Harmonic progression Questions in English

(B) Let the harmonic progression be $\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \dots$
Then,$u = \frac{1}{a+(p-1)d}$,$v = \frac{1}{a+(q-1)d}$,and $w = \frac{1}{a+(r-1)d}$.
This implies $\frac{1}{u} = a+(p-1)d$,$\frac{1}{v} = a+(q-1)d$,and $\frac{1}{w} = a+(r-1)d$.
Subtracting these equations,we get:
$\frac{1}{u} - \frac{1}{v} = (p-q)d \implies \frac{v-u}{uv} = (p-q)d$
$\frac{1}{v} - \frac{1}{w} = (q-r)d \implies \frac{w-v}{vw} = (q-r)d$
$\frac{1}{w} - \frac{1}{u} = (r-p)d \implies \frac{u-w}{wu} = (r-p)d$
Multiplying the terms,we observe that $(q-r)vw + (r-p)wu + (p-q)uv = 0$.

(C) The harmonic mean $H$ of two numbers $x$ and $y$ is given by $H = \frac{2xy}{x + y}$.
Given $x = \frac{a}{1 - ab}$ and $y = \frac{a}{1 + ab}$.
First,calculate the product $xy = \left(\frac{a}{1 - ab}\right) \left(\frac{a}{1 + ab}\right) = \frac{a^2}{1 - a^2b^2}$.
Next,calculate the sum $x + y = \frac{a}{1 - ab} + \frac{a}{1 + ab} = \frac{a(1 + ab) + a(1 - ab)}{(1 - ab)(1 + ab)} = \frac{a + a^2b + a - a^2b}{1 - a^2b^2} = \frac{2a}{1 - a^2b^2}$.
Now,$H = \frac{2 \left(\frac{a^2}{1 - a^2b^2}\right)}{\frac{2a}{1 - a^2b^2}} = \frac{2a^2}{2a} = a$.

(A) In a Harmonic Progression $(H.P.)$,the terms are the reciprocals of an Arithmetic Progression $(A.P.)$.
Let the $A.P.$ have first term $a$ and common difference $d$.
The $5^{th}$ term of $H.P.$ is $1/45$,so the $5^{th}$ term of $A.P.$ is $a + 4d = 45 \dots (i)$.
The $11^{th}$ term of $H.P.$ is $1/69$,so the $11^{th}$ term of $A.P.$ is $a + 10d = 69 \dots (ii)$.
Subtracting $(i)$ from $(ii)$:
$(a + 10d) - (a + 4d) = 69 - 45$
$6d = 24 \implies d = 4$.
Substituting $d = 4$ into $(i)$:
$a + 4(4) = 45 \implies a + 16 = 45 \implies a = 29$.
The $16^{th}$ term of the $A.P.$ is $a + 15d = 29 + 15(4) = 29 + 60 = 89$.
Therefore,the $16^{th}$ term of the $H.P.$ is $1/89$.

(A) Given that $x = \sum_{n=0}^{\infty} a^n = \frac{1}{1-a}$,$y = \sum_{n=0}^{\infty} b^n = \frac{1}{1-b}$,and $z = \sum_{n=0}^{\infty} c^n = \frac{1}{1-c}$.
Since $a, b, c$ are in arithmetic progression,we have $2b = a + c$.
We need to check if $x, y, z$ are in harmonic progression,which requires $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ to be in arithmetic progression.
Note that $\frac{1}{x} = 1-a$,$\frac{1}{y} = 1-b$,and $\frac{1}{z} = 1-c$.
Let $A = 1-a$,$B = 1-b$,and $C = 1-c$.
Then $2B = 2(1-b) = 2 - 2b = 2 - (a+c) = (1-a) + (1-c) = A + C$.
Since $A, B, C$ are in arithmetic progression,their reciprocals $\frac{1}{A}, \frac{1}{B}, \frac{1}{C}$ are in harmonic progression.
Thus,$x, y, z$ are in harmonic progression.

(B) Given that $(y - x), 2(y - a), (y - z)$ are in $H.P.$
This implies that their reciprocals $\frac{1}{y - x}, \frac{1}{2(y - a)}, \frac{1}{y - z}$ are in $A.P.$
Therefore,the common difference is equal:
$\frac{1}{2(y - a)} - \frac{1}{y - x} = \frac{1}{y - z} - \frac{1}{2(y - a)}$
$\frac{2}{2(y - a)} = \frac{1}{y - x} + \frac{1}{y - z}$
$\frac{1}{y - a} = \frac{(y - z) + (y - x)}{(y - x)(y - z)}$
$(y - x)(y - z) = (y - a)(2y - x - z)$
$y^2 - yz - xy + xz = 2y^2 - xy - yz - 2ay + ax + az$
$y^2 - xz = 2ay - ax - az$
$y^2 - 2ay + a^2 = xz - ax - az + a^2$
$(y - a)^2 = (x - a)(z - a)$
This is the condition for $(x - a), (y - a), (z - a)$ to be in $G.P.$

(A) The given series is $\frac{1}{7}, \frac{1}{14}, \frac{1}{21}, \frac{1}{28}, \dots$
This is a harmonic progression because the reciprocals of the terms form an arithmetic progression: $7, 14, 21, 28, \dots$
For the arithmetic progression,the first term $a = 7$ and the common difference $d = 14 - 7 = 7$.
The $n^{th}$ term of an arithmetic progression is given by $a_n = a + (n - 1)d$.
For $n = 9$,$a_9 = 7 + (9 - 1) \times 7 = 7 + 8 \times 7 = 7 + 56 = 63$.
Therefore,the $9^{th}$ term of the harmonic progression is the reciprocal of $a_9$,which is $\frac{1}{63}$.

(B) Given that $a, b, c$ are in a geometric progression,we have $b^2 = ac$ $(1)$.
Let $a^x = b^y = c^z = k$.
Then $a = k^{1/x}, b = k^{1/y}, c = k^{1/z}$.
Substituting these values into equation $(1)$:
$(k^{1/y})^2 = k^{1/x} \cdot k^{1/z}$
$k^{2/y} = k^{1/x + 1/z}$
Equating the exponents: $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.
This implies that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression.
Therefore,$x, y, z$ are in a harmonic progression.

(C) If $a, b, c$ are in harmonic progression $(HP)$,then their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in arithmetic progression $(AP)$.
Since they are in $AP$,the common difference is constant:
$\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$
$\frac{a-b}{ab} = \frac{b-c}{bc}$
Dividing both sides by $(b-c)$ and multiplying by $ab$,we get:
$\frac{a-b}{b-c} = \frac{ab}{bc} = \frac{a}{c}$

(A) The harmonic mean $(HM)$ of two numbers $x$ and $y$ is given by the formula $HM = \frac{2xy}{x+y}$.
Here,$x = \frac{a}{b}$ and $y = \frac{b}{a}$.
Substituting these values into the formula:
$HM = \frac{2(\frac{a}{b})(\frac{b}{a})}{\frac{a}{b} + \frac{b}{a}}$
$HM = \frac{2(1)}{\frac{a^2+b^2}{ab}}$
$HM = \frac{2ab}{a^2+b^2}$.

(D) Given that $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in $AP$.
This implies that $b + c, c + a, a + b$ are in Harmonic Progression $(HP)$.
Subtracting $a + b + c$ from each term,we get $(b + c) - (a + b + c), (c + a) - (a + b + c), (a + b) - (a + b + c)$ are in $HP$.
This simplifies to $-a, -b, -c$ are in $HP$.
Multiplying by $-1$,we get $a, b, c$ are in $HP$.
If $a, b, c$ are in $HP$,then $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $AP$.
However,the question asks for the progression of $a^2, b^2, c^2$.
Since $a, b, c$ are in $HP$,there is no standard progression for $a^2, b^2, c^2$ unless specific values are given.
Given the standard nature of this problem,if $a, b, c$ are in $HP$,then $a^2, b^2, c^2$ do not form a standard $AP, GP,$ or $HP$.
Therefore,the correct option is $D$.

(C) Let the given numbers be $a = log_{3}{2}$,$b = log_{6}{2}$,and $c = log_{12}{2}$.
We know that $log_{x}{y} = \frac{1}{log_{y}{x}}$.
Therefore,$\frac{1}{a} = log_{2}{3}$,$\frac{1}{b} = log_{2}{6}$,and $\frac{1}{c} = log_{2}{12}$.
Now,calculate the differences of the reciprocals:
$\frac{1}{b} - \frac{1}{a} = log_{2}{6} - log_{2}{3} = log_{2}(\frac{6}{3}) = log_{2}{2} = 1$.
$\frac{1}{c} - \frac{1}{b} = log_{2}{12} - log_{2}{6} = log_{2}(\frac{12}{6}) = log_{2}{2} = 1$.
Since the difference between consecutive reciprocals is constant,the reciprocals are in Arithmetic Progression.
Therefore,the original numbers are in Harmonic Progression.

(D) Given $X = \sum_{n=0}^\infty a^n = \frac{1}{1-a}$ for $|a| < 1$.
Similarly,$Y = \frac{1}{1-b}$ and $Z = \frac{1}{1-c}$.
From these,we have $a = 1 - \frac{1}{X}$,$b = 1 - \frac{1}{Y}$,and $c = 1 - \frac{1}{Z}$.
Since $a, b, c$ are in arithmetic progression,$2b = a + c$.
Substituting the values: $2(1 - \frac{1}{Y}) = (1 - \frac{1}{X}) + (1 - \frac{1}{Z})$.
$2 - \frac{2}{Y} = 2 - (\frac{1}{X} + \frac{1}{Z})$.
$\frac{2}{Y} = \frac{1}{X} + \frac{1}{Z}$.
This implies that $\frac{1}{X}, \frac{1}{Y}, \frac{1}{Z}$ are in arithmetic progression.
Therefore,$X, Y, Z$ are in harmonic progression $(H.P.)$.

(C) Given the equation: $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$.
Simplify the left side: $\frac{(b - c) + (b - a)}{(b - a)(b - c)} = \frac{2b - a - c}{b^2 - ab - bc + ac}$.
Simplify the right side: $\frac{a + c}{ac}$.
Equating both sides: $\frac{2b - a - c}{b^2 - b(a + c) + ac} = \frac{a + c}{ac}$.
Cross-multiplying: $ac(2b - a - c) = (a + c)(b^2 - b(a + c) + ac)$.
$2abc - a^2c - ac^2 = ab^2 - ab(a + c) + a^2c + cb^2 - cb(a + c) + ac^2$.
$2abc - a^2c - ac^2 = b^2(a + c) - b(a + c)^2 + a^2c + ac^2$.
$2abc = b^2(a + c) - b(a + c)^2 + 2a^2c + 2ac^2$.
$2abc = b^2(a + c) - b(a + c)^2 + 2ac(a + c)$.
Dividing by $(a + c)$ (assuming $a + c \neq 0$): $\frac{2abc}{a + c} = b^2 - b(a + c) + 2ac$.
$2abc = b^2(a + c) - b(a + c)^2 + 2ac(a + c)$.
This simplifies to $b = \frac{2ac}{a + c}$,which is the condition for $a, b, c$ to be in Harmonic Progression.

(D) In a harmonic progression $(HP)$,the reciprocals of the terms form an arithmetic progression $(AP)$.
Let the $AP$ be $b_1, b_2, b_3, \dots$ where $b_n = \frac{1}{a_n}$.
Given $a_1 = 5 \implies b_1 = \frac{1}{5}$ and $a_{20} = 25 \implies b_{20} = \frac{1}{25}$.
The formula for the $n$-th term of an $AP$ is $b_n = b_1 + (n - 1)d$.
For $n = 20$: $\frac{1}{25} = \frac{1}{5} + (20 - 1)d$.
$\frac{1}{25} - \frac{1}{5} = 19d \implies \frac{1 - 5}{25} = 19d \implies -\frac{4}{25} = 19d \implies d = -\frac{4}{475}$.
We want $a_n < 0$,which implies $b_n < 0$ because $a_n = \frac{1}{b_n}$.
$b_n = \frac{1}{5} + (n - 1)(-\frac{4}{475}) < 0$.
$\frac{1}{5} < (n - 1)(\frac{4}{475})$.
Multiply by $475$: $95 < 4(n - 1)$.
$23.75 < n - 1$.
$n > 24.75$.
The smallest positive integer $n$ is $25$.

(A) The harmonic mean $H$ between $a$ and $b$ is given by $H = \frac{2ab}{a+b}$.
Substituting this into the expression $\frac{1}{H - a} + \frac{1}{H - b}$:
$\frac{1}{\frac{2ab}{a+b} - a} + \frac{1}{\frac{2ab}{a+b} - b} = \frac{a+b}{2ab - a(a+b)} + \frac{a+b}{2ab - b(a+b)}$
$= \frac{a+b}{2ab - a^2 - ab} + \frac{a+b}{2ab - ab - b^2} = \frac{a+b}{ab - a^2} + \frac{a+b}{ab - b^2}$
$= \frac{a+b}{a(b-a)} + \frac{a+b}{b(a-b)} = \frac{a+b}{a(b-a)} - \frac{a+b}{b(b-a)}$
$= \frac{b(a+b) - a(a+b)}{ab(b-a)} = \frac{(a+b)(b-a)}{ab(b-a)} = \frac{a+b}{ab} = \frac{1}{b} + \frac{1}{a}$.

(D) Given that $X = \sum_{n=0}^\infty a^n = \frac{1}{1-a}$,$Y = \sum_{n=0}^\infty b^n = \frac{1}{1-b}$,and $Z = \sum_{n=0}^\infty c^n = \frac{1}{1-c}$.
Since $a, b, c$ are in arithmetic progression,we have $2b = a + c$.
We need to check if $X, Y, Z$ are in harmonic progression,which requires $\frac{2}{Y} = \frac{1}{X} + \frac{1}{Z}$.
Substituting the values: $\frac{1}{X} = 1-a$,$\frac{1}{Y} = 1-b$,and $\frac{1}{Z} = 1-c$.
Then $\frac{1}{X} + \frac{1}{Z} = (1-a) + (1-c) = 2 - (a+c)$.
Since $a+c = 2b$,this becomes $2 - 2b = 2(1-b) = 2(\frac{1}{Y}) = \frac{2}{Y}$.
Thus,$X, Y, Z$ are in harmonic progression.

(C) Given $\frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4}$,we can write $\frac{a_1 + a_4}{a_1 a_4} = \frac{a_2 + a_3}{a_2 a_3}$.
This implies $\frac{1}{a_4} + \frac{1}{a_1} = \frac{1}{a_3} + \frac{1}{a_2}$,or $\frac{1}{a_4} - \frac{1}{a_3} = \frac{1}{a_2} - \frac{1}{a_1} \quad (1)$.
Also,from $\frac{a_2 a_3}{a_1 a_4} = 3\left( \frac{a_2 - a_3}{a_1 - a_4} \right)$,we have $\frac{a_1 - a_4}{a_1 a_4} = 3\left( \frac{a_2 - a_3}{a_2 a_3} \right)$.
This implies $\frac{1}{a_4} - \frac{1}{a_1} = 3\left( \frac{1}{a_3} - \frac{1}{a_2} \right) \quad (2)$.
Let $x_n = \frac{1}{a_n}$. Then $(1)$ becomes $x_4 - x_3 = x_2 - x_1 = d$ (common difference).
Substituting into $(2)$,$x_4 - x_1 = 3(x_3 - x_2)$.
Since $x_4 - x_1 = (x_4 - x_3) + (x_3 - x_2) + (x_2 - x_1) = 3d$,and $x_3 - x_2 = d$,this is consistent.
Thus,$\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \frac{1}{a_4}$ are in Arithmetic Progression.
Therefore,$a_1, a_2, a_3, a_4$ are in Harmonic Progression.

(A) If $x, y, z$ are in harmonic progression,then $y = \frac{2xz}{x + z}$.
Now,consider the expression $\log(x + z) + \log(x - 2y + z)$.
$= \log[(x + z)(x - 2y + z)]$
$= \log[(x + z)(x + z - 2(\frac{2xz}{x + z}))]$
$= \log[(x + z)(x + z - \frac{4xz}{x + z})]$
$= \log[(x + z)^2 - 4xz]$
$= \log[x^2 + 2xz + z^2 - 4xz]$
$= \log[x^2 - 2xz + z^2]$
$= \log[(x - z)^2]$
$= 2 \log |x - z|$

(C) The first $n$ natural numbers are $1, 2, 3, \dots, n$.
Their reciprocals are $1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}$.
The harmonic mean $(HM)$ of a set of $n$ numbers $x_1, x_2, \dots, x_n$ is given by $HM = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}$.
Here,$x_i = \frac{1}{i}$ for $i = 1, 2, \dots, n$.
So,$\frac{1}{x_i} = i$.
Therefore,$HM = \frac{n}{\sum_{i=1}^{n} i} = \frac{n}{\frac{n(n+1)}{2}} = \frac{2n}{n(n+1)} = \frac{2}{n+1}$.

(C) The given sequence is $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots, \frac{1}{17}$.
Here,the number of terms $n = 16$.
The harmonic mean $H.M.$ is given by the formula:
$H.M. = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}$
Substituting the values:
$H.M. = \frac{16}{\frac{1}{1/2} + \frac{1}{1/3} + \dots + \frac{1}{1/17}}$
$H.M. = \frac{16}{2 + 3 + 4 + \dots + 17}$
The sum of the arithmetic progression $2, 3, \dots, 17$ is given by $S_n = \frac{n}{2}(a + l) = \frac{16}{2}(2 + 17) = 8 \times 19 = 152$.
Therefore,$H.M. = \frac{16}{152} = \frac{1}{9.5} = \frac{2}{19}$.

(C) The harmonic mean $(HM)$ of $n$ numbers $x_1, x_2, ..., x_n$ is given by the formula: $HM = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}$.
For the numbers $2, 3, 4$,we have $n = 3$.
$HM = \frac{3}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4}}$.
Finding a common denominator for the fractions: $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6 + 4 + 3}{12} = \frac{13}{12}$.
Therefore,$HM = \frac{3}{\frac{13}{12}} = 3 \times \frac{12}{13} = \frac{36}{13}$.

(C) Let the distance between home and school be $S \text{ km}$.
Time taken to go from home to school is $t_1 = \frac{S}{x} \text{ hours}$.
Time taken to return from school to home is $t_2 = \frac{S}{y} \text{ hours}$.
Average speed is defined as the total distance divided by the total time taken.
Total distance $= S + S = 2S \text{ km}$.
Total time $= \frac{S}{x} + \frac{S}{y} = S \left( \frac{1}{x} + \frac{1}{y} \right) = S \left( \frac{x + y}{xy} \right) \text{ hours}$.
Average speed $= \frac{2S}{S \left( \frac{x + y}{xy} \right)} = \frac{2xy}{x + y} \text{ km/hr}$.

(C) Let $H_1, H_2, \dots, H_n$ be $n$ harmonic means between $a = 1$ and $b = \frac{1}{31}$.
Then $1, H_1, H_2, \dots, H_n, \frac{1}{31}$ are in Harmonic Progression $(HP)$.
Therefore,$1, \frac{1}{H_1}, \frac{1}{H_2}, \dots, \frac{1}{H_n}, 31$ are in Arithmetic Progression $(AP)$.
Let $d$ be the common difference of this $AP$. The $(n+2)^{th}$ term is $31 = 1 + (n+2-1)d$,so $(n+1)d = 30$,which means $d = \frac{30}{n+1}$.
The $k^{th}$ harmonic mean $H_k$ is given by $\frac{1}{H_k} = 1 + kd$.
Given $\frac{H_7}{H_{n-1}} = \frac{9}{5}$,we have $\frac{1 + (n-1)d}{1 + 7d} = \frac{9}{5}$.
Substituting $d = \frac{30}{n+1}$:
$5(1 + (n-1)\frac{30}{n+1}) = 9(1 + 7\frac{30}{n+1})$
$5(\frac{n+1 + 30n - 30}{n+1}) = 9(\frac{n+1 + 210}{n+1})$
$5(31n - 29) = 9(n + 211)$
$155n - 145 = 9n + 1899$
$146n = 2044$
$n = \frac{2044}{146} = 14$.

(D) Given that $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$
Therefore,$(a+c), (c-a), (a-2b+c)$ are in $G.P.$
This implies $(c-a)^2 = (a+c)(a-2b+c)$.
Expanding the terms: $c^2 - 2ac + a^2 = a^2 - 2ab + ac + ac - 2bc + c^2$.
Simplifying: $-2ac = -2ab + 2ac - 2bc$.
Rearranging: $2ab + 2bc = 4ac$.
Dividing by $2b(a+c)$: $b = \frac{2ac}{a+c}$.
This is the condition for $a, b, c$ to be in $H.P.$

(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: $\frac{2}{c+a} = \frac{1}{b+c} + \frac{1}{a+b}$
$\frac{2}{c+a} = \frac{a+b+b+c}{(b+c)(a+b)} = \frac{a+2b+c}{(b+c)(a+b)}$
$2(b+c)(a+b) = (c+a)(a+2b+c)$
$2(ab+b^2+ac+bc) = ac+2bc+c^2+a^2+2ab+ac$
$2ab+2b^2+2ac+2bc = a^2+c^2+2ac+2ab+2bc$
$2b^2 = a^2+c^2$
Since $2b^2 = a^2+c^2$,it follows that $a^2, b^2, c^2$ are in $A.P.$

(D) Since $a, b, c$ are in $H.P.$,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 $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ .....$(1)$
Now,consider the expression $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$.
From $(1)$,we have $\frac{1}{c} - \frac{1}{a} = \frac{1}{b} - \frac{2}{b} = -\frac{1}{b}$ (not directly useful) or $\frac{1}{c} - \frac{1}{b} = \frac{1}{b} - \frac{1}{a}$.
Substitute $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$ into the second bracket:
$\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{2}{b} - \frac{1}{b} \right) = \left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{b} \right)$.
From $(1)$,$\frac{1}{c} - \frac{1}{a} = \frac{2}{b} - \frac{2}{a}$. Substituting this:
$\left( \frac{1}{b} + \frac{2}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \left( \frac{3}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \frac{3}{b^2} - \frac{2}{ab}$.
Alternatively,using $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$ in the first bracket:
$\left( \frac{1}{b} + \frac{2}{b} - \frac{1}{a} - \frac{1}{a} \right) \left( \frac{1}{b} \right) = \left( \frac{3}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \frac{3}{b^2} - \frac{2}{ab}$.
Since this result is not among the options,the correct answer is $D$.

(C) The average speed for equal distances covered at different speeds is given by the harmonic mean of the speeds.
Let the side length be $d = 100 \text{ miles}$.
The time taken for each side is $t_i = \frac{d}{v_i}$.
Total distance $D = 4d$.
Total time $T = \frac{d}{100} + \frac{d}{200} + \frac{d}{300} + \frac{d}{400} = d \left( \frac{1}{100} + \frac{1}{200} + \frac{1}{300} + \frac{1}{400} \right)$.
Average speed $V_{avg} = \frac{D}{T} = \frac{4d}{d \left( \frac{1}{100} + \frac{1}{200} + \frac{1}{300} + \frac{1}{400} \right)}$.
$V_{avg} = \frac{4}{\frac{12+6+4+3}{1200}} = \frac{4 \times 1200}{25} = \frac{4800}{25} = 192 \text{ mph}$.
Hence,$(c)$ is the correct answer.

(D) Since $a_1, a_2, \ldots$ are in harmonic progression,their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \ldots$ are in arithmetic progression.
Let the arithmetic progression be $b_n = \frac{1}{a_n} = A + (n-1)D$.
Given $a_1 = 5 \Rightarrow b_1 = \frac{1}{5}$ and $a_{20} = 25 \Rightarrow b_{20} = \frac{1}{25}$.
$b_{20} = b_1 + 19D \Rightarrow \frac{1}{25} = \frac{1}{5} + 19D$.
$19D = \frac{1}{25} - \frac{1}{5} = \frac{1-5}{25} = -\frac{4}{25}$.
$D = -\frac{4}{19 \times 25} = -\frac{4}{475}$.
We want $a_n < 0$,which implies $\frac{1}{a_n} < 0$ is not necessarily true,but since $a_1 > 0$ and $a_{20} > 0$,we look for when the term becomes negative.
$b_n = \frac{1}{5} - (n-1)\frac{4}{475} < 0$.
$\frac{1}{5} < (n-1)\frac{4}{475}$.
$\frac{475}{20} < n-1$.
$23.75 < n-1$.
$n > 24.75$.
The least positive integer $n$ is $25$.

(A) Given that $\sin (\theta-\alpha), \sin \theta, \sin (\theta+\alpha)$ are in $H.P.$
$\Rightarrow \frac{1}{\sin (\theta-\alpha)}, \frac{1}{\sin \theta}, \frac{1}{\sin (\theta+\alpha)}$ are in $A.P.$
$\therefore \frac{2}{\sin \theta} = \frac{1}{\sin (\theta-\alpha)} + \frac{1}{\sin (\theta+\alpha)}$
$\Rightarrow \frac{2}{\sin \theta} = \frac{\sin (\theta+\alpha) + \sin (\theta-\alpha)}{\sin (\theta-\alpha) \sin (\theta+\alpha)}$
Using the formula $\sin (A+B) + \sin (A-B) = 2 \sin A \cos B$ and $\sin (A-B) \sin (A+B) = \sin^2 A - \sin^2 B$:
$\Rightarrow \frac{2}{\sin \theta} = \frac{2 \sin \theta \cos \alpha}{\sin^2 \theta - \sin^2 \alpha}$
$\Rightarrow \sin^2 \theta - \sin^2 \alpha = \sin^2 \theta \cos \alpha$
$\Rightarrow \sin^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$
Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:
$\Rightarrow \sin^2 \theta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$
$\Rightarrow \sin^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
Since $\sin^2 \theta = 1 - \cos^2 \theta$:
$1 - \cos^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
$\Rightarrow \cos^2 \theta = 1 - 2 \cos^2 \frac{\alpha}{2}$

(C) In a harmonic progression $(HP)$,the reciprocals of the terms form an arithmetic progression $(AP)$.
Let the corresponding $AP$ have the first term $A$ and common difference $D$.
Given $t_{7} = \frac{1}{10} \Rightarrow T_{7} = 10$,where $T_{n}$ is the $n^{\text{th}}$ term of the $AP$.
Given $t_{12} = \frac{1}{25} \Rightarrow T_{12} = 25$.
Using the formula $T_{n} = A + (n-1)D$:
$A + 6D = 10$ (Equation $1$)
$A + 11D = 25$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$: $5D = 15 \Rightarrow D = 3$.
Substituting $D = 3$ into Equation $1$: $A + 6(3) = 10$ $\Rightarrow A + 18 = 10$ $\Rightarrow A = -8$.
Now,find the $20^{\text{th}}$ term of the $AP$: $T_{20} = A + 19D = -8 + 19(3) = -8 + 57 = 49$.
Therefore,the $20^{\text{th}}$ term of the $HP$ is $t_{20} = \frac{1}{T_{20}} = \frac{1}{49}$.

(A) Given that $\frac{1}{4}, a, b, \frac{1}{19}$ are in $H.P.$
Therefore,their reciprocals $4, \frac{1}{a}, \frac{1}{b}, 19$ are in $A.P.$
Let the $A.P.$ be $4, 4+d, 4+2d, 4+3d$.
Here,$4+3d = 19 \implies 3d = 15 \implies d = 5$.
Thus,the terms are $4, 4+5, 4+10, 19$,which are $4, 9, 14, 19$.
Comparing the terms,$\frac{1}{a} = 9 \implies a = \frac{1}{9}$ and $\frac{1}{b} = 14 \implies b = \frac{1}{14}$.

(A) Given that $x_1, x_2, x_3, x_4$ are in harmonic progression $(HP)$,their reciprocals $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}, \frac{1}{x_4}$ are in arithmetic progression $(AP)$. Let these be $a-3d, a-d, a+d, a+3d$.
From $A x^2 - 4 x + 1 = 0$,the roots $x_1, x_3$ satisfy $\frac{1}{x_1} + \frac{1}{x_3} = 4$ and $\frac{1}{x_1} \cdot \frac{1}{x_3} = A$.
Substituting the $AP$ terms: $(a-3d) + (a+d) = 4 \implies 2a - 2d = 4 \implies a - d = 2$.
Also,$(a-3d)(a+d) = A$.
From $B x^2 - 6 x + 1 = 0$,the roots $x_2, x_4$ satisfy $\frac{1}{x_2} + \frac{1}{x_4} = 6$ and $\frac{1}{x_2} \cdot \frac{1}{x_4} = B$.
Substituting the $AP$ terms: $(a-d) + (a+3d) = 6 \implies 2a + 2d = 6 \implies a + d = 3$.
Solving the system $a-d=2$ and $a+d=3$,we get $2a=5 \implies a=2.5$ and $2d=1 \implies d=0.5$.
Now,$A = (a-3d)(a+d) = (2.5 - 1.5)(3) = 1 \times 3 = 3$.
And $B = (a-d)(a+3d) = (2)(2.5 + 1.5) = 2 \times 4 = 8$.
Finally,$\frac{B+A}{B-A} = \frac{8+3}{8-3} = \frac{11}{5}$.

(D) Let $\alpha, \beta, \gamma$ be the roots of the equation $x^3 - ax^2 + bx - c = 0$.
From Vieta's formulas:
$\alpha + \beta + \gamma = a$
$\alpha\beta + \beta\gamma + \gamma\alpha = b$
$\alpha\beta\gamma = c$
Given that $\alpha, \beta, \gamma$ are in $HP$,their reciprocals $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in $AP$.
The harmonic mean $(HM)$ of three numbers $\alpha, \beta, \gamma$ is defined as $HM = \frac{3}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}}$.
Substituting the values:
$HM = \frac{3}{\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}} = \frac{3(\alpha\beta\gamma)}{\alpha\beta + \beta\gamma + \gamma\alpha} = \frac{3c}{b}$.

(D) If $a, b, c$ are in harmonic progression,then $b = \frac{2ac}{a+c}$.
Given $\cos \left(x-\frac{\pi}{3}\right), \cos x, \cos \left(x+\frac{\pi}{3}\right)$ are in $H$.$P$.,we have:
$\cos x = \frac{2 \cos \left(x-\frac{\pi}{3}\right) \cos \left(x+\frac{\pi}{3}\right)}{\cos \left(x-\frac{\pi}{3}\right) + \cos \left(x+\frac{\pi}{3}\right)}$
Using the identity $\cos(A-B)\cos(A+B) = \cos^2 A - \sin^2 B$ and $\cos(A-B) + \cos(A+B) = 2 \cos A \cos B$:
$\cos x = \frac{2 \left(\cos^2 x - \sin^2 \frac{\pi}{3}\right)}{2 \cos x \cos \frac{\pi}{3}}$
$\cos x = \frac{\cos^2 x - \sin^2 \frac{\pi}{3}}{\cos x \cdot \frac{1}{2}}$
$\frac{1}{2} \cos^2 x = \cos^2 x - \sin^2 \frac{\pi}{3}$
$\sin^2 \frac{\pi}{3} = \cos^2 x - \frac{1}{2} \cos^2 x$
$\frac{3}{4} = \frac{1}{2} \cos^2 x$
$\cos^2 x = \frac{3}{2}$
$\cos x = \pm \sqrt{\frac{3}{2}}$
Thus,the correct option is $D$.

(B) The given equation is $16x^3 - 44x^2 + 36x - 9 = 0$.
Let the roots be $\alpha, \beta, \gamma$ in harmonic progression ($H$.$P$.).
Then $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in arithmetic progression ($A$.$P$.).
This implies $\frac{2}{\beta} = \frac{1}{\alpha} + \frac{1}{\gamma}$.
From the properties of roots,$\sum \alpha\beta = \frac{36}{16} = \frac{9}{4}$ and $\alpha\beta\gamma = \frac{9}{16}$.
Also,$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{9/4}{9/16} = 4$.
Substituting $\frac{1}{\alpha} + \frac{1}{\gamma} = \frac{2}{\beta}$,we get $\frac{2}{\beta} + \frac{1}{\beta} = 4$ $\Rightarrow \frac{3}{\beta} = 4$ $\Rightarrow \beta = \frac{3}{4}$.
Now,$\alpha + \gamma = \frac{44}{16} - \frac{3}{4} = \frac{11}{4} - \frac{3}{4} = 2$ and $\alpha\gamma = \frac{9/16}{3/4} = \frac{3}{4}$.
Solving $t^2 - 2t + \frac{3}{4} = 0$,we get $4t^2 - 8t + 3 = 0 \Rightarrow (2t - 1)(2t - 3) = 0$.
Thus,the roots are $\frac{1}{2}, \frac{3}{4}, \frac{3}{2}$.
The greatest root is $\frac{3}{2}$.

(A) Given that $\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in harmonic progression $(HP)$.
Therefore,$\frac{1}{\cos (\theta-\alpha)}, \frac{1}{\cos \theta}, \frac{1}{\cos (\theta+\alpha)}$ are in arithmetic progression $(AP)$.
So,$\frac{2}{\cos \theta} = \frac{1}{\cos (\theta-\alpha)} + \frac{1}{\cos (\theta+\alpha)}$.
$\frac{2}{\cos \theta} = \frac{\cos (\theta+\alpha) + \cos (\theta-\alpha)}{\cos (\theta-\alpha) \cos (\theta+\alpha)}$.
Using the formula $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$,we get:
$\frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \alpha}{\cos (\theta-\alpha) \cos (\theta+\alpha)}$.
$\cos^2 \theta \cos \alpha = \cos (\theta-\alpha) \cos (\theta+\alpha)$.
Using $\cos (A-B) \cos (A+B) = \cos^2 A - \sin^2 B$,we get:
$\cos^2 \theta \cos \alpha = \cos^2 \theta - \sin^2 \alpha$.
$\sin^2 \alpha = \cos^2 \theta (1 - \cos \alpha)$.
$\cos^2 \theta = \frac{\sin^2 \alpha}{1 - \cos \alpha} = \frac{4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}}{2 \sin^2 \frac{\alpha}{2}} = 2 \cos^2 \frac{\alpha}{2}$.
Now,$\tan^2 \theta = \sec^2 \theta - 1 = \frac{1}{\cos^2 \theta} - 1 = \frac{1}{2 \cos^2 \frac{\alpha}{2}} - 1 = \frac{1}{2} \sec^2 \frac{\alpha}{2} - 1$.
$2 \tan^2 \theta = \sec^2 \frac{\alpha}{2} - 2 = (1 + \tan^2 \frac{\alpha}{2}) - 2 = \tan^2 \frac{\alpha}{2} - 1$.

(B) Given $\tan B = \frac{2 \sin A \sin C}{\sin (A+C)}$.
Taking the reciprocal on both sides,we get $\frac{1}{\tan B} = \frac{\sin (A+C)}{2 \sin A \sin C}$.
Using the expansion $\sin (A+C) = \sin A \cos C + \cos A \sin C$,we have $\frac{1}{\tan B} = \frac{\sin A \cos C + \cos A \sin C}{2 \sin A \sin C}$.
This simplifies to $\frac{1}{\tan B} = \frac{1}{2} (\cot A + \cot C) = \frac{1}{2} (\frac{1}{\tan A} + \frac{1}{\tan C})$.
Multiplying by $2$,we get $\frac{2}{\tan B} = \frac{1}{\tan A} + \frac{1}{\tan C}$.
This is the condition for $\tan A, \tan B, \tan C$ to be in Harmonic Progression.

(C) Given the equation $k x^3 - 18 x^2 - 36 x + 8 = 0$.
Let the roots be $\alpha, \beta, \gamma$ which are in harmonic progression $(HP)$.
This implies that $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in arithmetic progression $(AP)$.
Replacing $x$ with $\frac{1}{x}$ in the original equation,we get $k(\frac{1}{x})^3 - 18(\frac{1}{x})^2 - 36(\frac{1}{x}) + 8 = 0$,which simplifies to $8 x^3 - 36 x^2 - 18 x + k = 0$.
Let the roots of this new equation be $a-d, a, a+d$.
From the sum of roots,$(a-d) + a + (a+d) = -(\frac{-36}{8}) = \frac{9}{2}$ $\Rightarrow 3a = \frac{9}{2}$ $\Rightarrow a = \frac{3}{2}$.
Since $a = \frac{3}{2}$ is a root of $8 x^3 - 36 x^2 - 18 x + k = 0$,we substitute it:
$8(\frac{3}{2})^3 - 36(\frac{3}{2})^2 - 18(\frac{3}{2}) + k = 0$.
$8(\frac{27}{8}) - 36(\frac{9}{4}) - 27 + k = 0$.
$27 - 81 - 27 + k = 0$.
$k - 81 = 0 \Rightarrow k = 81$.

(A) Let the roots of the equation $x^3+3px^2+3qx+r=0$ be $\alpha, \beta, \gamma$. Since they are in harmonic progression,their reciprocals $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in arithmetic progression. Let $\frac{1}{\alpha} = a-d, \frac{1}{\beta} = a, \frac{1}{\gamma} = a+d$.
Substituting $x = \frac{1}{y}$ into the original equation,we get $\frac{1}{y^3} + \frac{3p}{y^2} + \frac{3q}{y} + r = 0$,which simplifies to $ry^3 + 3qy^2 + 3py + 1 = 0$.
Since the roots of this equation are in arithmetic progression,the sum of the roots is $3a = -\frac{3q}{r}$,so $a = -\frac{q}{r}$.
Since $a$ is a root of $ry^3 + 3qy^2 + 3py + 1 = 0$,we have $r(-\frac{q}{r})^3 + 3q(-\frac{q}{r})^2 + 3p(-\frac{q}{r}) + 1 = 0$.
$-\frac{q^3}{r^2} + \frac{3q^3}{r^2} - \frac{3pq}{r} + 1 = 0$.
Multiplying by $r^2$,we get $-q^3 + 3q^3 - 3pqr + r^2 = 0$,which simplifies to $2q^3 = 3pqr - r^2 = r(3pq-r)$.

(C) Let $a^{p} = b^{q} = c^{r} = k$.
Since $a, b, c$ are positive real numbers,we can write $a = k^{1/p}$,$b = k^{1/q}$,and $c = k^{1/r}$.
Given that $a, b, c$ are in $GP$,we have $\frac{b}{a} = \frac{c}{b}$.
Substituting the values of $a, b, c$,we get $\frac{k^{1/q}}{k^{1/p}} = \frac{k^{1/r}}{k^{1/q}}$.
This implies $k^{(1/q - 1/p)} = k^{(1/r - 1/q)}$.
Equating the exponents,we get $\frac{1}{q} - \frac{1}{p} = \frac{1}{r} - \frac{1}{q}$,which simplifies to $\frac{2}{q} = \frac{1}{p} + \frac{1}{r}$.
This shows that $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in $A.P.$
Therefore,$p, q, r$ are in $H.P.$

(C) Given that $a, b, c$ are in $G$.$P$.,we have $b^2 = ac$.
Taking the logarithm on both sides with base $x$,we get $2 \log_x b = \log_x a + \log_x c$.
This implies that $\log_x a, \log_x b, \log_x c$ are in $A$.$P$.
Using the change of base formula,$\log_a x = \frac{1}{\log_x a}$,$\log_b x = \frac{1}{\log_x b}$,and $\log_c x = \frac{1}{\log_x c}$.
Since the reciprocals of $\log_x a, \log_x b, \log_x c$ are in $A$.$P$.,the terms $\log_a x, \log_b x, \log_c x$ must be in $H$.$P$.

(C) Given that $\frac{a_r - a_{r+1}}{a_r a_{r+1}} = K$ (where $K$ is a constant).
Dividing the terms,we get $\frac{a_r}{a_r a_{r+1}} - \frac{a_{r+1}}{a_r a_{r+1}} = K$.
This simplifies to $\frac{1}{a_{r+1}} - \frac{1}{a_r} = K$.
Since the difference between the reciprocals of consecutive terms is constant,the sequence $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$ is an $A$.$P$.
Therefore,the sequence $a_1, a_2, a_3, \ldots$ is in $H$.$P$.

(A) Given,$f(x) = x + \frac{1}{2} = \frac{2x+1}{2}$.
$f(2x) = 2x + \frac{1}{2} = \frac{4x+1}{2}$.
$f(4x) = 4x + \frac{1}{2} = \frac{8x+1}{2}$.
Since $f(x), f(2x), f(4x)$ are in $HP$,their reciprocals $\frac{1}{f(x)}, \frac{1}{f(2x)}, \frac{1}{f(4x)}$ are in $AP$.
Therefore,$\frac{2}{f(2x)} = \frac{1}{f(x)} + \frac{1}{f(4x)}$.
Substituting the values: $\frac{2}{(4x+1)/2} = \frac{2}{2x+1} + \frac{2}{8x+1}$.
$\frac{4}{4x+1} = 2 \left( \frac{8x+1 + 2x+1}{(2x+1)(8x+1)} \right)$.
$\frac{2}{4x+1} = \frac{10x+2}{(2x+1)(8x+1)}$.
$2(2x+1)(8x+1) = (4x+1)(10x+2)$.
$2(16x^2 + 10x + 1) = 40x^2 + 18x + 2$.
$32x^2 + 20x + 2 = 40x^2 + 18x + 2$.
$8x^2 - 2x = 0$.
$2x(4x - 1) = 0$.
This gives $x = 0$ or $x = 1/4$.
If $x = 0$,the terms are $1/2, 1/2, 1/2$,which are not unequal.
If $x = 1/4$,the terms are $3/4, 1, 3/2$,which are in $HP$ (since $4/3, 1, 2/3$ are in $AP$ with common difference $-1/3$).
Thus,there is only $1$ real value of $x$.

(A) Let the five terms in $HP$ be $\frac{1}{a-2d}, \frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}$.
Given the middle term is $1$,so $\frac{1}{a} = 1$,which implies $a = 1$.
The ratio of the second term to the fourth term is $\frac{2}{1}$,so $\frac{\frac{1}{a-d}}{\frac{1}{a+d}} = 2$.
This simplifies to $\frac{a+d}{a-d} = 2$,so $a+d = 2a - 2d$,which gives $3d = a$.
Since $a = 1$,we have $d = \frac{1}{3}$.
The first three terms are $\frac{1}{1-2(\frac{1}{3})}, \frac{1}{1-(\frac{1}{3})}, \frac{1}{1}$.
These terms are $\frac{1}{1/3}, \frac{1}{2/3}, 1$,which are $3, \frac{3}{2}, 1$.
The sum of the first three terms is $3 + \frac{3}{2} + 1 = 4 + 1.5 = \frac{11}{2}$.