Progression and Sequence Questions in English

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{(n + a)(n + 1 + a)}$.
Using the method of partial fractions,the $n^{th}$ term $T_n$ can be written as:
$T_n = \frac{1}{(n + a)(n + 1 + a)} = \frac{1}{n + a} - \frac{1}{n + 1 + a}$.
Now,we write the sum of the first $n$ terms $S_n$:
$S_n = T_1 + T_2 + T_3 + \dots + T_n$
$S_n = \left( \frac{1}{1 + a} - \frac{1}{2 + a} \right) + \left( \frac{1}{2 + a} - \frac{1}{3 + a} \right) + \left( \frac{1}{3 + a} - \frac{1}{4 + a} \right) + \dots + \left( \frac{1}{n + a} - \frac{1}{n + 1 + a} \right)$.
This is a telescoping series where intermediate terms cancel out:
$S_n = \frac{1}{1 + a} - \frac{1}{n + 1 + a}$.
To find the sum to infinity,we take the limit as $n \to \infty$:
$S_{\infty} = \lim_{n \to \infty} \left( \frac{1}{1 + a} - \frac{1}{n + 1 + a} \right) = \frac{1}{1 + a} - 0 = \frac{1}{1 + a}$.

(B) Let the four angles of the quadrilateral be $x, x+10^o, x+20^o,$ and $x+30^o$ in $A.P.$
The sum of the interior angles of a quadrilateral is $360^o$.
Therefore,$x + (x+10^o) + (x+20^o) + (x+30^o) = 360^o$.
$4x + 60^o = 360^o$.
$4x = 300^o$.
$x = 75^o$.
The angles are $75^o, 75^o+10^o=85^o, 75^o+20^o=95^o,$ and $75^o+30^o=105^o$.
Thus,the angles are $75^o, 85^o, 95^o,$ and $105^o$.

(A) Let the first term be $a$ and the common difference be $d$.
The sum of the first $n$ terms is $S_n = \frac{n}{2} \{2a + (n - 1)d\}$.
The sum of the first $m$ terms is $S_m = \frac{m}{2} \{2a + (m - 1)d\}$.
Given that $S_n = S_m$,we have:
$\frac{n}{2} \{2a + (n - 1)d\} = \frac{m}{2} \{2a + (m - 1)d\}$
$n \{2a + (n - 1)d\} = m \{2a + (m - 1)d\}$
$2an + n(n - 1)d = 2am + m(m - 1)d$
$2a(n - m) + d(n^2 - n - m^2 + m) = 0$
$2a(n - m) + d((n^2 - m^2) - (n - m)) = 0$
$2a(n - m) + d((n - m)(n + m) - (n - m)) = 0$
Since $m \ne n$,we can divide by $(n - m)$:
$2a + d(n + m - 1) = 0$
Now,the sum of the first $(m + n)$ terms is:
$S_{m+n} = \frac{m+n}{2} \{2a + (m + n - 1)d\}$
Substituting $2a + (m + n - 1)d = 0$ into the equation:
$S_{m+n} = \frac{m+n}{2} \times 0 = 0$.

(A) Given that $p, q, r$ are in $A.P.$ and are positive.
Since they are in $A.P.$,we have $q = \frac{p + r}{2}$ ......$(i)$
For the quadratic equation $px^2 + qx + r = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = q^2 - 4pr \ge 0$
Substituting $q = \frac{p + r}{2}$ into the inequality:
$\left( \frac{p + r}{2} \right)^2 - 4pr \ge 0$
$\frac{p^2 + 2pr + r^2}{4} - 4pr \ge 0$
$p^2 + 2pr + r^2 - 16pr \ge 0$
$p^2 - 14pr + r^2 \ge 0$
Dividing by $p^2$ (since $p > 0$):
$1 - 14\left( \frac{r}{p} \right) + \left( \frac{r}{p} \right)^2 \ge 0$
Let $x = \frac{r}{p}$. Then $x^2 - 14x + 1 \ge 0$.
Completing the square: $(x - 7)^2 - 49 + 1 \ge 0$
$(x - 7)^2 \ge 48$
$|x - 7| \ge \sqrt{48} = 4\sqrt{3}$
Thus,$\left| \frac{r}{p} - 7 \right| \ge 4\sqrt{3}$.

(B) The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Given $a = 1$ for all three $A.P.'s$ and common differences $d_1 = 1, d_2 = 2, d_3 = 3$.
For $S_1$: $S_1 = \frac{n}{2}[2(1) + (n - 1)1] = \frac{n}{2}[n + 1]$.
For $S_2$: $S_2 = \frac{n}{2}[2(1) + (n - 1)2] = \frac{n}{2}[2n] = n^2$.
For $S_3$: $S_3 = \frac{n}{2}[2(1) + (n - 1)3] = \frac{n}{2}[3n - 1]$.
Now,calculating $S_1 + S_3$:
$S_1 + S_3 = \frac{n}{2}[n + 1 + 3n - 1] = \frac{n}{2}[4n] = 2n^2$.
Since $S_2 = n^2$,we have $2S_2 = 2n^2$.
Therefore,$S_1 + S_3 = 2S_2$.

(A) The given equation is: $\log_a x + \log_{a^{1/2}} x + \log_{a^{1/3}} x + \dots + \log_{a^{1/a}} x = \frac{a(a+1)}{2}$.
Using the property $\log_{a^k} x = \frac{1}{k} \log_a x$,we can rewrite the terms:
$\log_a x + 2 \log_a x + 3 \log_a x + \dots + a \log_a x = \frac{a(a+1)}{2}$.
Factor out $\log_a x$:
$\log_a x (1 + 2 + 3 + \dots + a) = \frac{a(a+1)}{2}$.
The sum of the first $a$ natural numbers is $\frac{a(a+1)}{2}$.
So,$\log_a x \cdot \frac{a(a+1)}{2} = \frac{a(a+1)}{2}$.
Dividing both sides by $\frac{a(a+1)}{2}$,we get $\log_a x = 1$.
Therefore,$x = a^1 = a$.

(C) The total cost of the house is Rs. $15000$. Jairam paid Rs. $5000$ upfront,so the remaining balance is $15000 - 5000 = 10000$.
He pays the remaining Rs. $10000$ in $10$ annual installments of Rs. $1000$ each,plus $10\%$ interest on the outstanding balance.
In the $1^{st}$ year,he pays: $1000 + (10\% \text{ of } 10000) = 1000 + 1000 = 2000$.
In the $2^{nd}$ year,the outstanding balance is $9000$,so he pays: $1000 + (10\% \text{ of } 9000) = 1000 + 900 = 1900$.
In the $3^{rd}$ year,the outstanding balance is $8000$,so he pays: $1000 + (10\% \text{ of } 8000) = 1000 + 800 = 1800$.
This forms an Arithmetic Progression $(AP)$ where the first term $a = 2000$,common difference $d = -100$,and number of terms $n = 10$.
The sum of these $10$ installments is $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_{10} = \frac{10}{2}[2(2000) + (10-1)(-100)] = 5[4000 - 900] = 5[3100] = 15500$.
The total amount paid by Jairam is the initial payment plus the sum of installments: $5000 + 15500 = 20500$.

(D) Let $x_n$ be the side length of square $S_n$. The diagonal of $S_{n+1}$ is $x_{n+1}\sqrt{2}$.
Given $x_n = x_{n+1}\sqrt{2}$,we have $x_{n+1} = \frac{x_n}{\sqrt{2}}$.
This forms a geometric progression for the side lengths: $x_n = x_1 \cdot (\frac{1}{\sqrt{2}})^{n-1}$.
The area of $S_n$ is $A_n = x_n^2 = x_1^2 \cdot (\frac{1}{2})^{n-1} = \frac{100}{2^{n-1}}$.
We want $A_n < 1$,so $\frac{100}{2^{n-1}} < 1$,which implies $2^{n-1} > 100$.
For $n=8$,$2^{8-1} = 2^7 = 128 > 100$.
For $n=9$,$2^{9-1} = 2^8 = 256 > 100$.
For $n=10$,$2^{10-1} = 2^9 = 512 > 100$.
Since all values $n=8, 9, 10$ satisfy the condition,the correct option is $(d)$.

(A) The sum of $n$ terms of an $A.P.$ is given by $S = \frac{n}{2}[2a + (n - 1)d]$.
For the $k$-th $A.P.$,the first term $a_k = k$ and the common difference $d_k = 2k - 1$.
Thus,$S_k = \frac{n}{2}[2k + (n - 1)(2k - 1)] = \frac{n}{2}[2k + 2kn - n - 2k + 1] = \frac{n}{2}[2kn - n + 1]$.
Now,we need to find the sum $\sum_{k=1}^{m} S_k = \sum_{k=1}^{m} \frac{n}{2}[2kn - n + 1]$.
$= \frac{n}{2} [2n \sum_{k=1}^{m} k + \sum_{k=1}^{m} (1 - n)]$.
$= \frac{n}{2} [2n \frac{m(m+1)}{2} + m(1 - n)]$.
$= \frac{n}{2} [nm(m+1) + m - mn] = \frac{n}{2} [nm^2 + nm + m - mn] = \frac{n}{2} [nm^2 + m] = \frac{mn(mn + 1)}{2}$.

(D) In an arithmetic progression $(A.P.)$,the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first and last terms.
Given: ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$.
We know that in an $A.P.$,${a_1} + {a_{24}} = {a_5} + {a_{20}} = {a_{10}} + {a_{15}}$.
Substituting these into the given equation:
$3({a_1} + {a_{24}}) = 225$
${a_1} + {a_{24}} = 75$.
The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}(a_1 + a_n)$.
For $n = 24$:
$S_{24} = \frac{24}{2}({a_1} + {a_{24}}) = 12 \times 75 = 900$.

(C) Let the roots of the cubic equation ${x^3} - 12{x^2} + 39x - 28 = 0$ be $a - d, a, a + d$,which are in $A.P.$
According to the properties of roots of a cubic equation:
Sum of roots = $(a - d) + a + (a + d) = -(-12)/1 = 12$
$3a = 12 \implies a = 4$
Product of roots = $(a - d) \cdot a \cdot (a + d) = -(-28)/1 = 28$
$a(a^2 - d^2) = 28$
Substitute $a = 4$ into the product equation:
$4(4^2 - d^2) = 28$
$16 - d^2 = 7$
$d^2 = 9$
$d = \pm 3$
Thus,the common difference is $\pm 3$.

(A) Let the first term of the $G.P.$ be $a_1 = 1$ and the common ratio be $r$.
Then the terms are $a_1 = 1$, $a_2 = r$, and $a_3 = r^2$.
We are given the expression $f(r) = 4a_2 + 5a_3 = 4r + 5r^2$.
To find the value of $r$ for which $f(r)$ is minimum, we differentiate $f(r)$ with respect to $r$ and set it to zero:
$f'(r) = \frac{d}{dr}(4r + 5r^2) = 4 + 10r$.
Setting $f'(r) = 0$, we get $4 + 10r = 0$, which implies $10r = -4$, so $r = -\frac{4}{10} = -\frac{2}{5}$.
Since the second derivative $f''(r) = 10 > 0$, the function has a minimum at $r = -\frac{2}{5}$.

(D) Let the $n$ terms of the $G.P.$ be $a, ar, ar^2, \dots, ar^{n-1}$.
The sum $S$ is given by $S = \frac{a(1 - r^n)}{1 - r}$ ......$(i)$
The product $P$ is given by $P = a \cdot (ar) \cdot (ar^2) \dots (ar^{n-1}) = a^n r^{0+1+2+\dots+(n-1)} = a^n r^{\frac{n(n-1)}{2}}$.
Squaring both sides,we get $P^2 = a^{2n} r^{n(n-1)}$ ......(ii)
The sum of the reciprocals $R$ is given by $R = \frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}}$.
This is a $G.P.$ with first term $\frac{1}{a}$ and common ratio $\frac{1}{r}$.
$R = \frac{\frac{1}{a}(1 - (\frac{1}{r})^n)}{1 - \frac{1}{r}} = \frac{\frac{1}{a}(\frac{r^n - 1}{r^n})}{\frac{r - 1}{r}} = \frac{r^n - 1}{a r^n} \cdot \frac{r}{r - 1} = \frac{r^n - 1}{a r^{n-1}(r - 1)} = \frac{1 - r^n}{a r^{n-1}(1 - r)}$ ......(iii)
Now,$\frac{S}{R} = \frac{a(1 - r^n)}{1 - r} \cdot \frac{a r^{n-1}(1 - r)}{1 - r^n} = a^2 r^{n-1}$.
Therefore,$(\frac{S}{R})^n = (a^2 r^{n-1})^n = a^{2n} r^{n(n-1)}$.
Comparing this with equation (ii),we get $P^2 = (\frac{S}{R})^n$.

(C) The given expression is a geometric series: $S = 1 + n + n^2 + \dots + n^{127} = \frac{n^{128} - 1}{n - 1}$.
We are given that $(n^m + 1)$ divides $S$.
Thus,$\frac{n^{128} - 1}{(n - 1)(n^m + 1)}$ must be an integer.
We know that $n^{128} - 1 = (n^{64} - 1)(n^{64} + 1)$.
Substituting this into the expression,we get $\frac{(n^{64} - 1)(n^{64} + 1)}{(n - 1)(n^m + 1)}$.
For this to be an integer for any $n > 1$,we can set $n^m + 1 = n^{64} + 1$,which implies $m = 64$.
If $m = 64$,the expression becomes $\frac{n^{64} - 1}{n - 1} = 1 + n + n^2 + \dots + n^{63}$,which is clearly an integer.
Thus,the largest integer $m$ is $64$.

(C) Let the $G.P.$ have $2n$ terms,with the first term $a$ and common ratio $r$.
The sum of all $2n$ terms is $S_{2n} = a\frac{(r^{2n} - 1)}{(r - 1)}$.
The terms at odd places are $a, ar^2, ar^4, \dots, ar^{2n-2}$. This is a $G.P.$ with $n$ terms,first term $a$,and common ratio $r^2$.
The sum of terms at odd places is $S_{odd} = a\frac{((r^2)^n - 1)}{(r^2 - 1)} = a\frac{(r^{2n} - 1)}{(r^2 - 1)}$.
According to the problem,$S_{2n} = 5 \times S_{odd}$.
Substituting the formulas: $a\frac{(r^{2n} - 1)}{(r - 1)} = 5 \times a\frac{(r^{2n} - 1)}{(r^2 - 1)}$.
Since $r \neq 1$ and $r^{2n} - 1 \neq 0$,we can divide both sides by $a\frac{(r^{2n} - 1)}{(r - 1)}$:
$1 = \frac{5}{r + 1}$.
$r + 1 = 5$.
$r = 4$.

(A) Given that $f(x + y) = f(x)f(y)$ for all $x, y \in N$.
For $x = 1$,$f(2) = f(1 + 1) = f(1)f(1) = 3^2 = 9$.
For $x = 2$,$f(3) = f(2 + 1) = f(2)f(1) = 3^2 \cdot 3 = 3^3 = 27$.
By induction,$f(x) = 3^x$.
Now,we are given $\sum_{x = 1}^n f(x) = 120$.
This is a geometric series: $3^1 + 3^2 + 3^3 + \dots + 3^n = 120$.
The sum of a geometric series is given by $S_n = \frac{a(r^n - 1)}{r - 1}$,where $a = 3$ and $r = 3$.
So,$\frac{3(3^n - 1)}{3 - 1} = 120$.
$\frac{3(3^n - 1)}{2} = 120$.
$3(3^n - 1) = 240$.
$3^n - 1 = 80$.
$3^n = 81$.
$3^n = 3^4$.
Therefore,$n = 4$.

(C) Given that $G$ is the geometric mean between $a$ and $b$,we have $G = (ab)^{1/2}$.
If $G_1, G_2, ..., G_n$ are $n$ geometric means between $a$ and $b$,then $a, G_1, G_2, ..., G_n, b$ form a geometric progression with $n+2$ terms.
Let $r$ be the common ratio. Then $b = a r^{n+1}$,which implies $r = (b/a)^{1/(n+1)}$.
The product of the $n$ geometric means is $P = G_1 \cdot G_2 \cdot ... \cdot G_n = (ar) \cdot (ar^2) \cdot ... \cdot (ar^n) = a^n r^{1+2+...+n} = a^n r^{n(n+1)/2}$.
Substituting $r = (b/a)^{1/(n+1)}$,we get:
$P = a^n \left( \frac{b}{a} \right)^{\frac{n(n+1)}{2(n+1)}} = a^n \left( \frac{b}{a} \right)^{n/2} = a^n \cdot \frac{b^{n/2}}{a^{n/2}} = a^{n/2} b^{n/2} = (ab)^{n/2}$.
Since $G = (ab)^{1/2}$,it follows that $G^n = ((ab)^{1/2})^n = (ab)^{n/2}$.
Therefore,$G_1 \cdot G_2 \cdot ... \cdot G_n = G^n$.

(C) Let the increasing $G.P.$ be $k, kr, kr^2, kr^3$ where $k > 0$ and $r > 1$.
Thus,$\alpha = k, \beta = kr, \gamma = kr^2, \delta = kr^3$.
From the first equation $x^2 - 3x + a = 0$,the sum of roots $\alpha + \beta = k(1 + r) = 3$ and the product $\alpha \beta = k^2r = a$.
From the second equation $x^2 - 12x + b = 0$,the sum of roots $\gamma + \delta = kr^2(1 + r) = 12$ and the product $\gamma \delta = k^2r^5 = b$.
Dividing the sum equations: $\frac{kr^2(1 + r)}{k(1 + r)} = \frac{12}{3} \implies r^2 = 4$. Since the $G.P.$ is increasing,$r = 2$.
Substituting $r = 2$ into $k(1 + r) = 3$,we get $k(3) = 3 \implies k = 1$.
Now,$a = k^2r = (1)^2(2) = 2$.
And $b = k^2r^5 = (1)^2(2^5) = 32$.
Therefore,$(a, b) = (2, 32)$.

(C) Given that $2.\overline{357} = 2.357357357...$
This can be written as $2 + 0.357357357...$
$= 2 + \frac{357}{10^3} + \frac{357}{10^6} + \frac{357}{10^9} + ...$
This is an infinite geometric series with the first term $a = \frac{357}{1000}$ and common ratio $r = \frac{1}{1000}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
$S = \frac{\frac{357}{1000}}{1 - \frac{1}{1000}} = \frac{\frac{357}{1000}}{\frac{999}{1000}} = \frac{357}{999}$.
Therefore,$2.\overline{357} = 2 + \frac{357}{999} = \frac{2 \times 999 + 357}{999} = \frac{1998 + 357}{999} = \frac{2355}{999}$.

(D) The given series is an infinite geometric series with the first term $a = 1$ and common ratio $r = \cos \alpha$.
For an infinite geometric series,the sum $S = \frac{a}{1 - r}$,provided $|r| < 1$.
Given $S = 2 - \sqrt{2}$,we have $\frac{1}{1 - \cos \alpha} = 2 - \sqrt{2}$.
Taking the reciprocal,$1 - \cos \alpha = \frac{1}{2 - \sqrt{2}}$.
Rationalizing the denominator: $\frac{1}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{2 + \sqrt{2}}{4 - 2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.
So,$1 - \cos \alpha = 1 + \frac{1}{\sqrt{2}}$,which implies $-\cos \alpha = \frac{1}{\sqrt{2}}$,or $\cos \alpha = -\frac{1}{\sqrt{2}}$.
Since $0 < \alpha < \pi$ and $\cos \alpha$ is negative,$\alpha$ must be in the second quadrant.
$\cos \alpha = -\cos(\pi/4) = \cos(\pi - \pi/4) = \cos(3\pi/4)$.
Therefore,$\alpha = 3\pi/4$.

(B) The sum $S$ of an infinite geometric progression with first term $a$ and common ratio $r$ is given by $S = \frac{a}{1 - r}$,where $|r| < 1$.
Given $a = x$ and $S = 5$,we have $5 = \frac{x}{1 - r}$.
Rearranging for $r$,we get $1 - r = \frac{x}{5}$,which implies $r = 1 - \frac{x}{5}$.
Since the condition for the sum of an infinite geometric progression to exist is $|r| < 1$,we have $|1 - \frac{x}{5}| < 1$.
This inequality can be written as $-1 < 1 - \frac{x}{5} < 1$.
Subtracting $1$ from all parts,we get $-2 < -\frac{x}{5} < 0$.
Multiplying by $-5$ (and reversing the inequality signs),we get $10 > x > 0$,which is $0 < x < 10$.

(C) Given that $a, b, c$ are in $H.P.$
Therefore,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$
This implies $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$,which simplifies to $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.
Let the expression be $E = \left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$.
Using the relation $\frac{1}{c} - \frac{1}{a} = \frac{1}{b} - \frac{2}{a}$ (from $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$),the first bracket becomes $\left( \frac{1}{b} + (\frac{2}{b} - \frac{1}{a}) - \frac{1}{a} \right) = \left( \frac{3}{b} - \frac{2}{a} \right)$.
Using the relation $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$,the second bracket becomes $\left( \frac{2}{b} - \frac{1}{b} \right) = \frac{1}{b}$.
Thus,$E = \left( \frac{3}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \frac{3}{b^2} - \frac{2}{ab}$.

(B) Given the equation $(1 - ab)m^2 - (a^2 + b^2)m - (1 + ab) = 0$.
Rearranging terms,we get $m(a^2 + b^2) + (m^2 + 1)ab = m^2 - 1$ ......$(i)$.
Let $H_1, H_2, \dots, H_m$ be the $m$ harmonic means between $a$ and $b$. The sequence $a, H_1, H_2, \dots, H_m, b$ is in Harmonic Progression ($H$.$P$.).
Thus,$\frac{1}{a}, \frac{1}{H_1}, \dots, \frac{1}{H_m}, \frac{1}{b}$ is in Arithmetic Progression ($A$.$P$.).
Let $d$ be the common difference of this $A$.$P$. Then $\frac{1}{b} = \frac{1}{a} + (m + 1)d$,so $d = \frac{a - b}{ab(m + 1)}$.
$H_1 = \frac{1}{\frac{1}{a} + d} = \frac{ab(m + 1)}{b(m + 1) + a - b} = \frac{ab(m + 1)}{mb + a}$.
$H_m = \frac{1}{\frac{1}{b} - d} = \frac{ab(m + 1)}{a(m + 1) - (a - b)} = \frac{ab(m + 1)}{ma + b}$.
Difference $H_m - H_1 = ab(m + 1) \left[ \frac{1}{ma + b} - \frac{1}{mb + a} \right] = ab(m + 1) \frac{mb + a - ma - b}{(ma + b)(mb + a)} = ab(m + 1) \frac{(m - 1)(b - a)}{m^2ab + ma^2 + mb^2 + ab} = \frac{ab(m^2 - 1)(b - a)}{m(a^2 + b^2) + ab(m^2 + 1)}$.
Using equation $(i)$,the denominator is $m^2 - 1$.
Thus,$H_m - H_1 = \frac{ab(m^2 - 1)(b - a)}{m^2 - 1} = ab(b - a)$.

(C) Let the distance between the home and the school be $d$ km.
The time taken to go to school is $t_1 = \frac{d}{x}$ hours.
The time taken to return home is $t_2 = \frac{d}{y}$ hours.
Average speed is defined as the total distance divided by the total time taken.
$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{d + d}{t_1 + t_2} = \frac{2d}{\frac{d}{x} + \frac{d}{y}}$.
Simplifying the expression: $\frac{2d}{d(\frac{1}{x} + \frac{1}{y})} = \frac{2}{\frac{x+y}{xy}} = \frac{2xy}{x+y}$.
This expression $\frac{2xy}{x+y}$ is the Harmonic Mean $(H.M.)$ of $x$ and $y$.

(C) If $a, b, c, d$ are in $H.P.$,then their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$
For any three consecutive terms in $H.P.$,the middle term is the harmonic mean of the other two. Thus,$b = \frac{2ac}{a+c}$ and $c = \frac{2bd}{b+d}$.
We know that for any two positive numbers,the Geometric Mean $(G.M.)$ is greater than the Harmonic Mean $(H.M.)$.
For $a$ and $c$,the $G.M.$ is $\sqrt{ac}$ and the $H.M.$ is $b$. Therefore,$\sqrt{ac} > b \Rightarrow ac > b^2$.
Similarly,for $b$ and $d$,the $G.M.$ is $\sqrt{bd}$ and the $H.M.$ is $c$. Therefore,$\sqrt{bd} > c \Rightarrow bd > c^2$.
Adding these two inequalities,we get $ac + bd > b^2 + c^2$.

(B) Since the Arithmetic Mean $(AM)$ is greater than the Geometric Mean $(GM)$ for positive real numbers,we have:
$\frac{a + b}{2} > \sqrt{ab}$
$\frac{b + c}{2} > \sqrt{bc}$
$\frac{c + a}{2} > \sqrt{ca}$
Multiplying these three inequalities together,we get:
$\left(\frac{a + b}{2}\right) \left(\frac{b + c}{2}\right) \left(\frac{c + a}{2}\right) > \sqrt{ab} \cdot \sqrt{bc} \cdot \sqrt{ca}$
$\frac{(a + b)(b + c)(c + a)}{8} > \sqrt{a^2 b^2 c^2}$
$\frac{(a + b)(b + c)(c + a)}{8} > abc$
Therefore,$(a + b)(b + c)(c + a) > 8abc$.

(A) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$.
Condition $I$: $\frac{a}{r} + a + ar = 14$
$\Rightarrow a(\frac{1}{r} + 1 + r) = 14$ ... $(i)$
Condition $II$: $\frac{a}{r} + 1, a + 1, ar - 1$ are in $A.P.$
Therefore,$2(a + 1) = (\frac{a}{r} + 1) + (ar - 1)$
$2a + 2 = \frac{a}{r} + ar$
$2a + 2 = a(\frac{1}{r} + r)$ ... (ii)
From $(i)$,$a(\frac{1}{r} + r) = 14 - a$. Substituting this into (ii):
$2a + 2 = 14 - a$
$3a = 12 \Rightarrow a = 4$.
Substituting $a = 4$ into $(i)$:
$4(\frac{1}{r} + 1 + r) = 14$
$\frac{1}{r} + r = \frac{14}{4} - 1 = 3.5 - 1 = 2.5 = \frac{5}{2}$
$2r^2 - 5r + 2 = 0$
$(2r - 1)(r - 2) = 0 \Rightarrow r = 2$ or $r = 0.5$.
If $r = 2$,the numbers are $\frac{4}{2}, 4, 4(2) \Rightarrow 2, 4, 8$.
If $r = 0.5$,the numbers are $\frac{4}{0.5}, 4, 4(0.5) \Rightarrow 8, 4, 2$.
In both cases,the set of numbers is ${2, 4, 8}$.
Thus,the greatest number is $8$.

(B) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Since $\log a - \log 2b, \log 2b - \log 3c, \log 3c - \log a$ are in $A.P.$,the sum of the first and third terms equals twice the middle term:
$(\log a - \log 2b) + (\log 3c - \log a) = 2(\log 2b - \log 3c)$
$\log 3c - \log 2b = 2\log 2b - 2\log 3c$
$3\log 3c = 3\log 2b \Rightarrow 3c = 2b \Rightarrow b = \frac{3}{2}c$.
Substituting $b = \frac{3}{2}c$ into $b^2 = ac$,we get $(\frac{3}{2}c)^2 = ac \Rightarrow \frac{9}{4}c^2 = ac \Rightarrow a = \frac{9}{4}c$.
Thus,the sides are $a = \frac{9}{4}c, b = \frac{3}{2}c, c = c$.
Multiplying by $\frac{4}{c}$,the sides are proportional to $9, 6, 4$.
Since $9^2 > 6^2 + 4^2$ $(81 > 36 + 16 = 52)$,the triangle is obtuse-angled.

(A) Let the two quantities be $a$ and $b$.
Since ${A_1}, {A_2}$ are $AMs$ between $a$ and $b$,$a, {A_1}, {A_2}, b$ are in $A.P.$
Thus,${A_1} - a = b - {A_2} \Rightarrow {A_1} + {A_2} = a + b$ ......$(i)$
Since ${G_1}, {G_2}$ are $GMs$ between $a$ and $b$,$a, {G_1}, {G_2}, b$ are in $G.P.$
Thus,$\frac{{G_1}}{a} = \frac{b}{{G_2}} \Rightarrow {G_1}{G_2} = ab$ ......$(ii)$
Since ${H_1}, {H_2}$ are $HMs$ between $a$ and $b$,$a, {H_1}, {H_2}, b$ are in $H.P.$
Thus,$\frac{1}{{H_1}} - \frac{1}{a} = \frac{1}{b} - \frac{1}{{H_2}} \Rightarrow \frac{1}{{H_1}} + \frac{1}{{H_2}} = \frac{1}{a} + \frac{1}{b}$
$\Rightarrow \frac{{H_1 + H_2}}{{H_1 H_2}} = \frac{{a + b}}{{ab}}$
Substituting $(i)$ and $(ii)$ into the expression,we get $\frac{{H_1 + H_2}}{{H_1 H_2}} = \frac{{A_1 + A_2}}{{G_1 G_2}}$
Therefore,$\frac{{G_1 G_2}}{{H_1 H_2}} = \frac{{A_1 + A_2}}{{H_1 + H_2}}$.

(A) Let the two numbers be $x$ and $y$.
The arithmetic mean is $A = \frac{x + y}{2}$ and the geometric mean is $G = \sqrt{xy}$.
The harmonic mean is given by $H = \frac{2xy}{x + y} = 4$.
From $H = 4$,we have $\frac{2xy}{x + y} = 4$,which implies $xy = 2(x + y)$.
Since $A = \frac{x + y}{2}$,we have $x + y = 2A$,so $xy = 2(2A) = 4A$.
Since $G^2 = xy$,we have $G^2 = 4A$.
Given the relation $2A + G^2 = 27$,substitute $G^2 = 4A$ into the equation:
$2A + 4A = 27 \Rightarrow 6A = 27 \Rightarrow A = \frac{27}{6} = 4.5$.
Now,$x + y = 2A = 2(4.5) = 9$ and $xy = 4A = 4(4.5) = 18$.
The quadratic equation with roots $x$ and $y$ is $t^2 - (x + y)t + xy = 0$,which is $t^2 - 9t + 18 = 0$.
Factoring the quadratic: $(t - 6)(t - 3) = 0$.
Thus,the numbers are $6$ and $3$.

(C) Let the two numbers be $a$ and $b$.
The Arithmetic Mean $(A.M.)$ is given by $\frac{a + b}{2}$.
The Geometric Mean $(G.M.)$ is given by $\sqrt{ab}$.
According to the problem,$A.M. = G.M. + 2$,so $\frac{a + b}{2} = \sqrt{ab} + 2$ ......$(i)$.
The ratio of the numbers is $4:1$,so $\frac{a}{b} = \frac{4}{1}$,which implies $a = 4b$ ......$(ii)$.
Substitute $a = 4b$ into equation $(i)$:
$\frac{4b + b}{2} = \sqrt{4b \cdot b} + 2$
$\frac{5b}{2} = \sqrt{4b^2} + 2$
$\frac{5b}{2} = 2b + 2$
Multiply by $2$: $5b = 4b + 4$,which gives $b = 4$.
Now,substitute $b = 4$ into equation $(ii)$:
$a = 4(4) = 16$.
Thus,the numbers are $16$ and $4$.

(C) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
The standard form of a quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Given the Arithmetic Mean $(A.M.)$ of the roots is $8$:
$\frac{\alpha + \beta}{2} = 8 \Rightarrow \alpha + \beta = 16$.
Given the Geometric Mean $(G.M.)$ of the roots is $5$:
$\sqrt{\alpha \beta} = 5 \Rightarrow \alpha \beta = 25$.
Substituting the sum and product of the roots into the standard form:
$x^2 - (16)x + (25) = 0$.
Thus,the required quadratic equation is $x^2 - 16x + 25 = 0$.

(B) For any two positive numbers,the relationship between Arithmetic Mean $(A.M.)$,Geometric Mean $(G.M.)$,and Harmonic Mean $(H.M.)$ is given by the inequality $A.M. \ge G.M. \ge H.M.$
Given values are $\frac{144}{15} = 9.6$,$15$,and $12$.
Comparing these values,we have $15 > 12 > 9.6$.
Therefore,$A.M. = 15$,$G.M. = 12$,and $H.M. = \frac{144}{15}$.
We are asked to find the values in the order $H.M., G.M., A.M.$
Substituting the identified values,we get $\frac{144}{15}, 12, 15$.

(B) Given that $a$ is the arithmetic mean of $b$ and $c$,we have $a = \frac{b+c}{2}$,so $b+c = 2a$.
Since $G_1, G_2$ are two geometric means between $b$ and $c$,the sequence $b, G_1, G_2, c$ is in Geometric Progression $(GP)$.
Let the common ratio be $r$. Then $G_1 = br$,$G_2 = br^2$,and $c = br^3$.
Thus,$r = (c/b)^{1/3}$.
$G_1 = b(c/b)^{1/3} = b^{2/3} c^{1/3}$ and $G_2 = b(c/b)^{2/3} = b^{1/3} c^{2/3}$.
Now,$G_1^3 + G_2^3 = (b^{2/3} c^{1/3})^3 + (b^{1/3} c^{2/3})^3 = b^2 c + b c^2 = bc(b+c)$.
Substituting $b+c = 2a$,we get $G_1^3 + G_2^3 = bc(2a)$.
Since $G_1 G_2 = (b^{2/3} c^{1/3})(b^{1/3} c^{2/3}) = bc$,we have $G_1^3 + G_2^3 = 2 a G_1 G_2$.

(C) Let the three numbers in $G.P.$ be $a, ar, ar^2$.
According to the first condition,$a, ar, ar^2 - 64$ are in $A.P.$
Therefore,$2(ar) = a + (ar^2 - 64) \implies a(r^2 - 2r + 1) = 64 \implies a(r - 1)^2 = 64$ .....$(i)$
According to the second condition,the second term of the $A.P.$ is decreased by $8$,so $a, ar - 8, ar^2 - 64$ are in $G.P.$
Therefore,$(ar - 8)^2 = a(ar^2 - 64) \implies a^2r^2 - 16ar + 64 = a^2r^2 - 64a \implies 16ar - 64a = 64 \implies ar - 4a = 4 \implies a(r - 4) = 4$ .....(ii)
Dividing $(i)$ by (ii): $\frac{a(r - 1)^2}{a(r - 4)} = \frac{64}{4} \implies \frac{(r - 1)^2}{r - 4} = 16 \implies r^2 - 2r + 1 = 16r - 64 \implies r^2 - 18r + 65 = 0$.
Solving the quadratic equation: $(r - 5)(r - 13) = 0$,so $r = 5$ or $r = 13$.
If $r = 5$,$a(5 - 4) = 4 \implies a = 4$. The numbers are $4, 4(5), 4(5^2) = 4, 20, 100$.
If $r = 13$,$a(13 - 4) = 4 \implies 9a = 4 \implies a = 4/9$. The numbers are $4/9, 52/9, 676/9$.
Checking option $(c)$: $4, 20, 100$. $3^{rd}$ term decreased by $64$ gives $4, 20, 36$,which is an $A.P.$ $(d=16)$. Decreasing the $2^{nd}$ term by $8$ gives $4, 12, 36$,which is a $G.P.$ $(r=3)$. Thus,the correct option is $(c)$.

(B) Given that $x, y, z$ are in $G.P.$,we have $y^2 = xz$.
Taking the natural logarithm on both sides,we get $2 \ln y = \ln x + \ln z$.
Adding $2$ to both sides,we get $2 + 2 \ln y = 2 + \ln x + \ln z$,which can be written as $2(1 + \ln y) = (1 + \ln x) + (1 + \ln z)$.
This implies that $(1 + \ln x), (1 + \ln y), (1 + \ln z)$ are in $A.P.$
Since the reciprocals of terms in an $A.P.$ form an $H.P.$,it follows that $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$ are in $H.P.$

(C) Given that $a, g, h$ are the arithmetic mean,geometric mean,and harmonic mean of two positive numbers $x$ and $y$ respectively.
We have $a = \frac{x + y}{2}$,$g = \sqrt{xy}$,and $h = \frac{2xy}{x + y}$.
Now,calculate $g^2$:
$g^2 = (\sqrt{xy})^2 = xy$ ...$(i)$
Next,calculate the product $ah$:
$ah = \left(\frac{x + y}{2}\right) \cdot \left(\frac{2xy}{x + y}\right) = xy$ ...(ii)
Comparing equations $(i)$ and (ii),we get $g^2 = ah$,which implies $g = \sqrt{ah}$.
This confirms that $g$ is the geometric mean between $a$ and $h$.

(D) Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:
$\frac{2^{\sin \theta} + 2^{\cos \theta}}{2} \ge \sqrt{2^{\sin \theta} \cdot 2^{\cos \theta}}$
$2^{\sin \theta} + 2^{\cos \theta} \ge 2 \cdot 2^{\frac{\sin \theta + \cos \theta}{2}}$
We know that $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})$.
The minimum value of $\sin(\theta + \frac{\pi}{4})$ is $-1$,so the minimum value of $\sin \theta + \cos \theta$ is $-\sqrt{2}$.
Thus,$2^{\sin \theta} + 2^{\cos \theta} \ge 2 \cdot 2^{\frac{-\sqrt{2}}{2}} = 2^1 \cdot 2^{-\frac{1}{\sqrt{2}}} = 2^{(1 - \frac{1}{\sqrt{2}})}$.

(A) Given $a, b, c, d > 0$ and $a + b + c + d = 2$.
Let $x = a + b$ and $y = c + d$. Then $x + y = 2$.
We want to find the range of $M = xy$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,we know that for positive real numbers $x$ and $y$,$\frac{x + y}{2} \ge \sqrt{xy}$.
Substituting the values,we get $\frac{2}{2} \ge \sqrt{M}$,which implies $1 \ge \sqrt{M}$.
Squaring both sides,we get $M \le 1$.
Since $a, b, c, d$ are positive,$x = a + b > 0$ and $y = c + d > 0$,so $M = xy > 0$.
Therefore,the relation satisfied is $0 < M \le 1$.

(D) Let the terms in $A.P.$ be $a = b - d$ and $c = b + d$,where $d > 0$ since $a < b < c$.
Given $a + b + c = \frac{3}{2}$,we have $(b - d) + b + (b + d) = \frac{3}{2}$,which implies $3b = \frac{3}{2}$,so $b = \frac{1}{2}$.
Thus,$a = \frac{1}{2} - d$ and $c = \frac{1}{2} + d$.
Since $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^2 = a^2 c^2$,which means $b^4 = (ac)^2$.
Taking the square root,$b^2 = \pm ac$.
If $b^2 = ac$,then $b^2 = (b - d)(b + d) = b^2 - d^2$,so $d^2 = 0$,which implies $d = 0$. This contradicts $a < b < c$.
Therefore,$b^2 = -ac$.
Substituting $b = \frac{1}{2}$,$a = \frac{1}{2} - d$,and $c = \frac{1}{2} + d$:
$(\frac{1}{2})^2 = -(\frac{1}{2} - d)(\frac{1}{2} + d)$
$\frac{1}{4} = -(\frac{1}{4} - d^2)$
$\frac{1}{4} = -\frac{1}{4} + d^2$
$d^2 = \frac{1}{2}$,so $d = \frac{1}{\sqrt{2}}$ (since $d > 0$).
Finally,$a = \frac{1}{2} - d = \frac{1}{2} - \frac{1}{\sqrt{2}}$.

(C) The given series is an Arithmetico-Geometric progression $(A.G.P.)$.
The numerator terms are $1, 4, 7, 10, \dots$,which form an Arithmetic Progression $(A.P.)$ with the first term $a = 1$ and common difference $d = 3$.
The $n^{th}$ term of this $A.P.$ is $T_n = a + (n - 1)d = 1 + (n - 1)3 = 3n - 2$.
The denominator terms are $1, 5, 5^2, 5^3, \dots$,which form a Geometric Progression $(G.P.)$ with the first term $a = 1$ and common ratio $r = 5$.
The $n^{th}$ term of this $G.P.$ is $G_n = ar^{n - 1} = 1 \cdot 5^{n - 1} = 5^{n - 1}$.
Therefore,the $n^{th}$ term of the given series is the ratio of the $n^{th}$ term of the $A.P.$ to the $n^{th}$ term of the $G.P.$,which is $\frac{3n - 2}{5^{n - 1}}$.

(B) Let $T_n$ be the $n^{th}$ term of the series.
$T_n = \frac{n}{{1 + n^2 + n^4}} = \frac{n}{{(1 + n^2)^2 - n^2}}$
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
$T_n = \frac{n}{{(n^2 + n + 1)(n^2 - n + 1)}}$
Using partial fractions:
$T_n = \frac{1}{2} \left[ \frac{1}{{n^2 - n + 1}} - \frac{1}{{n^2 + n + 1}} \right]$
Note that $n^2 - n + 1 = 1 + n(n - 1)$ and $n^2 + n + 1 = 1 + n(n + 1)$.
Thus,$T_n = \frac{1}{2} \left[ \frac{1}{{1 + (n - 1)n}} - \frac{1}{{1 + n(n + 1)}} \right]$.
Summing from $r=1$ to $n$:
$S_n = \sum_{r=1}^n T_r = \frac{1}{2} \left[ \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{7} \right) + \dots + \left( \frac{1}{{1 + (n-1)n}} - \frac{1}{{1 + n(n+1)}} \right) \right]$
This is a telescoping series,so all intermediate terms cancel out:
$S_n = \frac{1}{2} \left[ 1 - \frac{1}{{1 + n(n + 1)}} \right] = \frac{1}{2} \left[ \frac{1 + n^2 + n - 1}{{n^2 + n + 1}} \right] = \frac{n(n + 1)}{{2(n^2 + n + 1)}}$.

(D) Given the series $S = {n^3} - {(n - 1)^3} + {(n - 2)^3} - {(n - 3)^3} + \dots + {1^3}$.
Since $n$ is odd,the terms with even indices are subtracted and terms with odd indices are added.
We can rewrite the sum as the sum of all cubes from $1^3$ to $n^3$ minus twice the sum of the cubes of the even terms:
$S = \sum_{k=1}^{n} k^3 - 2 \sum_{j=1}^{(n-1)/2} (2j)^3$.
Using the formula $\sum_{k=1}^{n} k^3 = [\frac{n(n+1)}{2}]^2$,we get:
$S = [\frac{n(n+1)}{2}]^2 - 2 \cdot 8 \sum_{j=1}^{(n-1)/2} j^3$.
$S = \frac{n^2(n+1)^2}{4} - 16 [\frac{(\frac{n-1}{2})(\frac{n-1}{2} + 1)}{2}]^2$.
$S = \frac{n^2(n+1)^2}{4} - 16 [\frac{(\frac{n-1}{2})(\frac{n+1}{2})}{2}]^2$.
$S = \frac{n^2(n+1)^2}{4} - 16 [\frac{(n-1)(n+1)}{8}]^2$.
$S = \frac{n^2(n+1)^2}{4} - 16 \cdot \frac{(n-1)^2(n+1)^2}{64}$.
$S = \frac{n^2(n+1)^2}{4} - \frac{(n-1)^2(n+1)^2}{4}$.
$S = \frac{(n+1)^2}{4} [n^2 - (n-1)^2]$.
$S = \frac{(n+1)^2}{4} [n^2 - (n^2 - 2n + 1)]$.
$S = \frac{(n+1)^2}{4} (2n - 1)$.

(D) The $k$-th term of the series is $a_k = \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}}$.
Rationalizing the denominator,we get $a_k = \frac{\sqrt{2k+1} - \sqrt{2k-1}}{(2k+1) - (2k-1)} = \frac{\sqrt{2k+1} - \sqrt{2k-1}}{2}$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} a_k = \frac{1}{2} \sum_{k=1}^{n} (\sqrt{2k+1} - \sqrt{2k-1})$.
Expanding the sum: $S_n = \frac{1}{2} [(\sqrt{3} - 1) + (\sqrt{5} - \sqrt{3}) + (\sqrt{7} - \sqrt{5}) + \dots + (\sqrt{2n+1} - \sqrt{2n-1})]$.
This is a telescoping series where intermediate terms cancel out,leaving $S_n = \frac{1}{2} (\sqrt{2n+1} - 1)$.

(C) The $n^{th}$ term $T_n$ is given by the sum of the first $n$ cubes in the numerator and the sum of the first $n$ odd numbers in the denominator.
Numerator: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$.
Denominator: The sum of the first $n$ odd numbers is $1 + 3 + 5 + \dots + (2n-1)$. This is an arithmetic progression with $a=1$,$d=2$,and $n$ terms. The sum is $\frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}[2 + 2n - 2] = n^2$.
Therefore,$T_n = \frac{\frac{n^2(n+1)^2}{4}}{n^2} = \frac{(n+1)^2}{4} = \frac{n^2 + 2n + 1}{4}$.

(D) The given series is $S = \sum_{k=1}^{n^2-1} \frac{1}{\sqrt{k} + \sqrt{k+1}}$.
To solve this,we rationalize each term by multiplying the numerator and denominator by the conjugate $(\sqrt{k+1} - \sqrt{k})$:
$\frac{1}{\sqrt{k+1} + \sqrt{k}} \times \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} = \frac{\sqrt{k+1} - \sqrt{k}}{(k+1) - k} = \sqrt{k+1} - \sqrt{k}$.
Applying this to the series:
$S = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + ... + (\sqrt{n^2} - \sqrt{n^2 - 1})$.
This is a telescoping series where intermediate terms cancel out:
$S = -\sqrt{1} + \sqrt{n^2} = -1 + n = n - 1$.

(D) 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 $\frac{S_p}{S_q} = \frac{p^2}{q^2}$,we have $\frac{\frac{p}{2} [2a_1 + (p-1)d]}{\frac{q}{2} [2a_1 + (q-1)d]} = \frac{p^2}{q^2}$.
Simplifying,we get $\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p}{q}$.
Dividing the numerator and denominator by $2$,we get $\frac{a_1 + (\frac{p-1}{2})d}{a_1 + (\frac{q-1}{2})d} = \frac{p}{q}$.
To find $\frac{a_6}{a_{21}}$,we need the index $n$ such that $\frac{n-1}{2} = 5$ (for $a_6$) and $\frac{n-1}{2} = 20$ (for $a_{21}$).
For $a_6$,$p-1 = 10 \Rightarrow p = 11$.
For $a_{21}$,$q-1 = 40 \Rightarrow q = 41$.
Substituting these values,$\frac{a_6}{a_{21}} = \frac{p}{q} = \frac{11}{41}$.

(C) Since $a_1, a_2, \dots, a_n$ are in $H.P.$, their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \dots, \frac{1}{a_n}$ are in $A.P.$
Let the common difference be $d = \frac{1}{a_{k+1}} - \frac{1}{a_k}$.
Then, $a_k a_{k+1} = \frac{1}{d} (a_k - a_{k+1})$.
The given sum is $S = \sum_{k=1}^{n-1} a_k a_{k+1} = \sum_{k=1}^{n-1} \frac{a_k - a_{k+1}}{d} = \frac{1}{d} (a_1 - a_n)$.
From the $A.P.$ property, $\frac{1}{a_n} = \frac{1}{a_1} + (n-1)d$, which implies $\frac{1}{a_n} - \frac{1}{a_1} = (n-1)d$.
Thus, $\frac{a_1 - a_n}{a_1 a_n} = (n-1)d$, so $\frac{a_1 - a_n}{d} = (n-1) a_1 a_n$.
Therefore, the sum is $(n-1) a_1 a_n$.

(D) We know the Taylor series expansion for $e^x$ is given by:
$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$
Substitute $x = -1$ into the expansion:
$e^{-1} = 1 + (-1) + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \dots$
Simplifying the terms:
$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$
Since $1 - 1 = 0$,we get:
$e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$
Thus,the sum of the series is $e^{-1}$.

(A) Let the terms of the geometric progression be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.
Given that each term is equal to the sum of the next two terms:
$a = ar + ar^2$
Since $a > 0$,we can divide both sides by $a$:
$1 = r + r^2$
Rearranging the terms,we get the quadratic equation:
$r^2 + r - 1 = 0$
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$
$r = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$
Since the terms of the geometric progression are positive,the common ratio $r$ must be positive.
Therefore,we take the positive root:
$r = \frac{\sqrt{5} - 1}{2}$