \Rightarrow A = 2$.
Now, $S_{20} = A + 19D = 2 + 19(2) = 2 + 38 = 40$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality, for positive terms $x_{20}, y_{20}, z_{20}, w_{20}$:
$\frac{x_{20} + y_{20} + z_{20} + w_{20}}{4} \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4}$.
$\frac{40}{4} \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4} \Rightarrow 10 \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4}$.
Raising both sides to the power of $4$, we get $10^4 \geq x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20}$.
Thus, the maximum value is $10^4$.

(D) If $a, b, c$ are in $A.P.$,then $2b = a + c$.
Given terms are $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$.
Applying the condition: $2 \log _{10} (2^x - 1) = \log _{10} 2 + \log _{10} (2^x + 3)$.
Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:
$\log _{10} (2^x - 1)^2 = \log _{10} [2(2^x + 3)]$.
Removing the logarithm from both sides:
$(2^x - 1)^2 = 2(2^x + 3)$.
Let $2^x = y$. Then $(y - 1)^2 = 2(y + 3)$.
$y^2 - 2y + 1 = 2y + 6$.
$y^2 - 4y - 5 = 0$.
$(y - 5)(y + 1) = 0$.
So,$y = 5$ or $y = -1$.
Since $y = 2^x$ must be positive,$y = 5$.
$2^x = 5 \Rightarrow x = \log _{2} 5$.

(C) Let the two numbers be $x_1$ and $x_2$.
The geometric mean $GM = \sqrt{x_1 x_2} = 18$,so $x_1 x_2 = 18^2 = 324$.
The harmonic mean $HM = \frac{2x_1 x_2}{x_1 + x_2} = 16\frac{8}{13} = \frac{216}{13}$.
Substituting $x_1 x_2 = 324$ into the $HM$ formula: $\frac{2(324)}{x_1 + x_2} = \frac{216}{13}$.
$x_1 + x_2 = \frac{648 \times 13}{216} = 3 \times 13 = 39$.
Now,we know $(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1 x_2$.
$(x_1 - x_2)^2 = (39)^2 - 4(324) = 1521 - 1296 = 225$.
Therefore,$|x_1 - x_2| = \sqrt{225} = 15$.

(A) The sum of interior angles of an $n$-sided polygon is given by $(n-2) \times 180^\circ$.
Since the angles are in $G.P.$ with first term $a = 120^\circ$ and common ratio $r = 2$,the sum of $n$ terms is $S_n = a(r^n - 1) / (r - 1)$.
Substituting the values: $S_n = 120(2^n - 1) / (2 - 1) = 120(2^n - 1)$.
Equating the two expressions: $120(2^n - 1) = (n - 2) \times 180$.
Dividing both sides by $60$: $2(2^n - 1) = 3(n - 2)$.
$2^{n+1} - 2 = 3n - 6$,which simplifies to $2^{n+1} + 4 = 3n$.
For $n=3$,$2^4 + 4 = 20$ and $3(3) = 9$ $(20 \neq 9)$.
For $n=4$,$2^5 + 4 = 36$ and $3(4) = 12$ $(36 \neq 12)$.
As $n$ increases,$2^{n+1}$ grows much faster than $3n$. Thus,there are no integer solutions for $n \ge 3$.

(C) Given that $2a, b, 4c$ are in $A.P.$,we have $2b = 2a + 4c$,which simplifies to $b = a + 2c$.
Given that $c, a, b$ are in $G.P.$,we have $a^2 = bc$.
Substituting $b = a + 2c$ into the second equation: $a^2 = (a + 2c)c = ac + 2c^2$.
Rearranging gives $a^2 - ac - 2c^2 = 0$.
Factoring the quadratic: $(a - 2c)(a + c) = 0$.
Since $a, c \in \mathbb{R}^+$,we must have $a = 2c$.
Now,check the options:
For option $B$: The sequence is $c, a, a + 2c$. Substituting $a = 2c$,we get $c, 2c, 2c + 2c$,which is $c, 2c, 4c$.
This is a $G.P.$ with common ratio $2$,not an $A.P.$
For option $C$: The sequence is $c, a, a + 2c$. Substituting $a = 2c$,we get $c, 2c, 4c$. This is a $G.P.$ with common ratio $2$.
Thus,$c, a, a + 2c$ are in $G.P.$

(A) Let the roots of the equation be $x_1, x_2, x_3 > 0$.
From Vieta's formulas:
$x_1 + x_2 + x_3 = 2a$
$x_1x_2 + x_2x_3 + x_3x_1 = 3b$
$x_1x_2x_3 = 8$
Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for the terms $x_1x_2, x_2x_3, x_3x_1$:
$\frac{x_1x_2 + x_2x_3 + x_3x_1}{3} \geq \sqrt[3]{(x_1x_2)(x_2x_3)(x_3x_1)}$
$\frac{3b}{3} \geq \sqrt[3]{(x_1x_2x_3)^2}$
$b \geq \sqrt[3]{8^2}$
$b \geq \sqrt[3]{64}$
$b \geq 4$
Therefore,the minimum value of $b$ is $4$.

(D) To find the maximum value of $xyz^2$ given $x + y + z = 4$ where $x, y, z \in R^+$,we use the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality.
We want to maximize the product $x \cdot y \cdot (z/2) \cdot (z/2)$.
According to the $AM$-$GM$ inequality,for positive real numbers,the arithmetic mean is greater than or equal to the geometric mean:
$\frac{x + y + z/2 + z/2}{4} \geq \sqrt[4]{x \cdot y \cdot (z/2) \cdot (z/2)}$
Substituting the given sum $x + y + z = 4$:
$\frac{4}{4} \geq \sqrt[4]{\frac{xyz^2}{4}}$
$1 \geq \sqrt[4]{\frac{xyz^2}{4}}$
Raising both sides to the power of $4$:
$1 \geq \frac{xyz^2}{4}$
$xyz^2 \leq 4$
Thus,the maximum possible value of $xyz^2$ is $4$.

(A) Let the general term be $a_n = \frac{6\sum_{i=0}^n i + 1}{6\sum_{j=0}^n (j-1) + 1}$.
Using the sum formulas $\sum_{i=0}^n i = \frac{n(n+1)}{2}$ and $\sum_{j=0}^n (j-1) = \frac{n(n+1)}{2} - (n+1) = \frac{(n+1)(n-2)}{2}$.
The numerator is $6 \cdot \frac{n(n+1)}{2} + 1 = 3n^2 + 3n + 1$.
The denominator is $6 \cdot \frac{(n+1)(n-2)}{2} + 1 = 3(n^2 - n - 2) + 1 = 3n^2 - 3n - 5$. Wait,let's re-evaluate the denominator sum: $\sum_{j=0}^n (j-1) = -1 + 0 + 1 + ... + (n-1) = \frac{n(n-1)}{2} - 1$. Actually,$\sum_{j=0}^n (j-1) = \sum_{j=1}^n j - \sum_{j=0}^n 1 = \frac{n(n+1)}{2} - (n+1) = \frac{(n+1)(n-2)}{2}$.
Correcting the expression: The term is $\frac{3n^2+3n+1}{3n^2-3n+1}$.
For $n=1$: $\frac{3+3+1}{3-3+1} = \frac{7}{1}$.
For $n=2$: $\frac{12+6+1}{12-6+1} = \frac{19}{7}$.
For $n=3$: $\frac{27+9+1}{27-9+1} = \frac{37}{19}$.
This is a telescoping product: $\prod_{n=1}^{10} \frac{f(n)}{f(n-1)}$ where $f(n) = 3n^2+3n+1$.
The product is $\frac{f(10)}{f(0)} = \frac{3(100)+3(10)+1}{3(0)+3(0)+1} = \frac{331}{1} = 331$.

(D) Given the equation $x^8 - kx^2 + 3 = 0$,we can rewrite it as $k = \frac{x^8 + 3}{x^2} = x^6 + \frac{3}{x^2}$.
To find the minimum value of $k$,we use the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.
We split the term $\frac{3}{x^2}$ into three equal parts: $\frac{1}{x^2} + \frac{1}{x^2} + \frac{1}{x^2}$.
Thus,$k = x^6 + \frac{1}{x^2} + \frac{1}{x^2} + \frac{1}{x^2}$.
Applying $AM \geq GM$ for these four positive terms:
$\frac{x^6 + \frac{1}{x^2} + \frac{1}{x^2} + \frac{1}{x^2}}{4} \geq \sqrt[4]{x^6 \cdot \frac{1}{x^2} \cdot \frac{1}{x^2} \cdot \frac{1}{x^2}} = \sqrt[4]{1} = 1$.
Therefore,$x^6 + \frac{3}{x^2} \geq 4$.
The minimum value of $k$ is $4$.

(C) Since $2a+b, a-b, a+3b$ are in $G.P.$,the square of the middle term equals the product of the extremes:
$(a-b)^2 = (2a+b)(a+3b)$
$a^2 - 2ab + b^2 = 2a^2 + 6ab + ab + 3b^2$
$a^2 + 9ab + 2b^2 = 0$
Dividing by $a^2$ (since $a \neq 0$):
$2(\frac{b}{a})^2 + 9(\frac{b}{a}) + 1 = 0$
Let $x = \frac{b}{a}$. Then $2x^2 + 9x + 1 = 0$. The roots are $x_1 = \frac{b_1}{a_1}$ and $x_2 = \frac{b_2}{a_2}$.
From the quadratic equation,the sum of roots is $\frac{b_1}{a_1} + \frac{b_2}{a_2} = -\frac{9}{2}$ and the product of roots is $\frac{b_1b_2}{a_1a_2} = \frac{1}{2}$.
We need to find $2(a_1b_2 + a_2b_1) + 9a_1a_2$.
Divide the expression by $a_1a_2$: $2(\frac{b_2}{a_2} + \frac{b_1}{a_1}) + 9 = 2(-\frac{9}{2}) + 9 = -9 + 9 = 0$.

(A) The given series is an infinite geometric progression with the first term $a = \frac{4}{3}$ and common ratio $r = -\frac{x}{3}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$,provided $|r| < 1$.
Substituting the values,we get $x = \frac{4/3}{1 - (-x/3)} = \frac{4/3}{1 + x/3} = \frac{4}{3+x}$.
Cross-multiplying gives $x(3+x) = 4$,which simplifies to $x^2 + 3x - 4 = 0$.
Factoring the quadratic equation,we get $(x+4)(x-1) = 0$,so $x = 1$ or $x = -4$.
For the infinite series to converge,we must satisfy $|r| < 1$,which means $|-x/3| < 1$,or $|x| < 3$.
Since $|-4| = 4 > 3$,$x = -4$ is rejected. Thus,$x = 1$ is the only valid solution.

(B) Given $x_n = \frac{2n^2 + n + 1}{2n^2 - 3n + 2}$.
Let $f(n) = 2n^2 - 3n + 2$. Then $f(n-1) = 2(n-1)^2 - 3(n-1) + 2 = 2(n^2 - 2n + 1) - 3n + 3 + 2 = 2n^2 - 4n + 2 - 3n + 5 = 2n^2 - 7n + 7$. This does not match the denominator directly.
Let's re-evaluate: $x_n = \frac{2n^2 + n + 1}{2(n-1)^2 + (n-1) + 1} = \frac{f(n)}{f(n-1)}$ where $f(n) = 2n^2 + n + 1$.
Then $\prod_{i=1}^r x_i = \frac{f(1)}{f(0)} \cdot \frac{f(2)}{f(1)} \cdots \frac{f(r)}{f(r-1)} = \frac{f(r)}{f(0)}$.
$f(0) = 2(0)^2 + 0 + 1 = 1$. So $\prod_{i=1}^r x_i = 2r^2 + r + 1$.
Also,$\sum_{i=1}^r (2i - 1) = r^2$.
Thus,the expression becomes $\sum_{r=1}^n [(2r^2 + r + 1) - 2(r^2)] = \sum_{r=1}^n (r + 1)$.
$= \sum_{r=1}^n r + \sum_{r=1}^n 1 = \frac{n(n+1)}{2} + n = \frac{n^2 + n + 2n}{2} = \frac{n(n+3)}{2}$.

(C) Let $z = \left(\frac{3}{a}-1\right)^2+\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{c}-1\right)^2+(3c-1)^2$.
For minimization, each term is non-negative, so the minimum occurs when all expressions are equal:
$\frac{3}{a} = \frac{a}{b} = \frac{b}{c} = 3c = k$
From $3c = k$, we get:
$c = \frac{k}{3}$
From $\frac{b}{c} = k$, we get:
$b = kc = \frac{k^2}{3}$
From $\frac{a}{b} = k$, we get:
$a = kb = \frac{k^3}{3}$
From $\frac{3}{a} = k$, we get:
$3 = ak = \frac{k^4}{3}$
$k^4 = 9 \Rightarrow k = \sqrt{3}$
Now substituting $k = \sqrt{3}$:
Each term becomes:
$\sqrt{3} - 1$
So,
$z = 4(\sqrt{3} - 1)^2$
$= 4(3 + 1 - 2\sqrt{3})$
$= 16 - 8\sqrt{3}$
Comparing with $p - q\sqrt{r}$, we get:
$p = 16,\ q = 8,\ r = 3$
Thus,
$p + q + r = 16 + 8 + 3 = 27$

(D) The $n^{th}$ term of a sequence is given by $T_n = S_n - S_{n-1}$ for $n > 1$.
For $n=1$,$T_1 = S_1 = 3(1)^2 + 4(1) + 15 = 3 + 4 + 15 = 22$.
For $n=3$,$T_3 = S_3 - S_2$.
$S_3 = 3(3)^2 + 4(3) + 15 = 3(9) + 12 + 15 = 27 + 12 + 15 = 54$.
$S_2 = 3(2)^2 + 4(2) + 15 = 3(4) + 8 + 15 = 12 + 8 + 15 = 35$.
Thus,$T_3 = 54 - 35 = 19$.
Finally,$T_3 - T_1 = 19 - 22 = -3$.

(C) Given that $a, a + 2b, 2a + b$ are in $A.P.$
Therefore,$2(a + 2b) = a + (2a + b)$
$2a + 4b = 3a + b$
$a = 3b$ --- $(1)$
Also,$(b + 1)^2, ab + 5, (a + 1)^2$ are in $G.P.$
Therefore,$(ab + 5)^2 = (b + 1)^2(a + 1)^2$ --- $(2)$
Substituting $(1)$ into $(2)$:
$(3b^2 + 5)^2 = (b + 1)^2(3b + 1)^2$
Taking the square root on both sides:
$3b^2 + 5 = \pm(b + 1)(3b + 1)$
Case $1$: $3b^2 + 5 = (b + 1)(3b + 1)$
$3b^2 + 5 = 3b^2 + 4b + 1$
$4b = 4 \Rightarrow b = 1$
Since $a = 3b$,$a = 3(1) = 3$.
Thus,$a + b = 3 + 1 = 4$.
Case $2$: $3b^2 + 5 = -(b + 1)(3b + 1)$
$3b^2 + 5 = -(3b^2 + 4b + 1)$
$6b^2 + 4b + 6 = 0$
$3b^2 + 2b + 3 = 0$
The discriminant $D = 2^2 - 4(3)(3) = 4 - 36 = -32 < 0$.
Since $D < 0$,there are no real solutions for $b$ in this case.
Therefore,the only real solution is $a = 3, b = 1$,and $a + b = 4$.

(C) The general term of the series is $T_n = \frac{n+2}{n! + (n+1)! + (n+2)!}$ for $n = 1, 2, \dots, 2006$.
$T_n = \frac{n+2}{n! [1 + (n+1) + (n+2)(n+1)]}$
$T_n = \frac{n+2}{n! [n+2 + (n+2)(n+1)]} = \frac{n+2}{n! (n+2) [1 + n+1]} = \frac{n+2}{n! (n+2) (n+2)} = \frac{1}{n! (n+2)}$
$T_n = \frac{n+1}{n! (n+1)(n+2)} = \frac{n+1}{(n+2)!} = \frac{n+2-1}{(n+2)!} = \frac{1}{(n+1)!} - \frac{1}{(n+2)!}$
Sum $S = \sum_{n=1}^{2006} \left( \frac{1}{(n+1)!} - \frac{1}{(n+2)!} \right)$
$S = \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{3!} - \frac{1}{4!} \right) + \dots + \left( \frac{1}{2007!} - \frac{1}{2008!} \right)$
$S = \frac{1}{2!} - \frac{1}{2008!} = \frac{1}{2} - \frac{1}{2008!} = \frac{2008! - 2}{2 \cdot 2008!}$

(D) Given that $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$
By the property of $A.P.$,if $x, y, z$ are in $A.P.$,then $2y = x + z$. Thus:
$2 \ln(c-a) = \ln(a+c) + \ln(a-2b+c)$
Using the logarithmic property $\ln(m) + \ln(n) = \ln(mn)$ and $n \ln(m) = \ln(m^n)$:
$\ln((c-a)^2) = \ln((a+c)(a-2b+c))$
Taking the exponential on both sides:
$(c-a)^2 = (a+c)(a-2b+c)$
$c^2 - 2ac + a^2 = a^2 - 2ab + ac + ac - 2bc + c^2$
$c^2 - 2ac + a^2 = a^2 - 2ab + 2ac - 2bc + c^2$
Subtracting $a^2 + c^2$ from both sides:
$-2ac = -2ab + 2ac - 2bc$
$2ab + 2bc = 4ac$
$ab + bc = 2ac$
$b(a+c) = 2ac$
$b = \frac{2ac}{a+c}$
This is the condition for $a, b, c$ to be in $H.P.$ (Harmonic Progression).

(D) Let the roots of the equation be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$.
From the sum of roots,$\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 40$.
From the sum of reciprocals,$\frac{r^2}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} = 10$.
This can be written as $\frac{1}{a} (r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2}) = 10$.
Also,the sum of roots is $\frac{a}{r^2} (1 + r + r^2 + r^3 + r^4) = 40$.
Dividing the sum of roots by the sum of reciprocals: $\frac{a(1+r+r^2+r^3+r^4)}{(1/a)(1+r+r^2+r^3+r^4)/r^2} = \frac{40}{10} = 4$.
This simplifies to $a^2 r^2 = 4$,so $ar = \pm 2$.
The product of the roots is $s = (\frac{a}{r^2})(\frac{a}{r})(a)(ar)(ar^2) = a^5$.
Since $a$ is the middle term,from the sum of roots: $a(1 + (r + \frac{1}{r}) + (r^2 + \frac{1}{r^2})) = 40$.
Using $ar = \pm 2$,we find $a = 2$ or $a = -2$.
If $a = 2$,$s = a^5 = 2^5 = 32$. If $a = -2$,$s = (-2)^5 = -32$.
Thus,$|s| = 32$.

(D) We are given $x + y + z = 4$,where $x, y, z > 0$. We want to maximize $xyz^2$.
Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for four positive numbers $x, y, z/2, z/2$:
$\frac{x + y + z/2 + z/2}{4} \geq \sqrt[4]{x \cdot y \cdot \frac{z}{2} \cdot \frac{z}{2}}$
Substituting $x + y + z = 4$:
$\frac{4}{4} \geq \sqrt[4]{\frac{xyz^2}{4}}$
$1 \geq \sqrt[4]{\frac{xyz^2}{4}}$
Raising both sides to the power of $4$:
$1 \geq \frac{xyz^2}{4}$
$xyz^2 \leq 4$
Thus,the maximum value is $4$.

(B) Since $a, b, c$ are in $GP$,we have $b^2 = ac$ .....$(1)$
Since $4a, 5b, 4c$ are in $AP$,the middle term is the average of the other two:
$2(5b) = 4a + 4c$
$10b = 4(a + c)$
$a + c = \frac{5b}{2}$ .....$(2)$
We are given $a + b + c = 70$. Substituting $(2)$ into this:
$\frac{5b}{2} + b = 70$
$\frac{7b}{2} = 70$
$b = 20$
Now,substitute $b = 20$ into $(2)$:
$a + c = \frac{5(20)}{2} = 50$
From $(1)$,$ac = b^2 = 20^2 = 400$.
We have $a + c = 50$ and $ac = 400$. These are roots of the quadratic equation $x^2 - 50x + 400 = 0$:
$(x - 40)(x - 10) = 0$
So,${a, c} = {10, 40}$.
Finally,calculate $a^3 + b^3 + c^3$:
$a^3 + b^3 + c^3 = 10^3 + 20^3 + 40^3$
$= 1000 + 8000 + 64000$
$= 73000$.

(B) We know that $1 + \omega^r + \omega^{2r}$ is a geometric series sum or can be evaluated based on the properties of cube roots of unity.
For any integer $r$,the expression $1 + \omega^r + \omega^{2r}$ follows the property:
$1 + \omega^r + \omega^{2r} = \begin{cases} 3, & \text{if } r \text{ is a multiple of } 3 \\ 0, & \text{if } r \text{ is not a multiple of } 3 \end{cases}$
In the given summation $\sum_{r=1}^5 (1 + \omega^r + \omega^{2r})$,we evaluate for $r = 1, 2, 3, 4, 5$:
For $r=1$: $1 + \omega + \omega^2 = 0$
For $r=2$: $1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$
For $r=3$: $1 + \omega^3 + \omega^6 = 1 + 1 + 1 = 3$
For $r=4$: $1 + \omega^4 + \omega^8 = 1 + \omega + \omega^2 = 0$
For $r=5$: $1 + \omega^5 + \omega^{10} = 1 + \omega^2 + \omega = 0$
Summing these values: $0 + 0 + 3 + 0 + 0 = 3$.

(A) Let the common difference of the $A.P.$ be $d$. Then $S_{n+1} - S_n = d$ for all $n$.
The given sum is $\sum_{n=1}^{100} \frac{1}{S_n S_{n+1}} = \frac{1}{d} \sum_{n=1}^{100} (\frac{1}{S_n} - \frac{1}{S_{n+1}}) = \frac{1}{d} (\frac{1}{S_1} - \frac{1}{S_{101}}) = \frac{1}{d} (\frac{S_{101} - S_1}{S_1 S_{101}}) = \frac{1}{6}$.
Since $S_{101} = S_1 + 100d$, we have $S_{101} - S_1 = 100d$.
Substituting this into the equation: $\frac{1}{d} (\frac{100d}{S_1 S_{101}}) = \frac{100}{S_1 S_{101}} = \frac{1}{6}$, so $S_1 S_{101} = 600$.
We are given $S_1 + S_{101} = 50$.
Let $x = S_1$ and $y = S_{101}$. Then $x + y = 50$ and $xy = 600$.
The difference $|x - y| = \sqrt{(x+y)^2 - 4xy} = \sqrt{50^2 - 4(600)} = \sqrt{2500 - 2400} = \sqrt{100} = 10$.

(B) The first series is $2, 5, 8, \dots$ with $50$ terms.
The $n$-th term is $a_n = 2 + (n-1)3 = 3n - 1$. The $50$-th term is $3(50) - 1 = 149$.
The second series is $3, 5, 7, 9, \dots$ with $60$ terms.
The $n$-th term is $b_n = 3 + (n-1)2 = 2n + 1$. The $60$-th term is $2(60) + 1 = 121$.
Common terms must satisfy $3n - 1 = 2m + 1$,which simplifies to $3n = 2m + 2$,or $3n = 2(m+1)$.
This implies $n$ must be even. Let $n = 2k$. Then $3(2k) = 2(m+1) \Rightarrow m = 3k - 1$.
The common terms form an arithmetic progression starting at $5$ (where $n=2, m=1$) with a common difference of $6$ ($LCM$ of $3$ and $2$).
The common series is $5, 11, 17, \dots, T_k$.
We require $T_k \leq 121$ (the limit of the second series).
$5 + (k-1)6 \leq 121 \Rightarrow 6k - 1 \leq 121 \Rightarrow 6k \leq 122 \Rightarrow k \leq 20.33$.
Since $k$ must be an integer,the maximum value is $k = 20$.

(C) The general term of the series is given by $T_r = r \cdot \frac{^nC_r}{^nC_{r-1}}$.
Using the identity $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we substitute this into the expression:
$T_r = r \cdot \left( \frac{n-r+1}{r} \right) = n - r + 1$.
Now,we sum the terms from $r=1$ to $n$:
$S_n = \sum_{r=1}^n (n - r + 1)$.
Expanding the summation:
$S_n = (n+1) + n + (n-1) + \dots + 1$.
This is the sum of the first $n$ natural numbers:
$S_n = \frac{n(n+1)}{2}$.

(D) We observe that the numerator $3k^2 + 3k + 1$ can be written as $(k+1)^3 - k^3$.
Thus,the general term of the series is $T_k = \frac{(k+1)^3 - k^3}{k^3(k+1)^3}$.
This simplifies to $T_k = \frac{(k+1)^3}{k^3(k+1)^3} - \frac{k^3}{k^3(k+1)^3} = \frac{1}{k^3} - \frac{1}{(k+1)^3}$.
This is a telescoping series.
The sum $S_n = \sum_{k=1}^n \left( \frac{1}{k^3} - \frac{1}{(k+1)^3} \right) = \left( \frac{1}{1^3} - \frac{1}{2^3} \right) + \left( \frac{1}{2^3} - \frac{1}{3^3} \right) + \dots + \left( \frac{1}{n^3} - \frac{1}{(n+1)^3} \right)$.
All intermediate terms cancel out,leaving $S_n = 1 - \frac{1}{(n+1)^3}$.
Taking the limit as $n \to \infty$,we get $S = \lim_{n \to \infty} \left( 1 - \frac{1}{(n+1)^3} \right) = 1 - 0 = 1$.

(C) Let the first term be $b_1$ and the common ratio be $r$.
Given $b_1 + b_2 = 1$,we have $b_1 + b_1 r = 1$,which implies $b_1(1 + r) = 1$,so $b_1 = \frac{1}{1 + r}$.
The sum of an infinite geometric series is given by $S = \frac{b_1}{1 - r} = 2$.
Substituting $b_1 = \frac{1}{1 + r}$ into the sum formula: $\frac{1}{(1 + r)(1 - r)} = 2$.
This simplifies to $\frac{1}{1 - r^2} = 2$,which means $1 - r^2 = \frac{1}{2}$,so $r^2 = \frac{1}{2}$,giving $r = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.
Since $b_2 = b_1 r < 0$ and $b_1 = \frac{1}{1 + r}$,if $r = \frac{\sqrt{2}}{2}$,then $b_1 = \frac{1}{1 + \frac{\sqrt{2}}{2}} = \frac{2}{2 + \sqrt{2}} = 2 - \sqrt{2}$. Then $b_2 = (2 - \sqrt{2})(\frac{\sqrt{2}}{2}) = \sqrt{2} - 1 > 0$,which contradicts $b_2 < 0$.
If $r = -\frac{\sqrt{2}}{2}$,then $b_1 = \frac{1}{1 - \frac{\sqrt{2}}{2}} = \frac{2}{2 - \sqrt{2}} = \frac{2(2 + \sqrt{2})}{4 - 2} = 2 + \sqrt{2}$.
Then $b_2 = b_1 r = (2 + \sqrt{2})(-\frac{\sqrt{2}}{2}) = -\sqrt{2} - 1 < 0$,which satisfies the condition.
Thus,$b_1 = 2 + \sqrt{2}$.

(A) Let $50, a_1, a_2, \dots, a_n, 100$ be in an arithmetic progression $(A.P.)$.
The common difference $d = \frac{100 - 50}{n + 1} = \frac{50}{n + 1}$.
Then $a_2 = 50 + 2d = 50 + \frac{100}{n + 1} = \frac{50n + 50 + 100}{n + 1} = \frac{50(n + 3)}{n + 1}$.
Let $50, h_1, h_2, \dots, h_n, 100$ be in a harmonic progression $(H.P.)$.
Then $\frac{1}{50}, \frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}, \frac{1}{100}$ are in an $A.P.$
The common difference $d' = \frac{\frac{1}{100} - \frac{1}{50}}{n + 1} = \frac{-1}{100(n + 1)}$.
The $(n-1)$-th harmonic mean $h_{n-1}$ corresponds to the $(n-1+1)$-th term in the $A.P.$ of reciprocals,which is the $n$-th term.
$\frac{1}{h_{n-1}} = \frac{1}{50} + (n-1)d' = \frac{1}{50} - \frac{n-1}{100(n + 1)} = \frac{2(n + 1) - (n - 1)}{100(n + 1)} = \frac{2n + 2 - n + 1}{100(n + 1)} = \frac{n + 3}{100(n + 1)}$.
Thus,$h_{n-1} = \frac{100(n + 1)}{n + 3}$.
Finally,$a_2 h_{n-1} = \left( \frac{50(n + 3)}{n + 1} \right) \cdot \left( \frac{100(n + 1)}{n + 3} \right) = 50 \cdot 100 = 5000$.

(C) The given series is: $\left(\frac{11}{7}\right)^{2} + \left(\frac{12}{7}\right)^{2} + \left(\frac{13}{7}\right)^{2} + \ldots$
The $n^{th}$ term is given by $\left(\frac{10+n}{7}\right)^{2}$.
The sum of the first $11$ terms is $S_{11} = \sum_{n=1}^{11} \left(\frac{10+n}{7}\right)^{2} = \frac{1}{49} \sum_{n=1}^{11} (10+n)^{2}$.
Expanding the sum: $S_{11} = \frac{1}{49} [11^2 + 12^2 + 13^2 + \ldots + 21^2]$.
Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$S_{11} = \frac{1}{49} \left[ \sum_{k=1}^{21} k^2 - \sum_{k=1}^{10} k^2 \right] = \frac{1}{49} \left[ \frac{21 \times 22 \times 43}{6} - \frac{10 \times 11 \times 21}{6} \right]$.
$S_{11} = \frac{1}{49} \times \frac{21}{6} [22 \times 43 - 10 \times 11] = \frac{1}{49} \times \frac{7}{2} [946 - 110] = \frac{1}{14} \times 836 = \frac{11}{7} \times 38$.
Comparing this with $\frac{11}{7}\lambda$,we get $\lambda = 38$.

(B) Given $T_n = (n^2 + 1)n!$.
We can rewrite $T_n$ as:
$T_n = (n^2 + n - n + 1)n! = (n(n+1) - (n-1))n! = n(n+1)! - (n-1)n!$.
Let $f(n) = n(n+1)!$. Then $T_n = f(n) - f(n-1)$.
Thus,$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (f(k) - f(k-1)) = f(n) - f(0)$.
Since $f(0) = 0(1!) = 0$,we have $S_n = n(n+1)!$.
Now,$\frac{T_{10}}{S_{10}} = \frac{(10^2 + 1)10!}{10(11)!} = \frac{101 \times 10!}{10 \times 11 \times 10!} = \frac{101}{110}$.
Here,$a = 101$ and $b = 110$,which are relatively prime.
Therefore,$b - a = 110 - 101 = 9$.

(B) Given the sum of the first $n$ terms $S_n = cn(n-1) = cn^2 - cn$.
The $n^{th}$ term $t_n$ is given by $S_n - S_{n-1}$.
$S_{n-1} = c(n-1)(n-2) = c(n^2 - 3n + 2)$.
$t_n = (cn^2 - cn) - (cn^2 - 3cn + 2c) = 2cn - 2c = 2c(n-1)$.
We need to find the sum of the squares of these terms,i.e.,$\sum_{k=1}^{n} (t_k)^2$.
$t_k = 2c(k-1)$.
$(t_k)^2 = 4c^2(k-1)^2 = 4c^2(k^2 - 2k + 1)$.
Sum $= \sum_{k=1}^{n} 4c^2(k^2 - 2k + 1) = 4c^2 [\sum k^2 - 2\sum k + \sum 1]$.
Using standard summation formulas:
Sum $= 4c^2 [\frac{n(n+1)(2n+1)}{6} - 2\frac{n(n+1)}{2} + n]$.
Sum $= 4c^2 [\frac{n(n+1)(2n+1) - 6n(n+1) + 6n}{6}]$.
Sum $= \frac{4c^2}{6} [n(2n^2 + 3n + 1 - 6n - 6 + 6)] = \frac{2c^2}{3} [n(2n^2 - 3n + 1)]$.
Sum $= \frac{2c^2}{3} [n(n-1)(2n-1)] = \frac{2}{3}c^2n(n-1)(2n-1)$.

(C) The given series is a geometric progression with first term $a = 1$ and common ratio $r = 1/2$.
The sum of $n$ terms of a geometric progression is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.
Substituting the values,$S_n = \frac{1(1 - (1/2)^n)}{1 - 1/2} = \frac{1 - (1/2)^n}{1/2} = 2(1 - \frac{1}{2^n}) = 2 - \frac{2}{2^n} = 2 - \frac{1}{2^{n-1}}$.
We are given the inequality $2 - S_n < \frac{1}{100}$.
Substituting $S_n$,we get $2 - (2 - \frac{1}{2^{n-1}}) < \frac{1}{100}$.
This simplifies to $\frac{1}{2^{n-1}} < \frac{1}{100}$.
Taking the reciprocal,$2^{n-1} > 100$.
We know that $2^6 = 64$ and $2^7 = 128$.
Therefore,$n - 1$ must be at least $7$,which means $n - 1 \geq 7$,so $n \geq 8$.
The least integral value of $n$ is $8$.

(D) The series is $S = 1 - 2 + 3 - 4 + \dots$ up to $n$ terms.
Case $1$: If $n$ is even,let $n = 2m$.
$S = (1 - 2) + (3 - 4) + \dots + ((2m - 1) - 2m)$
$S = (-1) + (-1) + \dots + (-1)$ ($m$ times)
$S = -m = -\frac{n}{2}$.
Case $2$: If $n$ is odd,let $n = 2m + 1$.
$S = (1 - 2) + (3 - 4) + \dots + ((2m - 1) - 2m) + (2m + 1)$
$S = -m + (2m + 1) = m + 1$.
Since $n = 2m + 1$,we have $m = \frac{n - 1}{2}$.
$S = \frac{n - 1}{2} + 1 = \frac{n + 1}{2}$.
Since the sum depends on whether $n$ is even or odd (Statement $-1$ is true) and the formula for even $n$ is $-\frac{n}{2}$ (Statement $-2$ is true),and the dependence on parity is the reason for the different formulas,Statement $-1$ is the correct explanation for Statement $-2$.

(D) Given that $x, y, z$ are in $G.P.$
Taking the natural logarithm of each term,we get $\ln x, \ln y, \ln z$ in $A.P.$
Let $a = \ln x, b = \ln y, c = \ln z$. Since they are in $A.P.$,we have $2b = a + c$.
The given terms are $T_1 = \frac{1+a}{2}, T_2 = \frac{1+b}{4}, T_3 = \frac{1+c}{8}$.
For these to be in $A.G.P.$,they must be of the form $ar^0, (a+d)r^1, (a+2d)r^2$. However,checking the ratio of differences or the structure,we observe that the terms follow the form of an Arithmetico-Geometric Progression where the linear part is in $A.P.$ and the denominator forms a $G.P.$
Thus,the terms are in $A.G.P.$

(B) The sum of the infinite geometric series is given by $S = \frac{a}{1-r} = 7 \quad \dots(1)$.
The terms involving odd powers of $r$ are $ar, ar^3, ar^5, \dots$. This is a new geometric series with first term $A = ar$ and common ratio $R = r^2$.
The sum of this series is given as $3$,so $\frac{ar}{1-r^2} = 3 \quad \dots(2)$.
From $(1)$,we have $a = 7(1-r)$. Substituting this into $(2)$:
$\frac{7(1-r)r}{(1-r)(1+r)} = 3$
$\frac{7r}{1+r} = 3 \Rightarrow 7r = 3 + 3r \Rightarrow 4r = 3 \Rightarrow r = \frac{3}{4}$.
Substituting $r = \frac{3}{4}$ into $(1)$:
$a = 7(1 - \frac{3}{4}) = 7(\frac{1}{4}) = \frac{7}{4}$.
Now,calculate $(a^2 - r^2)$:
$a^2 - r^2 = (\frac{7}{4})^2 - (\frac{3}{4})^2 = \frac{49}{16} - \frac{9}{16} = \frac{40}{16} = \frac{5}{2}$.

(C) The $n^{th}$ term of a sequence can be calculated using the formula $T_n = S_n - S_{n-1}$ for $n > 1$.
Given $S_n = 4n^2 + 6n$.
Then $S_{n-1} = 4(n-1)^2 + 6(n-1) = 4(n^2 - 2n + 1) + 6n - 6 = 4n^2 - 8n + 4 + 6n - 6 = 4n^2 - 2n - 2$.
Now,$T_n = (4n^2 + 6n) - (4n^2 - 2n - 2) = 4n^2 + 6n - 4n^2 + 2n + 2 = 8n + 2$.
To find the $15^{th}$ term $(T_{15})$,substitute $n = 15$ into the expression for $T_n$.
$T_{15} = 8(15) + 2 = 120 + 2 = 122$.

(D) Given the recurrence relation $a_{n+1} - a_n = n - 2$.
We can write this for values from $n = 1$ to $n = 50$:
$a_2 - a_1 = 1 - 2 = -1$
$a_3 - a_2 = 2 - 2 = 0$
$a_4 - a_3 = 3 - 2 = 1$
...
$a_{51} - a_{50} = 50 - 2 = 48$
Summing these equations,the intermediate terms cancel out:
$a_{51} - a_1 = (-1) + 0 + 1 + 2 + \dots + 48$
The sum on the right is an arithmetic progression with $n = 50$ terms,first term $a = -1$,and last term $l = 48$.
Sum $S = \frac{n}{2}(a + l) = \frac{50}{2}(-1 + 48) = 25 \times 47 = 1175$.
Thus,$a_{51} - 5 = 1175$,which gives $a_{51} = 1180$.

(A) Given that $(b+c), (c+a), (a+b)$ are in $H.P.$
Therefore,their reciprocals $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in $A.P.$
This implies: $\frac{2}{c+a} = \frac{1}{b+c} + \frac{1}{a+b}$
$\frac{2}{c+a} = \frac{a+b+b+c}{(b+c)(a+b)} = \frac{a+2b+c}{ab+b^2+ac+bc}$
$2(ab+b^2+ac+bc) = (c+a)(a+2b+c)$
$2ab+2b^2+2ac+2bc = ac+2bc+c^2+a^2+2ab+ac$
$2ab+2b^2+2ac+2bc = a^2+c^2+2ac+2ab+2bc$
Subtracting $(2ab+2ac+2bc)$ from both sides,we get:
$2b^2 = a^2+c^2$
Since $2b^2 = a^2+c^2$,it follows that $a^2, b^2, c^2$ are in $A.P.$

(A) Since $a, b, c$ are in $H.P.$,their reciprocals $1/a, 1/b, 1/c$ are in $A.P.$
This implies $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$,which simplifies to $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ .....$(1)$
Now,consider the expression:
$E = \left( {\frac{1}{b} + \frac{1}{c} - \frac{1}{a}} \right)\left( {\frac{1}{c} + \frac{1}{a} - \frac{1}{b}} \right)$
From equation $(1)$,we have $\frac{1}{c} - \frac{1}{a} = \frac{2}{b} - \frac{2}{a}$ (not directly useful) or $\frac{1}{c} - \frac{1}{b} = \frac{1}{b} - \frac{1}{a}$.
Let's substitute $\frac{1}{a} = \frac{2}{b} - \frac{1}{c}$ into the first bracket:
$\left( \frac{1}{b} + \frac{1}{c} - (\frac{2}{b} - \frac{1}{c}) \right) = \left( \frac{2}{c} - \frac{1}{b} \right)$
Substitute $\frac{1}{a} = \frac{2}{b} - \frac{1}{c}$ into the second bracket:
$\left( \frac{1}{c} + (\frac{2}{b} - \frac{1}{c}) - \frac{1}{b} \right) = \left( \frac{1}{b} \right)$
Multiplying these results:
$E = \left( \frac{2}{c} - \frac{1}{b} \right) \left( \frac{1}{b} \right) = \frac{2}{bc} - \frac{1}{b^2}$.