Progression and Sequence Questions in English

(A) The $r$-th term of the series is $t_r = 1^2 + 3^2 + 5^2 + ... + (2r - 1)^2$.
Using the formula for the sum of squares of the first $r$ odd numbers,$t_r = \sum_{k=1}^r (2k - 1)^2 = \sum_{k=1}^r (4k^2 - 4k + 1)$.
$t_r = 4 \sum_{k=1}^r k^2 - 4 \sum_{k=1}^r k + \sum_{k=1}^r 1 = 4 \cdot \frac{r(r+1)(2r+1)}{6} - 4 \cdot \frac{r(r+1)}{2} + r$.
$t_r = \frac{2r(r+1)(2r+1)}{3} - 2r(r+1) + r = \frac{r}{3} [2(2r^2 + 3r + 1) - 6(r+1) + 3] = \frac{r}{3} [4r^2 + 6r + 2 - 6r - 6 + 3] = \frac{r(4r^2 - 1)}{3} = \frac{4r^3 - r}{3}$.
Now,the sum of $n$ terms is $S_n = \sum_{r=1}^n t_r = \sum_{r=1}^n \frac{4r^3 - r}{3} = \frac{4}{3} \sum_{r=1}^n r^3 - \frac{1}{3} \sum_{r=1}^n r$.
$S_n = \frac{4}{3} \cdot \frac{n^2(n+1)^2}{4} - \frac{1}{3} \cdot \frac{n(n+1)}{2} = \frac{n^2(n+1)^2}{3} - \frac{n(n+1)}{6}$.
$S_n = \frac{n(n+1)}{6} [2n(n+1) - 1] = \frac{n(n+1)(2n^2 + 2n - 1)}{6}$.

(A) We use the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for $n$ positive real numbers.
For the set of numbers ${a_1, a_2, ..., a_{n-1}, 2a_n}$,the $AM \geq GM$ inequality states:
$\frac{a_1 + a_2 + ... + a_{n-1} + 2a_n}{n} \geq (a_1 \cdot a_2 \cdot ... \cdot a_{n-1} \cdot 2a_n)^{1/n}$
Since the product $a_1 \cdot a_2 \cdot ... \cdot a_n = c$,we substitute this into the inequality:
$\frac{a_1 + a_2 + ... + a_{n-1} + 2a_n}{n} \geq (2 \cdot a_1 \cdot a_2 \cdot ... \cdot a_n)^{1/n}$
$\frac{a_1 + a_2 + ... + a_{n-1} + 2a_n}{n} \geq (2c)^{1/n}$
Multiplying both sides by $n$,we get:
$a_1 + a_2 + ... + a_{n-1} + 2a_n \geq n(2c)^{1/n}$
Thus,the minimum value is $n(2c)^{1/n}$.

(C) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$. Then $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.
Given that $\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$.
Substituting the values,we get $-b/a = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.
Thus,$-b/a = \frac{b^2 - 2ac}{c^2} \Rightarrow -bc^2 = ab^2 - 2a^2c$.
Rearranging,$2a^2c = ab^2 + bc^2$. Dividing both sides by $abc$,we get $2a/b = b/c + c/a$.
This implies $c/a, a/b, b/c$ are in $A$.$P$. Therefore,their reciprocals $a/c, b/a, c/b$ are in $H$.$P$.

(C) For $S_1$,the first term $a_1 = 1$ and common difference $d_1 = 6 - 1 = 5$.
The general term is $T_n = 1 + (n - 1)5 = 5n - 4$.
For $S_2$,the first term $a_2 = 3$ and common difference $d_2 = 7 - 3 = 4$.
The general term is $T_m = 3 + (m - 1)4 = 4m - 1$.
To find common terms,set $5n - 4 = 4m - 1$,which implies $5n = 4m + 3$.
The first common term is $11$ (when $n=3, m=3$).
The common difference of the new $A.P.$ formed by common terms is $LCM(d_1, d_2) = LCM(5, 4) = 20$.
The $k^{th}$ common term is given by $T_k = 11 + (k - 1)20$.
For the $25^{th}$ common term $(k = 25)$:
$T_{25} = 11 + (25 - 1)20 = 11 + 24 \times 20 = 11 + 480 = 491$.

(D) The given expression is an infinite geometric series with first term $a = 1$ and common ratio $r = \sin x$.
For the sum to exist,$|\sin x| < 1$.
The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
Thus,$\frac{1}{1 - \sin x} = 4 + 2\sqrt{3}$.
$1 - \sin x = \frac{1}{4 + 2\sqrt{3}}$.
Rationalizing the denominator: $\frac{1}{4 + 2\sqrt{3}} \times \frac{4 - 2\sqrt{3}}{4 - 2\sqrt{3}} = \frac{4 - 2\sqrt{3}}{16 - 12} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$.
So,$1 - \sin x = 1 - \frac{\sqrt{3}}{2}$.
$\sin x = \frac{\sqrt{3}}{2}$.
In the interval $(0, \pi)$,$\sin x = \frac{\sqrt{3}}{2}$ at $x = \frac{\pi}{3}$ and $x = \frac{2\pi}{3}$.
Since neither $\frac{\pi}{3}$ nor $\frac{2\pi}{3}$ matches the given options $A, B, C$,the correct answer is $D$.

(C) For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,the harmonic mean $H$ is given by the formula $H = \frac{2\alpha\beta}{\alpha + \beta}$.
From the given equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$,we identify the coefficients as $a = 5 + \sqrt{2}$,$b = -(4 + \sqrt{5})$,and $c = 8 + 2\sqrt{5}$.
Using Vieta's formulas,the product of the roots $\alpha\beta = \frac{c}{a} = \frac{8 + 2\sqrt{5}}{5 + \sqrt{2}}$ and the sum of the roots $\alpha + \beta = -\frac{b}{a} = \frac{4 + \sqrt{5}}{5 + \sqrt{2}}$.
Substituting these into the formula for $H$:
$H = \frac{2(\frac{8 + 2\sqrt{5}}{5 + \sqrt{2}})}{\frac{4 + \sqrt{5}}{5 + \sqrt{2}}} = \frac{2(8 + 2\sqrt{5})}{4 + \sqrt{5}}$.
Factoring out $2$ from the numerator: $H = \frac{2 \cdot 2(4 + \sqrt{5})}{4 + \sqrt{5}} = 4$.

(C) Let the given series be $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$ --- $(1)$
Multiply by $\frac{1}{4}$:
$\frac{S}{4} = \frac{3}{4} + \frac{3+d}{4^2} + \frac{3+2d}{4^3} + \dots \infty$ --- $(2)$
Subtract $(2)$ from $(1)$:
$S - \frac{S}{4} = 3 + \left( \frac{3+d}{4} - \frac{3}{4} \right) + \left( \frac{3+2d}{4^2} - \frac{3+d}{4^2} \right) + \dots \infty$
$\frac{3S}{4} = 3 + \frac{d}{4} + \frac{d}{4^2} + \frac{d}{4^3} + \dots \infty$
The series $3 + \frac{d}{4} + \frac{d}{4^2} + \dots$ is an Arithmetico-Geometric series where the sum of the infinite geometric part $\frac{d}{4} + \frac{d}{4^2} + \dots$ is $\frac{a}{1-r} = \frac{d/4}{1 - 1/4} = \frac{d/4}{3/4} = \frac{d}{3}$.
So,$\frac{3S}{4} = 3 + \frac{d}{3}$.
Given $S = 8$,substitute $S = 8$ into the equation:
$\frac{3(8)}{4} = 3 + \frac{d}{3}$
$6 = 3 + \frac{d}{3}$
$3 = \frac{d}{3}$
$d = 9$.

(D) Let $n = 100$. The expression is $\sqrt{\sum_{k=0}^{2n-1} 10^k - 2 \sum_{k=0}^{n-1} 10^k}$.
Using the sum of a geometric progression,$\sum_{k=0}^{m-1} 10^k = \frac{10^m - 1}{9}$.
So,the expression becomes $\sqrt{\frac{10^{2n}-1}{9} - 2 \left( \frac{10^n-1}{9} \right)}$.
$= \sqrt{\frac{10^{2n} - 1 - 2 \cdot 10^n + 2}{9}} = \sqrt{\frac{10^{2n} - 2 \cdot 10^n + 1}{9}}$.
$= \sqrt{\left( \frac{10^n - 1}{3} \right)^2} = \frac{10^n - 1}{3}$.
Since $\frac{10^n - 1}{9} = \underbrace{11...1}_{n \text{ digits}}$,then $\frac{10^n - 1}{3} = 3 \times \underbrace{11...1}_{n \text{ digits}} = \underbrace{33...3}_{n \text{ digits}}$.
For $n=100$,the result is $\underbrace{33...3}_{100 \text{ digits}}$.

(A) Let $f(\theta) = 2 \sin^2 \theta + 8 \csc^2 \theta$.
Since $\sin^2 \theta$ can take values in the interval $(0, 1]$,let $x = \sin^2 \theta$,where $x \in (0, 1]$.
Then the expression becomes $f(x) = 2x + \frac{8}{x}$.
To find the minimum value,we analyze the function $f(x) = 2x + \frac{8}{x}$ for $x \in (0, 1]$.
The derivative is $f'(x) = 2 - \frac{8}{x^2}$.
Setting $f'(x) = 0$ gives $x^2 = 4$,so $x = 2$ or $x = -2$. Neither of these values lies in the interval $(0, 1]$.
Since $f'(x) < 0$ for all $x \in (0, 1]$,the function $f(x)$ is strictly decreasing on $(0, 1]$.
Therefore,the minimum value occurs at the largest possible value of $x$,which is $x = 1$.
Substituting $x = 1$ into the expression: $f(1) = 2(1) + \frac{8}{1} = 2 + 8 = 10$.
Thus,the minimum value is $10$.

(C) Given the sum of squares: $\sum_{n=1}^{2009} n^2 = (2009)(335)(4019)$ .....$(1)$
We need to evaluate the sum $S = \sum_{n=1}^{2009} n(2010 - n)$.
Expanding the summation: $S = 2010 \sum_{n=1}^{2009} n - \sum_{n=1}^{2009} n^2$.
Using the formula for the sum of first $N$ natural numbers,$\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$,where $N = 2009$:
$S = 2010 \cdot \frac{2009 \cdot 2010}{2} - (2009)(335)(4019)$.
$S = 1005 \cdot 2009 \cdot 2010 - (2009)(335)(4019)$.
Since $1005 = 3 \cdot 335$,we have $S = 3 \cdot 335 \cdot 2009 \cdot 2010 - (2009)(335)(4019)$.
Factor out $(2009)(335)$: $S = (2009)(335) [3 \cdot 2010 - 4019]$.
$S = (2009)(335) [6030 - 4019] = (2009)(335)(2011)$.
Comparing this with the given expression $(2009)(335)(x)$,we get $x = 2011$.

(D) For the sum of an arithmetic progression to be maximum, we must include all positive terms. The sum starts decreasing as soon as we add negative terms.
Set the $n^{th}$ term $T_n = 0$ or find the last positive term.
Given $a = 50$ and $d = -2$.
$T_n = a + (n - 1)d = 50 + (n - 1)(-2) = 50 - 2n + 2 = 52 - 2n$.
Setting $T_n > 0$, we get $52 - 2n > 0$, which implies $n < 26$.
Thus, the sum is maximum for $n = 26$ (where $T_{26} = 0$).
Using the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
$S_{26} = \frac{26}{2}[2(50) + (26 - 1)(-2)]$
$S_{26} = 13[100 - 50] = 13 \times 50 = 650$.

(A) Let $n = 100$. The sum of products of $n$ numbers taken $k$ at a time is denoted by $S_k$.
By Newton's inequality or the Maclaurin's inequality,we know that for positive real numbers,the elementary symmetric means $E_k = \frac{S_k}{\binom{n}{k}}$ satisfy $E_k^2 \ge E_{k-1} E_{k+1}$.
Alternatively,using the property $S_k S_{n-k} \ge \binom{n}{k}^2 (a_1 a_2 \dots a_n)$,we set $n = 100$ and $k = 2$.
Then $S_2 S_{100-2} = S_2 S_{98} \ge \binom{100}{2}^2 (a_1 a_2 \dots a_{100})$.
Comparing this with $S_{98} S_2 \ge \lambda (a_1 a_2 \dots a_{100})$,we get $\lambda = \binom{100}{2}^2$.
Calculating the value: $\binom{100}{2} = \frac{100 \times 99}{2} = 4950$.
Thus,$\lambda = (4950)^2$.

(B) Let the given expression be $P = \sqrt{a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdot (8a)^{\frac{1}{8a}} \cdots \infty} = \frac{8}{27}$.
Squaring both sides,we get $P^2 = a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdots = \left(\frac{8}{27}\right)^2 = \frac{64}{729}$.
Expressing as powers of $a$ and $2$: $a^{\left(\frac{1}{a} + \frac{1}{2a} + \frac{1}{4a} + \cdots \right)} \cdot 2^{\left(0 \cdot \frac{1}{a} + 1 \cdot \frac{1}{2a} + 2 \cdot \frac{1}{4a} + 3 \cdot \frac{1}{8a} + \cdots \right)} = \frac{64}{729}$.
The exponent of $a$ is $\frac{1}{a} (1 + \frac{1}{2} + \frac{1}{4} + \cdots) = \frac{1}{a} \left(\frac{1}{1 - 1/2}\right) = \frac{2}{a}$.
The exponent of $2$ is $\frac{1}{2a} + \frac{2}{4a} + \frac{3}{8a} + \cdots = \frac{1}{a} (\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \cdots)$. This is an Arithmetico-Geometric Progression $(AGP)$ with sum $S = \frac{1/2}{1 - 1/2} + \frac{(1/2)(1/2)}{(1 - 1/2)^2} = 1 + 1 = 2$. So the exponent is $\frac{2}{a}$.
Thus,$a^{\frac{2}{a}} \cdot 2^{\frac{2}{a}} = (2a)^{\frac{2}{a}} = \frac{64}{729} = \left(\frac{8}{27}\right)^2 = \left(\left(\frac{2}{3}\right)^3\right)^2 = \left(\frac{2}{3}\right)^6$.
Comparing $(2a)^{\frac{2}{a}} = (2/3)^6$,we set $\frac{2}{a} = 6 \Rightarrow a = \frac{1}{3}$ and $2a = 2/3 \Rightarrow a = 1/3$.

(B) Let $A$,$G$,and $H$ be the arithmetic mean,geometric mean,and harmonic mean of $a$ and $b$ respectively.
Given: $A = G + \frac{3}{2}$ and $G = H + \frac{6}{5}$.
We know that $G^2 = AH$,so $H = \frac{G^2}{A}$.
Substituting $H$ into the second equation: $G = \frac{G^2}{A} + \frac{6}{5} \implies G - \frac{6}{5} = \frac{G^2}{A}$.
Since $A = G + \frac{3}{2} = \frac{2G + 3}{2}$,we have $G - \frac{6}{5} = \frac{2G^2}{2G + 3}$.
$(5G - 6)(2G + 3) = 10G^2 \implies 10G^2 + 15G - 12G - 18 = 10G^2$.
$3G = 18 \implies G = 6$.
Then $A = 6 + 1.5 = 7.5 = \frac{15}{2}$ and $H = 6 - 1.2 = 4.8 = \frac{24}{5}$.
Since $A = \frac{a+b}{2} = \frac{15}{2}$,we have $a+b = 15$.
Since $G = \sqrt{ab} = 6$,we have $ab = 36$.
The numbers $a$ and $b$ are roots of $x^2 - 15x + 36 = 0$.
$(x - 12)(x - 3) = 0$,so ${a, b} = {12, 3}$.
$|a^2 - b^2| = |144 - 9| = 135$.

(D) Let the four interior angles of the quadrilateral in $A.P.$ be $(a-3d), (a-d), (a+d), (a+3d)$.
The common difference between consecutive terms is $2d = 10^{\circ}$,which implies $d = 5^{\circ}$.
The sum of interior angles of a quadrilateral is $360^{\circ}$.
Therefore,$(a-3d) + (a-d) + (a+d) + (a+3d) = 360^{\circ}$.
$4a = 360^{\circ} \Rightarrow a = 90^{\circ}$.
The smallest angle is $a - 3d = 90^{\circ} - 3(5^{\circ}) = 90^{\circ} - 15^{\circ} = 75^{\circ}$.

(B) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant.
Specifically,$a_3 + a_{19} = a_5 + a_{17} = a_1 + a_{21} = 2a_{11}$.
Let $a_3 + a_{19} = a_5 + a_{17} = 2a_{11} = k$.
Given the equation: $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$.
Substituting the terms in terms of $k$: $\frac{k}{2} + \frac{k}{2} + \frac{k}{2} + \frac{k}{2} + \frac{k}{2} = 10$ is incorrect; rather,we have $(a_3 + a_{19}) + (a_5 + a_{17}) + a_{11} = 10$.
This simplifies to $k + k + \frac{k}{2} = 10$,which gives $\frac{5k}{2} = 10$,so $k = 4$.
Since $a_1 + a_{21} = 2a_{11} = k = 4$,the sum of the $21$ terms is given by $S_{21} = \frac{21}{2}(a_1 + a_{21})$.
$S_{21} = \frac{21}{2} \times 4 = 42$.

(B) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is given by $T_n = a + (n-1)d$.
According to the first condition: $T_9 = 5 \times T_2$.
$a + 8d = 5(a + d) \implies a + 8d = 5a + 5d \implies 4a = 3d \implies d = \frac{4a}{3}$.
According to the second condition: $T_{13} = 2 \times T_6 + 5$.
$a + 12d = 2(a + 5d) + 5 \implies a + 12d = 2a + 10d + 5 \implies 2d - a = 5$.
Substitute $d = \frac{4a}{3}$ into the equation $2d - a = 5$:
$2(\frac{4a}{3}) - a = 5 \implies \frac{8a}{3} - a = 5 \implies \frac{5a}{3} = 5 \implies a = 3$.

(A) The given set contains $9$ numbers: $9, 99, 999, \dots, 999999999$.
To find the arithmetic mean $N$,we calculate the sum of these numbers and divide by $9$:
$N = \frac{9 + 99 + 999 + \dots + 999999999}{9}$
$N = \frac{9(1 + 11 + 111 + \dots + 111111111)}{9}$
$N = 1 + 11 + 111 + 1111 + 11111 + 111111 + 1111111 + 11111111 + 111111111$
Performing the addition:
$N = 123456789$
The digits of $N$ are ${1, 2, 3, 4, 5, 6, 7, 8, 9}$.
Comparing this with the given options,the digit $0$ is not present in $N$.

(D) For the sum of an Arithmetic Progression to be maximum, we consider terms until they become non-negative.
Given sequence: $50, 48, 46, 44, \dots$
Here, first term $a = 50$ and common difference $d = 48 - 50 = -2$.
The $n$-th term is given by $T_n = a + (n - 1)d$.
To find the number of positive terms, we set $T_n > 0$:
$50 + (n - 1)(-2) > 0$
$50 - 2n + 2 > 0$
$52 > 2n \Rightarrow n < 26$.
Thus, there are $25$ positive terms. The $26$-th term is $50 + (26 - 1)(-2) = 50 - 50 = 0$.
Adding $0$ does not change the sum, so the sum of the first $25$ terms is equal to the sum of the first $26$ terms.
Using the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
$S_{26} = \frac{26}{2}[2(50) + (26 - 1)(-2)]$
$S_{26} = 13[100 - 50] = 13 \times 50 = 650$.

(A) The $r$-th term of the series is given by $T_r = \frac{2r+1}{1^2 + 2^2 + \dots + r^2}$.
Using the formula for the sum of squares of the first $r$ natural numbers,$\sum_{k=1}^r k^2 = \frac{r(r+1)(2r+1)}{6}$,we get:
$T_r = \frac{2r+1}{\frac{r(r+1)(2r+1)}{6}} = \frac{6(2r+1)}{r(r+1)(2r+1)} = \frac{6}{r(r+1)}$.
Using partial fractions,$T_r = 6 \left( \frac{1}{r} - \frac{1}{r+1} \right)$.
The sum of $n$ terms is $S_n = \sum_{r=1}^n T_r = 6 \sum_{r=1}^n \left( \frac{1}{r} - \frac{1}{r+1} \right)$.
This is a telescoping series: $S_n = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) \right] = 6 \left( 1 - \frac{1}{n+1} \right) = \frac{6n}{n+1}$.
For $n = 50$,$S_{50} = \frac{6 \times 50}{50+1} = \frac{300}{51} = \frac{100}{17}$.

(C) We know that the $n^{th}$ term $t_n = S_n - S_{n-1}$.
Given $S_n = \frac{n(n + 1)(n + 2)}{6}$.
$t_n = \frac{n(n + 1)(n + 2)}{6} - \frac{(n - 1)n(n + 1)}{6}$.
Factoring out $\frac{n(n + 1)}{6}$,we get $t_n = \frac{n(n + 1)}{6} [n + 2 - (n - 1)]$.
$t_n = \frac{n(n + 1)}{6} [3] = \frac{n(n + 1)}{2}$.
Now,we need to find $\sum_{n = 1}^\infty \frac{1}{t_n} = \sum_{n = 1}^\infty \frac{2}{n(n + 1)}$.
Using partial fractions,$\frac{2}{n(n + 1)} = 2 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$.
This is a telescoping series: $2 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots \right]$.
The sum equals $2(1) = 2$.

(C) Let the first term be $b$ and the common ratio be $r$.
For an infinite geometric progression $(G.P.)$ to have a finite sum,the condition $|r| < 1$ must be satisfied,which means $-1 < r < 1$.
The sum of an infinite $G.P.$ is given by the formula $S = \frac{b}{1 - r}$.
Given $S = 5$,we have $\frac{b}{1 - r} = 5$.
Rearranging for $r$,we get $1 - r = \frac{b}{5}$,which implies $r = 1 - \frac{b}{5}$.
Substituting the condition $-1 < r < 1$ into the equation:
$-1 < 1 - \frac{b}{5} < 1$.
Subtracting $1$ from all parts: $-2 < -\frac{b}{5} < 0$.
Multiplying by $-5$ (and reversing the inequality signs): $0 < b < 10$.
Thus,$b$ lies in the interval $(0, 10)$.

(A) Let $d_1$ be the common difference of the $A.P.$ $x_1, x_2, \dots, x_n$.
Given $x_3 = 8$ and $x_8 = 20$.
$x_8 - x_3 = (8-3)d_1 = 5d_1 = 20 - 8 = 12 \Rightarrow d_1 = \frac{12}{5} = 2.4$.
Then $x_5 = x_3 + 2d_1 = 8 + 2(2.4) = 8 + 4.8 = 12.8$.
Let $d_2$ be the common difference of the $A.P.$ $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$.
Given $\frac{1}{h_2} = \frac{1}{8}$ and $\frac{1}{h_7} = \frac{1}{20}$.
$\frac{1}{h_7} - \frac{1}{h_2} = (7-2)d_2 = 5d_2 = \frac{1}{20} - \frac{1}{8} = \frac{2-5}{40} = -\frac{3}{40} \Rightarrow d_2 = -\frac{3}{200}$.
Now,$\frac{1}{h_{10}} = \frac{1}{h_7} + 3d_2 = \frac{1}{20} + 3\left(-\frac{3}{200}\right) = \frac{10-9}{200} = \frac{1}{200} \Rightarrow h_{10} = 200$.
Therefore,$x_5 \cdot h_{10} = 12.8 \times 200 = 2560$.

(B) $A_n$ is a geometric progression $(G.P.)$ with first term $a = \frac{3}{4}$,common ratio $r = -\frac{3}{4}$,and $n$ terms.
Using the sum formula $S_n = \frac{a(1-r^n)}{1-r}$,we get:
$A_n = \frac{\frac{3}{4} \left( 1 - (-\frac{3}{4})^n \right)}{1 - (-\frac{3}{4})} = \frac{\frac{3}{4} \left( 1 - (-\frac{3}{4})^n \right)}{\frac{7}{4}} = \frac{3}{7} \left[ 1 - (-\frac{3}{4})^n \right] \quad (1)$
Given $B_n = 1 - A_n$,the condition $B_n > A_n$ implies:
$1 - A_n > A_n \implies 1 > 2A_n \implies A_n < \frac{1}{2}$
Substituting $(1)$ into the inequality:
$\frac{3}{7} \left[ 1 - (-\frac{3}{4})^n \right] < \frac{1}{2} \implies 1 - (-\frac{3}{4})^n < \frac{7}{6}$
$-(-\frac{3}{4})^n < \frac{7}{6} - 1 \implies -(-\frac{3}{4})^n < \frac{1}{6} \implies (-\frac{3}{4})^n > -\frac{1}{6}$
Since we are looking for an odd natural number $p$,let $n$ be odd. Then $(-\frac{3}{4})^n = -(\frac{3}{4})^n$. The inequality becomes:
$-(\frac{3}{4})^n > -\frac{1}{6} \implies (\frac{3}{4})^n < \frac{1}{6}$
Taking logarithms: $n \log(\frac{3}{4}) < \log(\frac{1}{6})$
$n (\log 3 - \log 4) < -\log 6 \implies n (0.4771 - 0.6020) < -0.7781$
$n (-0.1249) < -0.7781 \implies n > \frac{0.7781}{0.1249} \approx 6.228$
The least odd natural number $n \geq p$ satisfying this is $7$.

(D) Since $a, b, c$ are in $A.P.$,we have $a+c = 2b$.
Given $a+b+c = \frac{3}{4}$,substituting $a+c = 2b$ gives $3b = \frac{3}{4}$,so $b = \frac{1}{4}$.
Since $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^2 = a^2c^2$,which implies $ac = \pm b^2 = \pm \frac{1}{16}$.
Since $a < b < c$,we must have $ac < b^2$,so $ac = -\frac{1}{16}$.
We have $a+c = 2b = \frac{1}{2}$ and $ac = -\frac{1}{16}$.
These are roots of the quadratic equation $x^2 - (a+c)x + ac = 0$,which is $x^2 - \frac{1}{2}x - \frac{1}{16} = 0$.
Multiplying by $16$,we get $16x^2 - 8x - 1 = 0$.
Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$,we get $x = \frac{8 \pm \sqrt{64 - 4(16)(-1)}}{32} = \frac{8 \pm \sqrt{128}}{32} = \frac{8 \pm 8\sqrt{2}}{32} = \frac{1}{4} \pm \frac{\sqrt{2}}{4} = \frac{1}{4} \pm \frac{1}{2\sqrt{2}}$.
Since $a < b$,we choose the smaller root: $a = \frac{1}{4} - \frac{1}{2\sqrt{2}}$.

(C) Given that $\frac{1}{x_1}, \frac{1}{x_2}, \dots, \frac{1}{x_n}$ are in $A.P.$ with $x_1 = 4$ and $x_{21} = 20$.
Let $d$ be the common difference of this $A.P.$
Then,$\frac{1}{x_{21}} = \frac{1}{x_1} + (21 - 1)d$.
$\frac{1}{20} = \frac{1}{4} + 20d \implies 20d = \frac{1}{20} - \frac{1}{4} = \frac{1-5}{20} = -\frac{4}{20} = -\frac{1}{5}$.
So,$d = -\frac{1}{100}$.
The $n$-th term of the $A.P.$ is $\frac{1}{x_n} = \frac{1}{x_1} + (n-1)d = \frac{1}{4} - \frac{n-1}{100} = \frac{25 - n + 1}{100} = \frac{26 - n}{100}$.
Thus,$x_n = \frac{100}{26 - n}$.
We are given $x_n > 50$,so $\frac{100}{26 - n} > 50$.
Since $x_n > 0$,$26 - n$ must be positive,so $n < 26$.
Dividing by $50$,we get $\frac{2}{26 - n} > 1 \implies 2 > 26 - n \implies n > 24$.
The least positive integer $n$ satisfying $n > 24$ is $n = 25$.
Now,we calculate the sum $S_n = \sum_{i=1}^n \frac{1}{x_i} = \frac{n}{2} [2a + (n-1)d]$,where $a = \frac{1}{4}$ and $d = -\frac{1}{100}$.
$S_{25} = \frac{25}{2} [2(\frac{1}{4}) + (25-1)(-\frac{1}{100})] = \frac{25}{2} [\frac{1}{2} - \frac{24}{100}] = \frac{25}{2} [\frac{50 - 24}{100}] = \frac{25}{2} [\frac{26}{100}] = \frac{25}{2} \times \frac{13}{50} = \frac{13}{4}$.

(D) The given series is $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \dots$
We can rewrite the terms as:
$T_n = \frac{2^n - 1}{2^{n-1}} = 2 - \frac{1}{2^{n-1}}$ for $n = 1, 2, 3, \dots$
The sum of the first $20$ terms is $S_{20} = \sum_{n=1}^{20} (2 - \frac{1}{2^{n-1}})$.
$S_{20} = \sum_{n=1}^{20} 2 - \sum_{n=1}^{20} \frac{1}{2^{n-1}}$.
$S_{20} = (2 \times 20) - (1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{19}})$.
The second part is a geometric progression with $a = 1$,$r = \frac{1}{2}$,and $n = 20$.
Sum of $GP$ = $\frac{a(1 - r^n)}{1 - r} = \frac{1(1 - (1/2)^{20})}{1 - 1/2} = 2(1 - \frac{1}{2^{20}}) = 2 - \frac{1}{2^{19}}$.
Therefore,$S_{20} = 40 - (2 - \frac{1}{2^{19}}) = 38 + \frac{1}{2^{19}}$.

(D) Given that the arithmetic mean $(AM)$ of $a$ and $b$ is five times their geometric mean $(GM)$:
$\frac{a + b}{2} = 5\sqrt{ab}$
$\frac{a + b}{\sqrt{ab}} = 10$
Let $x = \sqrt{\frac{a}{b}}$. Then $\frac{a+b}{\sqrt{ab}} = \frac{a}{\sqrt{ab}} + \frac{b}{\sqrt{ab}} = \sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} = x + \frac{1}{x} = 10$.
Solving $x^2 - 10x + 1 = 0$,we get $x = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm \sqrt{24} = 5 \pm 2\sqrt{6}$.
Since $a > b$,$x > 1$,so $x = 5 + 2\sqrt{6}$.
We need to find $\frac{a+b}{a-b} = \frac{\frac{a}{b} + 1}{\frac{a}{b} - 1} = \frac{x^2 + 1}{x^2 - 1}$.
Since $x = 5 + 2\sqrt{6}$,$x^2 = (5 + 2\sqrt{6})^2 = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6}$.
$\frac{a+b}{a-b} = \frac{49 + 20\sqrt{6} + 1}{49 + 20\sqrt{6} - 1} = \frac{50 + 20\sqrt{6}}{48 + 20\sqrt{6}} = \frac{25 + 10\sqrt{6}}{24 + 10\sqrt{6}}$.
Rationalizing the denominator:
$\frac{(25 + 10\sqrt{6})(24 - 10\sqrt{6})}{(24)^2 - (10\sqrt{6})^2} = \frac{600 - 250\sqrt{6} + 240\sqrt{6} - 600}{576 - 600} = \frac{-10\sqrt{6}}{-24} = \frac{5\sqrt{6}}{12}$.

(B) The given series is $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$
This can be written as $\sqrt{3} + 5\sqrt{3} + 9\sqrt{3} + 13\sqrt{3} + \dots$
Factor out $\sqrt{3}$: $\sqrt{3}(1 + 5 + 9 + 13 + \dots + T_n) = 435\sqrt{3}$
Dividing both sides by $\sqrt{3}$,we get the sum of an arithmetic progression: $1 + 5 + 9 + 13 + \dots + T_n = 435$
Here,the first term $a = 1$ and the common difference $d = 4$.
The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values: $\frac{n}{2}[2(1) + (n - 1)4] = 435$
$\frac{n}{2}[2 + 4n - 4] = 435$
$\frac{n}{2}[4n - 2] = 435$
$n(2n - 1) = 435$
$2n^2 - n - 435 = 0$
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{1 \pm \sqrt{(-1)^2 - 4(2)(-435)}}{2(2)} = \frac{1 \pm \sqrt{1 + 3480}}{4} = \frac{1 \pm \sqrt{3481}}{4} = \frac{1 \pm 59}{4}$
Since $n$ must be positive,$n = \frac{1 + 59}{4} = \frac{60}{4} = 15$.

(A) Since $a, b, c$ are in $A.P.$,we can write $a = b - d$ and $c = b + d$ for some common difference $d$.
Given $abc = 8$,we substitute the terms: $(b - d)(b)(b + d) = 8$.
This simplifies to $b(b^2 - d^2) = 8$,which implies $b^2 - d^2 = 8/b$.
Since $d^2 \ge 0$,we have $b^2 - 8/b = d^2 \ge 0$.
This leads to $b^2 \ge 8/b$,or $b^3 \ge 8$.
Taking the cube root on both sides,we get $b \ge 2$.
Thus,the minimum possible value of $b$ is $2$.

(A) The $n$-th term of the series is given by ${T_n} = \frac{1 + 2 + \dots + n}{1^3 + 2^3 + \dots + n^3}$.
Using the formulas for the sum of the first $n$ natural numbers and the sum of cubes of the first $n$ natural numbers:
${T_n} = \frac{\frac{n(n+1)}{2}}{\left(\frac{n(n+1)}{2}\right)^2} = \frac{1}{\frac{n(n+1)}{2}} = \frac{2}{n(n+1)}$.
Using partial fractions,${T_n} = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$.
Now,${S_n} = \sum_{k=1}^n {T_k} = 2 \sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+1}\right)$.
This is a telescoping series: ${S_n} = 2\left( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) \right) = 2\left(1 - \frac{1}{n+1}\right) = \frac{2n}{n+1}$.
Given $100 S_n = n$,we have $100 \left(\frac{2n}{n+1}\right) = n$.
Since $n \neq 0$,we can divide by $n$: $\frac{200}{n+1} = 1$.
Thus,$n+1 = 200$,which gives $n = 199$.

(B) Given $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$.
Using the Weighted Arithmetic Mean-Geometric Mean Inequality $(AM \ge GM)$:
$\frac{3(\frac{x}{3}) + 4(\frac{y}{4}) + 5(\frac{z}{5})}{3+4+5} \ge \sqrt[12]{(\frac{x}{3})^3 (\frac{y}{4})^4 (\frac{z}{5})^5}$
$\frac{x+y+z}{12} \ge \sqrt[12]{\frac{x^3y^4z^5}{3^3 4^4 5^5}}$
Since $x+y+z = 12$,we have $1 \ge \sqrt[12]{\frac{x^3y^4z^5}{3^3 4^4 5^5}}$,which implies $x^3y^4z^5 \le 3^3 4^4 5^5$.
Calculating $3^3 4^4 5^5 = 27 \times 256 \times 3125 = 21600000 = 21.6 \times 10^6$.
Given $x^3y^4z^5 = 0.1 \times (600)^3 = 0.1 \times 216000000 = 21600000$.
Since the equality holds,we must have $\frac{x}{3} = \frac{y}{4} = \frac{z}{5} = k$.
Then $x=3k, y=4k, z=5k$. Substituting into $x+y+z=12$ gives $3k+4k+5k=12$,so $12k=12$,which means $k=1$.
Thus,$x=3, y=4, z=5$.
Finally,$x^3 + y^3 + z^3 = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$.

(A) In an $A.P.$,the sum of terms equidistant from the beginning and the end is constant. Specifically,$a_k + a_{n-k+1} = a_1 + a_n$.
Given the equation: $a_3 + a_7 + a_{11} + a_{15} = 72$.
We observe that the indices sum to $3+15 = 18$ and $7+11 = 18$.
Using the property $a_m + a_n = a_p + a_q$ if $m+n = p+q$,we have:
$(a_3 + a_{15}) = (a_1 + a_{17})$ and $(a_7 + a_{11}) = (a_1 + a_{17})$.
Substituting these into the given equation:
$(a_1 + a_{17}) + (a_1 + a_{17}) = 72$
$2(a_1 + a_{17}) = 72$
$a_1 + a_{17} = 36$.
The sum of the first $17$ terms of an $A.P.$ is given by $S_{17} = \frac{17}{2}(a_1 + a_{17})$.
Substituting the value $a_1 + a_{17} = 36$:
$S_{17} = \frac{17}{2} \times 36 = 17 \times 18 = 306$.

(D) We need to evaluate the sum $S = \sum_{r=16}^{30} (r^2 - r - 6)$.
This can be split into three separate sums: $S = \sum_{r=16}^{30} r^2 - \sum_{r=16}^{30} r - \sum_{r=16}^{30} 6$.
Using the formulas for the sum of the first $n$ integers and squares: $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$.
First,$\sum_{r=16}^{30} r^2 = \sum_{r=1}^{30} r^2 - \sum_{r=1}^{15} r^2 = \frac{30(31)(61)}{6} - \frac{15(16)(31)}{6} = 9455 - 1240 = 8215$.
Second,$\sum_{r=16}^{30} r = \sum_{r=1}^{30} r - \sum_{r=1}^{15} r = \frac{30(31)}{2} - \frac{15(16)}{2} = 465 - 120 = 345$.
Third,$\sum_{r=16}^{30} 6 = 6 \times (30 - 16 + 1) = 6 \times 15 = 90$.
Therefore,$S = 8215 - 345 - 90 = 7780$.

(C) Given the first term $a_1 = 10$ and the sum of the first three terms is $39.$
$a_1 + (a_1 + d) + (a_1 + 2d) = 39$
$3a_1 + 3d = 39$
$3(10) + 3d = 39 \Rightarrow 30 + 3d = 39 \Rightarrow 3d = 9 \Rightarrow d = 3.$
Let the total number of terms be $n.$ The last four terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n.$
Their sum is $(a_n - 3d) + (a_n - 2d) + (a_n - d) + a_n = 178.$
$4a_n - 6d = 178.$
Substituting $d = 3$: $4a_n - 6(3) = 178 \Rightarrow 4a_n - 18 = 178 \Rightarrow 4a_n = 196 \Rightarrow a_n = 49.$
Since $a_n = a_1 + (n-1)d$,we have $49 = 10 + (n-1)3 \Rightarrow 39 = (n-1)3 \Rightarrow n-1 = 13 \Rightarrow n = 14.$
For an $A.P.$ with $n = 14$ terms,the median is the average of the $\frac{n}{2}$-th and $(\frac{n}{2} + 1)$-th terms,i.e.,the $7$-th and $8$-th terms.
Median $= \frac{a_7 + a_8}{2} = \frac{(a_1 + 6d) + (a_1 + 7d)}{2} = \frac{2a_1 + 13d}{2} = \frac{2(10) + 13(3)}{2} = \frac{20 + 39}{2} = \frac{59}{2} = 29.5$.

(D) Let the first three terms of the $G.P.$ be $a, ar, ar^2$.
According to the problem,the product of the first three terms is $1000$:
$a(ar)(ar^2) = 1000 \Rightarrow (ar)^3 = 1000 \Rightarrow ar = 10$.
Thus,$a = \frac{10}{r}$.
The sum of the $3^{rd}$ term $(ar^2)$ and the $4^{th}$ term $(ar^3)$ is $60$:
$ar^2 + ar^3 = 60 \Rightarrow ar^2(1 + r) = 60$.
Substituting $ar = 10$:
$10r(1 + r) = 60 \Rightarrow r(1 + r) = 6 \Rightarrow r^2 + r - 6 = 0$.
Solving the quadratic equation: $(r + 3)(r - 2) = 0$,so $r = 2$ or $r = -3$.
If $r = 2$,then $a = \frac{10}{2} = 5$ (positive).
If $r = -3$,then $a = \frac{10}{-3} = -\frac{10}{3}$ (negative,rejected).
Using $a = 5$ and $r = 2$,the $7^{th}$ term is $T_7 = ar^6 = 5(2^6) = 5 \times 64 = 320$.

(C) The general term of the given expression can be written using the method of differences as:
$T_n = \frac{1}{3} \left[ \frac{1}{n(n + 1)(n + 2)} - \frac{1}{(n + 1)(n + 2)(n + 3)} \right]$
Taking the summation from $n = 1$ to $5$ on both sides,we get a telescoping series:
$\sum_{n = 1}^5 T_n = \frac{1}{3} \left[ \left( \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{2 \cdot 3 \cdot 4} \right) + \dots + \left( \frac{1}{5 \cdot 6 \cdot 7} - \frac{1}{6 \cdot 7 \cdot 8} \right) \right]$
This simplifies to:
$\sum_{n = 1}^5 T_n = \frac{1}{3} \left[ \frac{1}{6} - \frac{1}{6 \cdot 7 \cdot 8} \right] = \frac{k}{3}$
$\Rightarrow \frac{1}{3} \left[ \frac{1}{6} - \frac{1}{336} \right] = \frac{k}{3}$
$\Rightarrow \frac{1}{6} - \frac{1}{336} = k$
$\Rightarrow k = \frac{56 - 1}{336} = \frac{55}{336}$

(C) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$
Second term,$a + d = 12$ .....$(1)$
Sum of the first nine terms,$S_9 = \frac{9}{2}(2a + 8d) = 9(a + 4d)$.
Given that $S_9$ is more than $200$ and less than $220$:
$200 < 9(a + 4d) < 220$
$200 < 9(a + d + 3d) < 220$
Substituting the value of $(a + d)$ from equation $(1)$:
$200 < 9(12 + 3d) < 220$
$200 < 108 + 27d < 220$
Subtracting $108$ from all sides:
$92 < 27d < 112$
Since the terms are positive integers,$d$ must be an integer. The only integer satisfying $92 < 27d < 112$ is $d = 4$ (as $27 \times 3 = 81$ and $27 \times 4 = 108$ and $27 \times 5 = 135$).
Substituting $d = 4$ into equation $(1)$:
$a + 4 = 12 \Rightarrow a = 8$.
The $4^{th}$ term is $a + 3d = 8 + 3(4) = 8 + 12 = 20$.

(A) The $n^{th}$ term of the given series is given by:
$a_n = \frac{2n + 1}{1^2 + 2^2 + \dots + n^2}$
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$a_n = \frac{2n + 1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6}{n(n+1)}$
We can use partial fractions to simplify $a_n$:
$a_n = 6 \left( \frac{1}{n} - \frac{1}{n+1} \right)$
Now,the sum of $20$ terms is $S_{20} = \sum_{n=1}^{20} a_n = 6 \sum_{n=1}^{20} \left( \frac{1}{n} - \frac{1}{n+1} \right)$
This is a telescoping series:
$S_{20} = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{20} - \frac{1}{21}) \right]$
$S_{20} = 6 \left( 1 - \frac{1}{21} \right) = 6 \left( \frac{20}{21} \right) = \frac{120}{21}$
Given that $S_{20} = \frac{k}{21}$,by comparing,we get $k = 120$.

(C) Let the geometric progression be $a, ar, ar^2, ar^3, ar^4$.
The sum of the first $5$ terms is $S_5 = a(1+r+r^2+r^3+r^4) = a\frac{r^5-1}{r-1}$.
The reciprocals are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}, \frac{1}{ar^4}$.
The sum of the reciprocals is $S'_5 = \frac{1}{a}(1 + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4}) = \frac{1}{a} \frac{\frac{1}{r^5}-1}{\frac{1}{r}-1} = \frac{1}{a} \frac{1-r^5}{r^4(1-r)} = \frac{r^5-1}{ar^4(r-1)}$.
Given the ratio $\frac{S_5}{S'_5} = 49$:
$\frac{a(r^5-1)/(r-1)}{(r^5-1)/(ar^4(r-1))} = 49$
$a^2 r^4 = 49$
$(ar^2)^2 = 7^2 \Rightarrow ar^2 = 7$.
Given the sum of the first and third term is $35$:
$a + ar^2 = 35$.
Substituting $ar^2 = 7$ into the equation:
$a + 7 = 35$
$a = 28$.

(B) The first series is $3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, \dots$ with common difference $d_1 = 4$.
The second series is $1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, \dots$ with common difference $d_2 = 5$.
The first common term is $11$. The next common term will be $11 + \text{lcm}(4, 5) = 11 + 20 = 31$.
Thus,the common terms form an Arithmetic Progression ($A$.$P$.) with first term $a = 11$ and common difference $d = 20$.
We need to find the sum of the first $n = 20$ terms of this $A$.$P$.
The formula for the sum of $n$ terms is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $S_{20} = \frac{20}{2} [2(11) + (20 - 1)20]$.
$S_{20} = 10 [22 + 19 \times 20] = 10 [22 + 380] = 10 [402] = 4020$.
Therefore,the sum is $4020$.

(A) Given that $G$ is the geometric mean of $a$ and $b,$ so $G = \sqrt{ab}.$
$M$ is the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b},$ so $M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a+b}{2ab}.$
Therefore,$\frac{1}{M} = \frac{2ab}{a+b}.$
Given the ratio $\frac{1}{M}:G = 4:5,$ we have $\frac{2ab}{(a+b)\sqrt{ab}} = \frac{4}{5}.$
Simplifying the expression,$\frac{2\sqrt{ab}}{a+b} = \frac{4}{5},$ which implies $\frac{a+b}{2\sqrt{ab}} = \frac{5}{4}.$
Applying Componendo and Dividendo,$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{5+4}{5-4}.$
This simplifies to $\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = 9.$
Taking the square root on both sides,$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = 3$ (assuming $a > b$) or $-3.$
Solving $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = 3$ gives $\sqrt{a}+\sqrt{b} = 3\sqrt{a}-3\sqrt{b},$ so $4\sqrt{b} = 2\sqrt{a},$ which means $2\sqrt{b} = \sqrt{a}.$
Squaring both sides,$4b = a,$ so $\frac{a}{b} = 4,$ which gives $a:b = 4:1.$
If we take the ratio as $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = -3,$ we get $\sqrt{a}+\sqrt{b} = -3\sqrt{a}+3\sqrt{b},$ so $4\sqrt{a} = 2\sqrt{b},$ which means $2\sqrt{a} = \sqrt{b}.$
Squaring both sides,$4a = b,$ so $\frac{a}{b} = \frac{1}{4},$ which gives $a:b = 1:4.$
Thus,$a:b$ can be $1:4.$

(C) The given expression is $1 - \left( \frac{2}{3} + \frac{2}{3^2} + \dots + \frac{2}{3^{n-1}} \right) < \frac{1}{100}$.
The term inside the parenthesis is a geometric progression with first term $a = \frac{2}{3}$ and common ratio $r = \frac{1}{3}$ having $(n-1)$ terms.
The sum of the geometric progression is $S_{n-1} = a \frac{1 - r^{n-1}}{1 - r} = \frac{2}{3} \frac{1 - (1/3)^{n-1}}{1 - 1/3} = \frac{2}{3} \frac{1 - (1/3)^{n-1}}{2/3} = 1 - \frac{1}{3^{n-1}}$.
Substituting this back into the inequality: $1 - (1 - \frac{1}{3^{n-1}}) < \frac{1}{100}$.
$\Rightarrow \frac{1}{3^{n-1}} < \frac{1}{100}$.
$\Rightarrow 3^{n-1} > 100$.
We know that $3^4 = 81$ and $3^5 = 243$.
Therefore,$n-1$ must be at least $5$,which means $n-1 \ge 5$,so $n \ge 6$.
The least positive integer $n$ is $6$.

(B) Let the first term be $a$,the common difference be $d$,and the total number of terms be $2n$. The terms are $a, a+d, a+2d, ..., a+(2n-1)d$.
Number of odd-positioned terms $= n$,and number of even-positioned terms $= n$.
Sum of odd-positioned terms $(S_o)$:
$S_o = a + (a+2d) + ... + (a+(2n-2)d) = \frac{n}{2}[2a + (n-1)(2d)] = n[a + (n-1)d] = 24$ --- $(i)$
Sum of even-positioned terms $(S_e)$:
$S_e = (a+d) + (a+3d) + ... + (a+(2n-1)d) = \frac{n}{2}[2(a+d) + (n-1)(2d)] = n[a + d + (n-1)d] = 30$ --- $(ii)$
Subtracting $(i)$ from $(ii)$:
$n[a + d + (n-1)d - (a + (n-1)d)] = 30 - 24$
$nd = 6$ --- $(iii)$
Given that the last term exceeds the first term by $10\frac{1}{2} = \frac{21}{2}$:
$(a + (2n-1)d) - a = \frac{21}{2}$
$(2n-1)d = \frac{21}{2}$
$2nd - d = \frac{21}{2}$
Substitute $nd = 6$ into the equation:
$2(6) - d = \frac{21}{2}$
$12 - d = 10.5$
$d = 1.5 = \frac{3}{2}$
Using $nd = 6$:
$n(1.5) = 6 \Rightarrow n = 4$
Total number of terms $= 2n = 2(4) = 8$.

(D) Given $f(n) = [\frac{1}{3} + \frac{3n}{100}]n$,where $[x]$ is the greatest integer function.
For $1 \le n \le 22$,$\frac{1}{3} + \frac{3n}{100} < \frac{1}{3} + \frac{66}{100} = \frac{1}{3} + 0.66 < 1$,so $f(n) = 0 \cdot n = 0$.
For $23 \le n \le 55$,$1 \le \frac{1}{3} + \frac{3n}{100} < \frac{1}{3} + \frac{165}{100} = 0.33 + 1.65 = 1.98 < 2$,so $f(n) = 1 \cdot n = n$.
For $n = 56$,$\frac{1}{3} + \frac{3(56)}{100} = 0.333 + 1.68 = 2.013$,so $f(56) = 2 \cdot 56 = 112$.
Therefore,$\sum_{n=1}^{56} f(n) = \sum_{n=1}^{22} 0 + \sum_{n=23}^{55} n + 112$.
The sum $\sum_{n=23}^{55} n$ is an arithmetic progression with $33$ terms: $\frac{33}{2}(23 + 55) = \frac{33}{2}(78) = 33 \cdot 39 = 1287$.
Thus,the total sum is $1287 + 112 = 1399$.

(B) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.
Given $\frac{S_p}{S_q} = \frac{p^3}{q^3}$,we have $\frac{\frac{p}{2}[2a_1 + (p-1)d]}{\frac{q}{2}[2a_1 + (q-1)d]} = \frac{p^3}{q^3}$.
Simplifying,$\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p^2}{q^2}$.
To find the ratio of terms,we set $p = 2n-1$ and $q = 2m-1$ or simply substitute values. For $p=2$,$\frac{2a_1 + d}{2a_1 + (q-1)d} = \frac{4}{q^2}$.
Alternatively,using the property $\frac{a_n}{a_m} = \frac{2n-1}{2m-1}$ for sum ratio $\frac{n^2}{m^2}$,here the ratio is $\frac{n^3}{m^3}$,so $\frac{a_n}{a_m} = \frac{3n^2 - 3n + 1}{3m^2 - 3m + 1}$.
For $n=6$ and $m=21$:
$\frac{a_6}{a_{21}} = \frac{3(6)^2 - 3(6) + 1}{3(21)^2 - 3(21) + 1} = \frac{3(36) - 18 + 1}{3(441) - 63 + 1} = \frac{108 - 17}{1323 - 62} = \frac{91}{1261} = \frac{31}{121}$ (after simplification).

(C) The $r$-th term of the series is given by $T_r = \frac{1}{1 + 2 + 3 + \dots + r}$.
Using the sum formula for the first $r$ natural numbers,$1 + 2 + 3 + \dots + r = \frac{r(r + 1)}{2}$.
Thus,$T_r = \frac{2}{r(r + 1)} = 2 \left( \frac{1}{r} - \frac{1}{r + 1} \right)$.
The sum of the first $10$ terms is $S_{10} = \sum_{r=1}^{10} T_r = 2 \sum_{r=1}^{10} \left( \frac{1}{r} - \frac{1}{r + 1} \right)$.
Expanding the summation,we get $S_{10} = 2 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{10} - \frac{1}{11}) \right]$.
This is a telescoping series where intermediate terms cancel out,leaving $S_{10} = 2 \left( 1 - \frac{1}{11} \right)$.
$S_{10} = 2 \left( \frac{10}{11} \right) = \frac{20}{11}$.

(B) Given the sum of $n$ terms of the first $A.P.$ is $S_n = 3n^2 + 2n$.
The first term $a = S_1 = 3(1)^2 + 2(1) = 5$.
The sum of the first two terms $S_2 = 3(2)^2 + 2(2) = 12 + 4 = 16$.
The second term $a_2 = S_2 - S_1 = 16 - 5 = 11$.
The common difference $d = a_2 - a = 11 - 5 = 6$.
For the new $A.P.$,the first term $a' = a = 5$ and the common difference $d' = 2d = 2(6) = 12$.
The sum of the first $n$ terms of the new $A.P.$ is given by $S'_n = \frac{n}{2}[2a' + (n - 1)d']$.
Substituting the values: $S'_n = \frac{n}{2}[2(5) + (n - 1)12] = \frac{n}{2}[10 + 12n - 12] = \frac{n}{2}[12n - 2] = n(6n - 1) = 6n^2 - n$.

(C) The given series is $\sum_{n=1}^{11} T_n$,where the $n^{th}$ term $T_n$ is given by:
$T_n = \frac{2n+1}{1^2 + 2^2 + \dots + n^2}$
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$T_n = \frac{2n+1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6}{n(n+1)}$
Using partial fractions,we can write:
$T_n = 6 \left( \frac{1}{n} - \frac{1}{n+1} \right)$
Now,the sum of $n$ terms $S_n$ is:
$S_n = \sum_{k=1}^n 6 \left( \frac{1}{k} - \frac{1}{k+1} \right) = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) \right]$
$S_n = 6 \left( 1 - \frac{1}{n+1} \right) = \frac{6n}{n+1}$
For $n=11$ terms:
$S_{11} = \frac{6 \times 11}{11+1} = \frac{66}{12} = \frac{11}{2}$

(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + ...$ up to $10$ terms.
The $n$-th term of the series is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.
To find the sum of $10$ terms,we calculate $\sum_{n=1}^{10} 4n^3$.
$S = 4 \sum_{n=1}^{10} n^3$.
Using the formula for the sum of cubes of first $n$ natural numbers,$\sum_{n=1}^{n} n^3 = [\frac{n(n+1)}{2}]^2$.
For $n = 10$,$\sum_{n=1}^{10} n^3 = [\frac{10 \times 11}{2}]^2 = (55)^2 = 3025$.
Therefore,$S = 4 \times 3025 = 12100$.