Harmonic progression Questions in English

(A) The given series is $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$
Rewriting the terms: $27, \frac{27}{3}, \frac{27}{5}, \frac{27}{7}, \dots$
This can be expressed as $\frac{27}{1}, \frac{27}{3}, \frac{27}{5}, \frac{27}{7}, \dots$
The denominator follows an arithmetic progression: $1, 3, 5, 7, \dots$ where the $n^{th}$ term is $a_n = 1 + (n-1)2 = 2n - 1$.
Thus,the $n^{th}$ term of the series is $T_n = \frac{27}{2n - 1}$.
For the $9^{th}$ term $(n = 9)$:
$T_9 = \frac{27}{2(9) - 1} = \frac{27}{18 - 1} = \frac{27}{17} = 1\frac{10}{17}$.

(C) Given $T_m = n$ and $T_n = m$ for a $H.P.$
Therefore,for the corresponding $A.P.$,the $m^{th}$ term is $\frac{1}{n}$ and the $n^{th}$ term is $\frac{1}{m}$.
Let $a$ and $d$ be the first term and common difference of this $A.P.$,then:
$a + (m - 1)d = \frac{1}{n}$ ---$(i)$
$a + (n - 1)d = \frac{1}{m}$ ---(ii)
Subtracting (ii) from $(i)$:
$(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$
$d = \frac{1}{mn}$
Substituting $d$ in $(i)$:
$a + (m - 1)\frac{1}{mn} = \frac{1}{n}$
$a = \frac{1}{n} - \frac{m - 1}{mn} = \frac{m - m + 1}{mn} = \frac{1}{mn}$
Now,the $r^{th}$ term of the corresponding $A.P.$ is:
$T_r = a + (r - 1)d = \frac{1}{mn} + (r - 1)\frac{1}{mn} = \frac{1 + r - 1}{mn} = \frac{r}{mn}$
Therefore,the $r^{th}$ term of the corresponding $H.P.$ is the reciprocal of the $A.P.$ term:
$T_r(H.P.) = \frac{mn}{r}$.

(D) Let the number to be added be $x$.
Then the new numbers are $(13 + x), (15 + x), (19 + x)$.
Since these numbers are in $H.P.$,their reciprocals must be in $A.P.$
Therefore,$\frac{1}{15 + x} - \frac{1}{13 + x} = \frac{1}{19 + x} - \frac{1}{15 + x}$.
This implies $\frac{(13 + x) - (15 + x)}{(15 + x)(13 + x)} = \frac{(15 + x) - (19 + x)}{(19 + x)(15 + x)}$.
$\frac{-2}{(15 + x)(13 + x)} = \frac{-4}{(19 + x)(15 + x)}$.
Canceling $(15 + x)$ from both sides (assuming $x \neq -15$),we get $\frac{-2}{13 + x} = \frac{-4}{19 + x}$.
$2(19 + x) = 4(13 + x)$.
$38 + 2x = 52 + 4x$.
$-14 = 2x$.
$x = -7$.

(D) The given series $2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ is in $H.P.$
Taking the reciprocals,the series $\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \dots$ is in $A.P.$
For this $A.P.$,the first term $a = \frac{1}{2}$ and the common difference $d = \frac{2}{5} - \frac{1}{2} = \frac{4-5}{10} = -\frac{1}{10}$.
The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$.
For $n = 5$,$T_5 = \frac{1}{2} + (5-1)\left(-\frac{1}{10}\right) = \frac{1}{2} - \frac{4}{10} = \frac{5-4}{10} = \frac{1}{10}$.
Since the $n^{th}$ term of $H.P.$ is the reciprocal of the $n^{th}$ term of the corresponding $A.P.$,the $5^{th}$ term of the $H.P.$ is $\frac{1}{1/10} = 10$.

(C) Since ${a_1}, {a_2}, {a_3}, \dots, {a_n}$ 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 of this $A.P.$ be $d$.
Then,$\frac{1}{a_{k+1}} - \frac{1}{a_k} = d$ for $k = 1, 2, \dots, n-1$.
This implies $\frac{a_k - a_{k+1}}{a_k a_{k+1}} = d$,or $a_k a_{k+1} = \frac{1}{d}(a_k - a_{k+1})$.
Summing this from $k=1$ to $n-1$:
$\sum_{k=1}^{n-1} a_k a_{k+1} = \frac{1}{d} \sum_{k=1}^{n-1} (a_k - a_{k+1}) = \frac{1}{d} (a_1 - a_n)$.
For the $A.P.$,the $n^{th}$ term is $\frac{1}{a_n} = \frac{1}{a_1} + (n-1)d$,so $d = \frac{a_1 - a_n}{(n-1)a_1 a_n}$.
Substituting $d$ into the sum:
$\sum_{k=1}^{n-1} a_k a_{k+1} = \frac{1}{\frac{a_1 - a_n}{(n-1)a_1 a_n}} (a_1 - a_n) = (n-1)a_1 a_n$.

(B) Given that $x, y, z$ are in $H.P.$,we have $y = \frac{2xz}{x + z}$.
Consider the expression $E = \log(x + z) + \log(x - 2y + z)$.
Using the property $\log(a) + \log(b) = \log(ab)$,we get $E = \log((x + z)(x - 2y + z))$.
Substitute $y = \frac{2xz}{x + z}$ into the expression:
$E = \log\left((x + z)\left(x + z - 2\left(\frac{2xz}{x + z}\right)\right)\right)$
$E = \log\left((x + z)\left(x + z - \frac{4xz}{x + z}\right)\right)$
$E = \log\left((x + z)^2 - 4xz\right)$
$E = \log(x^2 + 2xz + z^2 - 4xz) = \log(x^2 - 2xz + z^2)$
$E = \log((x - z)^2) = 2\log(x - z)$.

(A) Let the corresponding $A.P.$ be $a, a+d, a+2d, \dots$
Since the $5^{th}$ term of the $H.P.$ is $\frac{1}{45}$,the $5^{th}$ term of the $A.P.$ is $45$.
So,$a + 4d = 45$ $(i)$
Since the $11^{th}$ term of the $H.P.$ is $\frac{1}{69}$,the $11^{th}$ term of the $A.P.$ is $69$.
So,$a + 10d = 69$ $(ii)$
Subtracting $(i)$ from $(ii)$,we get $(a + 10d) - (a + 4d) = 69 - 45$,which gives $6d = 24$,so $d = 4$.
Substituting $d = 4$ into $(i)$,we get $a + 4(4) = 45$,so $a + 16 = 45$,which gives $a = 29$.
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 $\frac{1}{89}$.

(B) In a harmonic progression $(H.P.)$,the reciprocals of the terms form an arithmetic progression $(A.P.)$.
Given the first term $a_1 = 1/7$ and the second term $a_2 = 1/9$,the corresponding terms of the $A.P.$ are $7$ and $9$.
The first term of the $A.P.$ is $A = 7$ and the common difference is $d = 9 - 7 = 2$.
The $n^{th}$ term of an $A.P.$ is given by $A_n = A + (n - 1)d$.
For $n = 12$,$A_{12} = 7 + (12 - 1) \times 2 = 7 + 11 \times 2 = 7 + 22 = 29$.
Therefore,the $12^{th}$ term of the $H.P.$ is the reciprocal of $A_{12}$,which is $1/29$.

(D) Given that $a, b, c$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$
Let $\frac{1}{a} = p - q, \frac{1}{b} = p, \frac{1}{c} = p + q$,where $p, q > 0$ and $p > q$.
Then $a = \frac{1}{p-q}, b = \frac{1}{p}, c = \frac{1}{p+q}$.
Substituting these into the expression $E = \frac{3a + 2b}{2a - b} + \frac{3c + 2b}{2c - b}$:
$E = \frac{\frac{3}{p-q} + \frac{2}{p}}{\frac{2}{p-q} - \frac{1}{p}} + \frac{\frac{3}{p+q} + \frac{2}{p}}{\frac{2}{p+q} - \frac{1}{p}}$
$E = \frac{3p + 2(p-q)}{2p - (p-q)} + \frac{3p + 2(p+q)}{2p - (p+q)} = \frac{5p - 2q}{p + q} + \frac{5p + 2q}{p - q}$
$E = \frac{(5p - 2q)(p - q) + (5p + 2q)(p + q)}{p^2 - q^2} = \frac{5p^2 - 7pq + 2q^2 + 5p^2 + 7pq + 2q^2}{p^2 - q^2} = \frac{10p^2 + 4q^2}{p^2 - q^2}$
$E = \frac{10(p^2 - q^2) + 14q^2}{p^2 - q^2} = 10 + \frac{14q^2}{p^2 - q^2}$.
Since $p > q > 0$,$p^2 - q^2 > 0$,so $E > 10$.

(A) Since $a, b, c, d$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$
Let the common difference be $k$. Then $\frac{1}{b} = \frac{1}{a} + k$,$\frac{1}{c} = \frac{1}{a} + 2k$,and $\frac{1}{d} = \frac{1}{a} + 3k$.
Alternatively,using the property of $H.P.$,$b = \frac{2ac}{a+c}$ and $c = \frac{2bd}{b+d}$.
From $b = \frac{2ac}{a+c}$,we get $ab + bc = 2ac$. From $c = \frac{2bd}{b+d}$,we get $bc + cd = 2bd$.
Using the property for four terms in $H.P.$,we have $ab + bc + cd = 3ad$.
Verification: Let $a=1, b=\frac{1}{2}, c=\frac{1}{3}, d=\frac{1}{4}$.
$ab + bc + cd = (1 \times \frac{1}{2}) + (\frac{1}{2} \times \frac{1}{3}) + (\frac{1}{3} \times \frac{1}{4}) = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{6+2+1}{12} = \frac{9}{12} = \frac{3}{4}$.
Since $3ad = 3(1 \times \frac{1}{4}) = \frac{3}{4}$,the result is $3ad$.

(D) Let the harmonic progression $(H.P.)$ be $H_1, H_2, \dots, H_n$. The corresponding arithmetic progression $(A.P.)$ is $A_1, A_2, \dots, A_n$ where $A_n = \frac{1}{H_n}$.
Given that the $7^{th}$ term of $H.P.$ is $8$,the $7^{th}$ term of $A.P.$ is $A_7 = \frac{1}{8}$.
Given that the $8^{th}$ term of $H.P.$ is $7$,the $8^{th}$ term of $A.P.$ is $A_8 = \frac{1}{7}$.
For an $A.P.$,$A_n = a + (n-1)d$. Thus:
$a + 6d = \frac{1}{8}$ (Equation $1$)
$a + 7d = \frac{1}{7}$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$ gives $d = \frac{1}{7} - \frac{1}{8} = \frac{8-7}{56} = \frac{1}{56}$.
Substituting $d$ into Equation $1$: $a + 6(\frac{1}{56}) = \frac{1}{8} \implies a = \frac{1}{8} - \frac{6}{56} = \frac{7-6}{56} = \frac{1}{56}$.
The $15^{th}$ term of the $A.P.$ is $A_{15} = a + 14d = \frac{1}{56} + 14(\frac{1}{56}) = \frac{15}{56}$.
Therefore,the $15^{th}$ term of the $H.P.$ is $H_{15} = \frac{1}{A_{15}} = \frac{56}{15}$.

(D) Let the $H.P.$ be the reciprocal of an $A.P.$ with first term $a$ and common difference $d$.
Given that the $7^{th}$ term of $H.P.$ is $\frac{1}{10}$,the $7^{th}$ term of the corresponding $A.P.$ is $10$.
So,$a + 6d = 10$ $(i)$
Given that the $12^{th}$ term of $H.P.$ is $\frac{1}{25}$,the $12^{th}$ term of the corresponding $A.P.$ is $25$.
So,$a + 11d = 25$ $(ii)$
Subtracting $(i)$ from $(ii)$:
$(a + 11d) - (a + 6d) = 25 - 10$
$5d = 15 \Rightarrow d = 3$
Substituting $d = 3$ in $(i)$:
$a + 6(3) = 10$ $\Rightarrow a + 18 = 10$ $\Rightarrow a = -8$
Now,the $20^{th}$ term of the $A.P.$ is $T_{20} = a + 19d = -8 + 19(3) = -8 + 57 = 49$.
Therefore,the $20^{th}$ term of the $H.P.$ is the reciprocal of $49$,which is $\frac{1}{49}$.

(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d.$
Since the terms of a $H.P.$ are the reciprocals of the terms of an $A.P.,$ we have:
$T_6$ of $H.P. = \frac{1}{61} \implies T_6$ of $A.P. = a + 5d = 61$ $(i)$
$T_{10}$ of $H.P. = \frac{1}{105} \implies T_{10}$ of $A.P. = a + 9d = 105$ $(ii)$
Subtracting $(i)$ from $(ii):$
$(a + 9d) - (a + 5d) = 105 - 61$
$4d = 44 \implies d = 11$
Substituting $d = 11$ into $(i):$
$a + 5(11) = 61$
$a + 55 = 61 \implies a = 6$
Thus,the first term of the $H.P.$ is $\frac{1}{a} = \frac{1}{6}$.

(B) Let the first term of the corresponding $A.P.$ be $A$ and the common difference be $D$. Since the terms of $H.P.$ are the reciprocals of the terms of $A.P.$,we have:
$T_p = \frac{1}{q} \implies A + (p - 1)D = \frac{1}{q} \quad (i)$
$T_q = \frac{1}{p} \implies A + (q - 1)D = \frac{1}{p} \quad (ii)$
Subtracting $(ii)$ from $(i)$:
$(p - q)D = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq}$
$D = \frac{1}{pq}$
Substituting $D$ in $(i)$:
$A + (p - 1)\frac{1}{pq} = \frac{1}{q}$
$A = \frac{1}{q} - \frac{p - 1}{pq} = \frac{p - p + 1}{pq} = \frac{1}{pq}$
Now,the $(pq)^{th}$ term of the $A.P.$ is:
$T_{pq} = A + (pq - 1)D = \frac{1}{pq} + (pq - 1)\frac{1}{pq} = \frac{1 + pq - 1}{pq} = 1$
Since the $(pq)^{th}$ term of the $A.P.$ is $1$,the $(pq)^{th}$ term of the $H.P.$ is the reciprocal of $1$,which is $1$.

(B) Let the $H.P.$ be $\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \dots$ where the corresponding $A.P.$ is $a, a+d, a+2d, \dots$
Given that the $4^{th}$ term of $H.P.$ is $\frac{3}{5},$ the $4^{th}$ term of $A.P.$ is $\frac{5}{3}.$ Thus,$a + 3d = \frac{5}{3} \dots (1)$
Given that the $8^{th}$ term of $H.P.$ is $\frac{1}{3},$ the $8^{th}$ term of $A.P.$ is $3.$ Thus,$a + 7d = 3 \dots (2)$
Subtracting $(1)$ from $(2),$ we get $(a + 7d) - (a + 3d) = 3 - \frac{5}{3} \implies 4d = \frac{4}{3} \implies d = \frac{1}{3}.$
Substituting $d = \frac{1}{3}$ into $(1),$ we get $a + 3(\frac{1}{3}) = \frac{5}{3} \implies a + 1 = \frac{5}{3} \implies a = \frac{2}{3}.$
The $6^{th}$ term of the $A.P.$ is $a + 5d = \frac{2}{3} + 5(\frac{1}{3}) = \frac{2+5}{3} = \frac{7}{3}.$
Therefore,the $6^{th}$ term of the $H.P.$ is the reciprocal of $\frac{7}{3},$ which is $\frac{3}{7}.$
Hence,the correct option is $B$.

(A) Given that $H$ is the harmonic mean between $p$ and $q$,we have $H = \frac{2pq}{p + q}$.
Now,we need to evaluate the expression $\frac{H}{p} + \frac{H}{q}$.
Substituting the value of $H$:
$\frac{H}{p} + \frac{H}{q} = \frac{1}{p} \left( \frac{2pq}{p + q} \right) + \frac{1}{q} \left( \frac{2pq}{p + q} \right)$
$= \frac{2q}{p + q} + \frac{2p}{p + q}$
$= \frac{2q + 2p}{p + q}$
$= \frac{2(p + q)}{p + q}$
$= 2$.

(A) Given that $H$ is the harmonic mean between $a$ and $b$,we have $H = \frac{2ab}{a + b}$.
Substituting this into the expression $\frac{1}{H - a} + \frac{1}{H - b}$:
$\frac{1}{\frac{2ab}{a + b} - a} + \frac{1}{\frac{2ab}{a + b} - b} = \frac{a + b}{2ab - a(a + b)} + \frac{a + b}{2ab - b(a + b)}$
$= \frac{a + b}{2ab - a^2 - ab} + \frac{a + b}{2ab - ab - b^2} = \frac{a + b}{ab - a^2} + \frac{a + b}{ab - b^2}$
$= \frac{a + b}{a(b - a)} + \frac{a + b}{b(a - b)} = \frac{a + b}{a(b - a)} - \frac{a + b}{b(b - a)}$
$= \frac{a + b}{b - a} \left( \frac{1}{a} - \frac{1}{b} \right) = \frac{a + b}{b - a} \left( \frac{b - a}{ab} \right)$
$= \frac{a + b}{ab} = \frac{1}{b} + \frac{1}{a}$.

(A) Let the two numbers be $a = 3$ and $b = \frac{6}{13}$.
The $n$-th $H.M.$ between $a$ and $b$ is given by the formula:
$H_n = \frac{(n+1)ab}{na + b}$
For the sixth $H.M.$,$n = 6$:
$H_6 = \frac{(6+1) \times 3 \times \frac{6}{13}}{6 \times 3 + \frac{6}{13}}$
$H_6 = \frac{7 \times 3 \times \frac{6}{13}}{18 + \frac{6}{13}}$
$H_6 = \frac{\frac{126}{13}}{\frac{18 \times 13 + 6}{13}}$
$H_6 = \frac{126}{234 + 6} = \frac{126}{240}$
Simplifying the fraction by dividing both numerator and denominator by $2$:
$H_6 = \frac{63}{120}$

(B) The harmonic mean between $a$ and $b$ is given by $\frac{2ab}{a + b}$.
Given that $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n} = \frac{2ab}{a + b}$.
Cross-multiplying,we get $(a^{n + 1} + b^{n + 1})(a + b) = 2ab(a^n + b^n)$.
Expanding both sides: $a^{n + 2} + a^{n + 1}b + ab^{n + 1} + b^{n + 2} = 2a^{n + 1}b + 2ab^{n + 1}$.
Rearranging the terms: $a^{n + 2} + b^{n + 2} - a^{n + 1}b - ab^{n + 1} = 0$.
$a^{n + 1}(a - b) - b^{n + 1}(a - b) = 0$.
$(a^{n + 1} - b^{n + 1})(a - b) = 0$.
Since $a \neq b$,we must have $a^{n + 1} = b^{n + 1}$.
This implies $(\frac{a}{b})^{n + 1} = 1 = (\frac{a}{b})^0$.
Therefore,$n + 1 = 0$,which gives $n = -1$.

(B) Given that $H$ is the harmonic mean between $a$ and $b$,we have $H = \frac{2ab}{a + b}$.
Consider the expression $E = \frac{H + a}{H - a} + \frac{H + b}{H - b}$.
Substituting $H = \frac{2ab}{a + b}$ into the first term:
$\frac{H + a}{H - a} = \frac{\frac{2ab}{a + b} + a}{\frac{2ab}{a + b} - a} = \frac{2ab + a(a + b)}{2ab - a(a + b)} = \frac{2ab + a^2 + ab}{2ab - a^2 - ab} = \frac{a^2 + 3ab}{ab - a^2} = \frac{a(a + 3b)}{a(b - a)} = \frac{a + 3b}{b - a}$.
Similarly,for the second term:
$\frac{H + b}{H - b} = \frac{\frac{2ab}{a + b} + b}{\frac{2ab}{a + b} - b} = \frac{2ab + b(a + b)}{2ab - b(a + b)} = \frac{2ab + ab + b^2}{2ab - ab - b^2} = \frac{3ab + b^2}{ab - b^2} = \frac{b(3a + b)}{b(a - b)} = \frac{3a + b}{a - b} = -\frac{3a + b}{b - a}$.
Adding the two terms:
$E = \frac{a + 3b}{b - a} - \frac{3a + b}{b - a} = \frac{a + 3b - 3a - b}{b - a} = \frac{2b - 2a}{b - a} = \frac{2(b - a)}{b - a} = 2$.

(C) Given that $a, b, c$ are in $H.P.$,we have $b = \frac{2ac}{a+c}$.
Since $a, b, c$ are positive real numbers,by the property of $A.M. > H.M.$,we have $\frac{a+c}{2} > b$.
Squaring both sides,we get $\left(\frac{a+c}{2}\right)^2 > b^2$,which implies $\frac{a^2 + 2ac + c^2}{4} > b^2$.
Since $b = \frac{2ac}{a+c}$,we have $ac = \frac{b(a+c)}{2}$.
Alternatively,using the power mean inequality,for $n=2$,$\frac{a^2+c^2}{2} > \left(\frac{a+c}{2}\right)^2$.
Since $\frac{a+c}{2} > b$,it follows that $\left(\frac{a+c}{2}\right)^2 > b^2$.
Therefore,$\frac{a^2+c^2}{2} > b^2$,which simplifies to $a^2 + c^2 > 2b^2$.

(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 four terms in $H.P.$,we know that $a, b, c$ are in $H.P.$,so $b = \frac{2ac}{a+c}$. Since $A.M. > H.M.$,we have $\frac{a+c}{2} > b$,which implies $a+c > 2b$.
Similarly,for $b, c, d$ in $H.P.$,we have $b+d > 2c$.
Adding these two inequalities: $(a+c) + (b+d) > 2b + 2c$,which simplifies to $a+d > b+c$. Thus,$(a)$ is true.
Also,for $a, b, c$ in $H.P.$,$ac > b^2$. For $b, c, d$ in $H.P.$,$bd > c^2$.
Multiplying these: $(ac)(bd) > (b^2)(c^2)$,which simplifies to $ad > bc$. Thus,$(b)$ is true.
Therefore,both $(a)$ and $(b)$ are correct.

(C) Given that $\log_a x, \log_b x, \log_c x$ are in $H.P.$
Using the change of base formula,we can write these as $\frac{\log x}{\log a}, \frac{\log x}{\log b}, \frac{\log x}{\log c}$ are in $H.P.$
Taking the reciprocal,we get $\frac{\log a}{\log x}, \frac{\log b}{\log x}, \frac{\log c}{\log x}$ are in $A.P.$
This simplifies to $\log_x a, \log_x b, \log_x c$ are in $A.P.$
By the property of logarithms,if $\log_x a, \log_x b, \log_x c$ are in $A.P.$,then $a, b, c$ must be in $G.P.$

(C) Given: $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$
Rearranging the terms: $\frac{1}{b - a} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b - c}$
$\frac{a - (b - a)}{a(b - a)} = \frac{(b - c) - c}{c(b - c)}$
$\frac{2a - b}{a(b - a)} = \frac{b - 2c}{c(b - c)}$
$(2a - b)(bc - c^2) = (b - 2c)(ab - a^2)$
$2abc - 2ac^2 - b^2c + bc^2 = ab^2 - a^2b - 2abc + 2a^2c$
$4abc - 2ac^2 - b^2c + bc^2 - ab^2 + a^2b - 2a^2c = 0$
This simplifies to $b(a+c) = 2ac$,which is the condition for $a, b, c$ to be in $H.P.$
Thus,$a, b, c$ are in $H.P.$

(C) Let ${a^x} = {b^y} = {c^z} = {d^u} = k$ (say).
Then $a = {k^{1/x}}, b = {k^{1/y}}, c = {k^{1/z}}, d = {k^{1/u}}$.
Since $a, b, c, d$ are in $G.P.$,their logarithms are in $A.P.$ or we can use the property of $G.P.$: ${b^2} = ac$ and ${c^2} = bd$.
Substituting the values,we get ${k^{2/y}} = {k^{1/x}} \cdot {k^{1/z}} = {k^{(1/x + 1/z)}}$.
This implies $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$,which means $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
Therefore,$x, y, z$ are in $H.P.$
Similarly,for $b, c, d$ in $G.P.$,we get $\frac{2}{z} = \frac{1}{y} + \frac{1}{u}$,implying $y, z, u$ are in $H.P.$
Thus,$x, y, z, u$ are in $H.P.$

(C) Given that $a, b, c$ are in $H.P.$
This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$
Divide each term by the sum $(a+b+c)$ or multiply by $(a+b+c)$,we get $\frac{a+b+c}{a}, \frac{a+b+c}{b}, \frac{a+b+c}{c}$ are in $A.P.$
Subtracting $1$ from each term: $\frac{a+b+c}{a} - 1, \frac{a+b+c}{b} - 1, \frac{a+b+c}{c} - 1$ are in $A.P.$
This simplifies to $\frac{b+c}{a}, \frac{a+c}{b}, \frac{a+b}{c}$ are in $A.P.$
Taking the reciprocal of each term,we get $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in $H.P.$

(B) Given that $\frac{x + y}{2}, y, \frac{y + z}{2}$ are in $H.P.$
Since the terms are in $H.P.$,their reciprocals $\frac{2}{x + y}, \frac{1}{y}, \frac{2}{y + z}$ are in $A.P.$
Therefore,$2 \times \frac{1}{y} = \frac{2}{x + y} + \frac{2}{y + z}$
$\frac{2}{y} = 2 \left( \frac{1}{x + y} + \frac{1}{y + z} \right)$
$\frac{1}{y} = \frac{y + z + x + y}{(x + y)(y + z)}$
$(x + y)(y + z) = y(x + 2y + z)$
$xy + xz + y^2 + yz = xy + 2y^2 + yz$
$xz = y^2$
Since $y^2 = xz$,the terms $x, y, z$ are in $G.P.$

(B) Since $a, b, c$ are in $H.P.$,we have $b = \frac{2ac}{a+c}$.
By the Power Mean Inequality,for $n > 1$ and $a \neq c$,we have $\frac{a^n + c^n}{2} > \left(\frac{a+c}{2}\right)^n$.
Since $A.M. > H.M.$,we have $\frac{a+c}{2} > b$.
Raising both sides to the power $n$ (where $n > 1$),we get $\left(\frac{a+c}{2}\right)^n > b^n$.
Combining these inequalities,we get $\frac{a^n + c^n}{2} > \left(\frac{a+c}{2}\right)^n > b^n$.
Therefore,$a^n + c^n > 2b^n$.

(C) Let ${x^a} = {x^{b/2}}{z^{b/2}} = {z^c} = \lambda$.
From this,we have $x = \lambda^{1/a}$,$z = \lambda^{1/c}$,and $xz = \lambda^{2/b}$.
Substituting the values of $x$ and $z$ into the expression for $xz$:
$\lambda^{1/a} \cdot \lambda^{1/c} = \lambda^{2/b}$.
Using the laws of exponents,we get $\lambda^{(1/a) + (1/c)} = \lambda^{2/b}$.
Equating the exponents,we have $\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$.
This is the condition for $a, b, c$ to be in $H.P.$

(C) Let the given terms be $a = \log _3 2, b = \log _6 2, c = \log _{12} 2$.
Taking the reciprocals of these terms:
$\frac{1}{a} = \log _2 3$
$\frac{1}{b} = \log _2 6 = \log _2 (2 \times 3) = \log _2 2 + \log _2 3 = 1 + \log _2 3$
$\frac{1}{c} = \log _2 12 = \log _2 (4 \times 3) = \log _2 4 + \log _2 3 = 2 + \log _2 3$
Let $x = \log _2 3$. Then the reciprocals are $x, 1+x, 2+x$.
Since the difference between consecutive terms is constant $(1)$,these reciprocals are in $A.P.$
Therefore,the original terms are in $H.P.$

(C) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac \dots (i)$.
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 $(i)$:
$k^{2/y} = k^{1/x} \cdot k^{1/z} = k^{(1/x + 1/z)}$.
Equating the exponents,we get $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.
This implies that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
Therefore,$x, y, z$ are in $H.P.$

(A) Given: $\log (x + z) + \log (x + z - 2y) = 2\log (x - z)$
Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:
$\log ((x + z)(x + z - 2y)) = \log ((x - z)^2)$
$(x + z)(x + z - 2y) = (x - z)^2$
$(x + z)^2 - 2y(x + z) = x^2 - 2xz + z^2$
$x^2 + 2xz + z^2 - 2xy - 2yz = x^2 - 2xz + z^2$
$2xz - 2xy - 2yz = -2xz$
$4xz = 2xy + 2yz$
$2xz = xy + yz$
Dividing both sides by $xyz$:
$\frac{2xz}{xyz} = \frac{xy}{xyz} + \frac{yz}{xyz}$
$\frac{2}{y} = \frac{1}{z} + \frac{1}{x}$
This implies that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
Therefore,$x, y, z$ are in $H.P.$

(C) Given that $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ are in $H.P.$
Taking the reciprocal,$\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in $A.P.$
Adding $1$ to each term,we get $\frac{b + c}{a} + 1, \frac{c + a}{b} + 1, \frac{a + b}{c} + 1$ are in $A.P.$
This simplifies to $\frac{a + b + c}{a}, \frac{a + b + c}{b}, \frac{a + b + c}{c}$ are in $A.P.$
Dividing each term by $(a + b + c)$,we get $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$
Therefore,$a, b, c$ are in $H.P.$

(B) 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 $A.P.$
Let the common difference of this $A.P.$ be $k$.
Then $\frac{1}{b} - \frac{1}{a} = k$,$\frac{1}{c} - \frac{1}{b} = k$,and $\frac{1}{d} - \frac{1}{c} = k$.
For $a, b, c, d$ to be in $H.P.$,the condition $a + d > b + c$ is satisfied when the terms are in harmonic progression.
Thus,$\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$

(D) Given that $a, b, c, d$ are in $A.P.$
Dividing each term by the product $abcd$,we get:
$\frac{a}{abcd}, \frac{b}{abcd}, \frac{c}{abcd}, \frac{d}{abcd}$ are in $A.P.$
This simplifies to:
$\frac{1}{bcd}, \frac{1}{acd}, \frac{1}{abd}, \frac{1}{abc}$ are in $A.P.$
By the definition of $H.P.$,the reciprocals of terms in $A.P.$ are in $H.P.$
Therefore,$bcd, acd, abd, abc$ are in $H.P.$
Reversing the order,$abc, abd, acd, bcd$ are also in $H.P.$

(A) Given that $b + c, c + a, a + b$ are in $H.P.$
This implies that their reciprocals $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in $A.P.$
Let $S = a + b + c$. Multiplying each term by $S$,we get $\frac{a + b + c}{b + c}, \frac{a + b + c}{c + a}, \frac{a + b + c}{a + b}$ are in $A.P.$
This can be written as $\frac{a}{b + c} + 1, \frac{b}{c + a} + 1, \frac{c}{a + b} + 1$ are in $A.P.$
Subtracting $1$ from each term,we get $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ are in $A.P.$

(A) Given that $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are in $H.P.$
Therefore,their reciprocals $\frac{b}{a}, \frac{c}{b}, \frac{a}{c}$ are in $A.P.$
This implies $2 \times (\frac{c}{b}) = \frac{b}{a} + \frac{a}{c}$.
Multiplying both sides by $abc$,we get $2ac^2 = b^2c + a^2b$.
This equation represents the condition for $a^2b, c^2a, b^2c$ to be in $A.P.$ because $2(c^2a) = a^2b + b^2c$ is equivalent to $2c^2a = a^2b + b^2c$.

(D) Given that $\ln(a + c)$,$\ln(c - a)$,and $\ln(a - 2b + c)$ are in $A.P.$
Therefore,$2\ln(c - a) = \ln(a + c) + \ln(a - 2b + c)$
Using the property $\ln(x) + \ln(y) = \ln(xy)$,we get:
$\ln((c - a)^2) = \ln((a + c)(a - 2b + c))$
Removing the logarithms:
$(c - a)^2 = (a + c)(a - 2b + c)$
$c^2 + a^2 - 2ac = a^2 - 2ab + ac + ac - 2bc + c^2$
$c^2 + a^2 - 2ac = a^2 + c^2 + 2ac - 2ab - 2bc$
$-2ac = 2ac - 2b(a + c)$
$2b(a + c) = 4ac$
$b = \frac{2ac}{a + c}$
This is the condition for $a, b, c$ to be in $H.P.$

(C) Given $a, b, c$ are in $G.P.$
Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Taking $\log$ on both sides,we get $2 \log b = \log a + \log c$.
This implies that $\log a, \log b, \log c$ are in $A.P.$
Therefore,$\frac{1}{\log a}, \frac{1}{\log b}, \frac{1}{\log c}$ are in $H.P.$
Multiplying by $\log x$,we get $\frac{\log x}{\log a}, \frac{\log x}{\log b}, \frac{\log x}{\log c}$ are in $H.P.$
Using the change of base formula $\log_a x = \frac{\log x}{\log a}$,we conclude that $\log_a x, \log_b x, \log_c x$ are in $H.P.$

(B) Given that $(y - x)$,$2(y - a)$,and $(y - z)$ are in $H.P.$
This implies that their reciprocals $\frac{1}{y - x}$,$\frac{1}{2(y - a)}$,and $\frac{1}{y - z}$ are in $A.P.$
Therefore,$\frac{1}{2(y - a)} - \frac{1}{y - x} = \frac{1}{y - z} - \frac{1}{2(y - a)}$.
Adding $\frac{1}{2(y - a)}$ to both sides,we get $\frac{1}{y - x} + \frac{1}{y - z} = \frac{2}{2(y - a)} = \frac{1}{y - a}$.
$\frac{(y - z) + (y - x)}{(y - x)(y - z)} = \frac{1}{y - a}$.
$(2y - x - z)(y - a) = (y - x)(y - z) = y^2 - yz - xy + xz$.
$2y^2 - 2ay - xy + ax - zy + az = y^2 - yz - xy + xz$.
$y^2 - 2ay + ax + az - xz = 0$.
This simplifies to $(y - a)^2 = (x - a)(z - a)$.
Thus,$(x - a)$,$(y - a)$,and $(z - a)$ are in $G.P.$

(A) Given the equation: $\frac{1}{a} + \frac{1}{c} + \frac{1}{a - b} + \frac{1}{c - b} = 0$
Rearranging the terms,we get: $\frac{1}{a} + \frac{1}{c - b} = -\frac{1}{c} - \frac{1}{a - b} = \frac{1}{b - a} - \frac{1}{c}$
$\frac{c - b + a}{a(c - b)} = \frac{c - (b - a)}{(b - a)c}$
$\frac{a + c - b}{ac - ab} = \frac{c - b + a}{bc - ac}$
Assuming $a + c - b \ne 0$,we can cancel the term $(a + c - b)$ from both sides:
$\frac{1}{ac - ab} = \frac{1}{bc - ac}$
$bc - ac = ac - ab$
$2ac = ab + bc$
Dividing both sides by $abc$: $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
This is the condition for $a, b, c$ to be in $H.P.$

(D) Given that $a, b, c$ are in $H.P.$,the condition for harmonic progression is $b = \frac{2ac}{a + c}$.
Let us check option $(a)$:
$\frac{1}{b - a} + \frac{1}{b - c} = \frac{b - c + b - a}{(b - a)(b - c)} = \frac{2b - (a + c)}{b^2 - b(a + c) + ac}$.
Substituting $b = \frac{2ac}{a + c}$,we find this does not equal $\frac{1}{b}$.
Let us check option $(b)$:
$\frac{ac}{a + c} = b$ $\Rightarrow \frac{ac}{a + c} = \frac{2ac}{a + c}$ $\Rightarrow 1 = 2$,which is false.
Let us check option $(c)$:
$\frac{b + a}{b - a} + \frac{b + c}{b - c} = \frac{(b + a)(b - c) + (b + c)(b - a)}{(b - a)(b - c)} = \frac{2b^2 - 2ac}{b^2 - b(a + c) + ac}$.
Substituting $b = \frac{2ac}{a + c}$,we find this does not equal $1$.
Therefore,none of the given options are true.

(C) Given that $a, b, c$ are in $H.P.$,the 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 means $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$.
Substituting $\frac{1}{c}$ into the expression $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$:
First term: $\left( \frac{1}{b} + (\frac{2}{b} - \frac{1}{a}) - \frac{1}{a} \right) = \left( \frac{3}{b} - \frac{2}{a} \right)$.
Second term: $\left( (\frac{2}{b} - \frac{1}{a}) + \frac{1}{a} - \frac{1}{b} \right) = \left( \frac{1}{b} \right)$.
Multiplying these: $\left( \frac{3}{b} - \frac{2}{a} \right) \left( \frac{1}{b} \right) = \frac{3}{b^2} - \frac{2}{ab}$.

(C) Let the distance of the school from home be $d$.
Let the time taken for the journey to school be $t_1$ and the time taken for the return journey be $t_2$.
Therefore,$t_1 = \frac{d}{x}$ and $t_2 = \frac{d}{y}$.
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 is the Harmonic Mean $(H.M.)$ of $x$ and $y$.

(C) Given $a, b, c, d$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$
For any three terms in $H.P.$,say $x, y, z$,we know that $y = \frac{2xz}{x+z}$. Since $A.M. > H.M.$,we have $\sqrt{xz} > y$,which implies $xz > y^2$.
Applying this to the consecutive terms of the $H.P.$:
$1$. For $a, b, c$ in $H.P.$,we have $ac > b^2$.
$2$. For $b, c, d$ in $H.P.$,we have $bd > c^2$.
Adding these two inequalities,we get:
$ac + bd > b^2 + c^2$.

(C) Given that $a, b, c$ are in harmonic progression $(H.P.)$,we have $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$,which implies $\frac{1}{a} + \frac{1}{c} - \frac{2}{b} = 0$ $... (i)$
The given equation of the line is $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$ $... (ii)$
We can rewrite equation $(ii)$ as $\frac{1}{a}(x) + \frac{1}{c}(1) + \frac{1}{b}(y) = 0$.
From equation $(i)$,we know $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$. Substituting this into the line equation:
$\frac{x}{a} + \frac{y}{b} + (\frac{2}{b} - \frac{1}{a}) = 0$
Rearranging the terms to group by $\frac{1}{a}$ and $\frac{1}{b}$:
$\frac{1}{a}(x - 1) + \frac{1}{b}(y + 2) = 0$
For this equation to hold for all $a, b, c$ in $H.P.$,the coefficients must be zero:
$x - 1 = 0 \Rightarrow x = 1$
$y + 2 = 0 \Rightarrow y = -2$
Thus,the fixed point is $(1, -2)$.

(D) The harmonic mean $(H.M.)$ of $n$ observations $x_1, x_2, ..., x_n$ is given by the formula:
$H.M. = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}} = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}}$
Here,$n = 5$ and the observations are $3, 7, 8, 10, 14$.
Substituting these values into the formula:
$H.M. = \frac{5}{\frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{10} + \frac{1}{14}}$
Thus,the correct option is $D$.

(C) The average speed is defined as the total distance divided by the total time taken.
Total distance $D = 120 \ km + 120 \ km + 120 \ km = 360 \ km$.
Time taken for the first leg $t_1 = \frac{120}{30} = 4 \ hr$.
Time taken for the return trip $t_2 = \frac{120}{25} = 4.8 \ hr$.
Time taken for the final leg $t_3 = \frac{120}{50} = 2.4 \ hr$.
Total time $T = t_1 + t_2 + t_3 = \frac{120}{30} + \frac{120}{25} + \frac{120}{50} \ hr$.
Average speed $v_{avg} = \frac{D}{T} = \frac{120 + 120 + 120}{\frac{120}{30} + \frac{120}{25} + \frac{120}{50}}$.
Dividing numerator and denominator by $120$,we get $v_{avg} = \frac{3}{\frac{1}{30} + \frac{1}{25} + \frac{1}{50}} \ km/hr$.

(C) The harmonic mean $(H.M.)$ of $n$ numbers $x_1, x_2, ..., x_n$ is given by $\frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}$.
For the numbers $4, 8, 16$,we have $n = 3$.
$H.M. = \frac{3}{\frac{1}{4} + \frac{1}{8} + \frac{1}{16}}$
$= \frac{3}{\frac{4+2+1}{16}}$
$= \frac{3}{\frac{7}{16}}$
$= \frac{3 \times 16}{7} = \frac{48}{7} \approx 6.857$.
Rounding to two decimal places,we get $6.86$. Given the options,the closest value is $6.85$.

(D) Given that $a_1, a_2, a_3, \dots, a_n$ are in harmonic progression $(HP)$.
Therefore,$\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \dots, \frac{1}{a_n}$ are in arithmetic progression $(AP)$.
Let the common difference be $d = \frac{1}{a_{k+1}} - \frac{1}{a_k}$.
Then,$a_k a_{k+1} = \frac{a_{k+1} - a_k}{d} = \frac{1}{d} (a_k - a_{k+1})$.
Sum $S = \sum_{k=1}^{n-1} a_k a_{k+1} = \frac{1}{d} \sum_{k=1}^{n-1} (a_k - a_{k+1}) = \frac{1}{d} (a_1 - a_n)$.
In the $AP$,$\frac{1}{a_n} = \frac{1}{a_1} + (n - 1)d$,so $d = \frac{a_1 - a_n}{(n - 1)a_1 a_n}$.
Substituting $d$ into the sum expression:
$S = \frac{a_1 - a_n}{\frac{a_1 - a_n}{(n - 1)a_1 a_n}} = (n - 1)a_1 a_n$.