Relation between A.P., G.P. Questions in English

(C) Given that $\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a+b}{2}$.
Cross-multiplying,we get:
$2(a^{n+1} + b^{n+1}) = (a+b)(a^n + b^n)$
Expanding the right side:
$2a^{n+1} + 2b^{n+1} = a^{n+1} + ab^n + ba^n + b^{n+1}$
Rearranging the terms:
$a^{n+1} - ab^n - ba^n + b^{n+1} = 0$
Factoring the expression:
$a^n(a - b) - b^n(a - b) = 0$
$(a^n - b^n)(a - b) = 0$
Since $a \neq b$,we must have $a^n - b^n = 0$,which implies $a^n = b^n$.
Dividing by $b^n$ (assuming $b \neq 0$):
$(\frac{a}{b})^n = 1 = (\frac{a}{b})^0$
Therefore,$n = 0$.

(D) Given that the product of $n$ positive numbers $x_1, x_2, \dots, x_n$ is $1$,i.e.,$x_1 \cdot x_2 \cdot \dots \cdot x_n = 1$.
By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,we know that for positive real numbers,the Arithmetic Mean is always greater than or equal to the Geometric Mean:
$\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 \cdot x_2 \cdot \dots \cdot x_n}$
Substituting the given product:
$\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{1} = 1$
Multiplying both sides by $n$:
$x_1 + x_2 + \dots + x_n \ge n$
Therefore,the sum of the $n$ positive numbers is never less than $n$.

(A) For any two distinct positive real numbers $a$ and $b$,the arithmetic mean $A = \frac{a+b}{2}$,geometric mean $G = \sqrt{ab}$,and harmonic mean $H = \frac{2ab}{a+b}$.
We know that $A \times H = \left(\frac{a+b}{2}\right) \times \left(\frac{2ab}{a+b}\right) = ab = G^2$.
Thus,$G^2 = AH$,which implies $G$ is the geometric mean of $A$ and $H$.
For distinct positive real numbers,$A > G > H$ always holds true.

(D) Let the two positive real numbers be $a$ and $b$.
The arithmetic mean is $A = \frac{a + b}{2}$.
The geometric mean is $G = \sqrt{ab}$.
The harmonic mean is $H = \frac{2ab}{a + b}$.
Now,calculate $G^2$:
$G^2 = (\sqrt{ab})^2 = ab$.
Next,calculate $AH$:
$AH = \left( \frac{a + b}{2} \right) \times \left( \frac{2ab}{a + b} \right) = ab$.
Since both $G^2$ and $AH$ are equal to $ab$,we conclude that $G^2 = AH$.

(B) Given that $a, b, c$ are in $A.P.$
So,$2b = a + c$ --- $(1)$
Given that $b, c, d$ are in $H.P.$
So,$c = \frac{2bd}{b + d}$ --- $(2)$
From $(1)$,we have $a + c = 2b$. Substituting this into the expression derived from $(2)$:
$c(b + d) = 2bd$
Since $2b = a + c$,we can write $2bd = (a + c)d$
So,$c(b + d) = (a + c)d$
$bc + cd = ad + cd$
Subtracting $cd$ from both sides,we get:
$bc = ad$

(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
The terms are $T_p = a + (p - 1)d, T_q = a + (q - 1)d, T_r = a + (r - 1)d, T_s = a + (s - 1)d$.
Since $T_p, T_q, T_r, T_s$ are in $G.P.$,the common ratio $R$ is given by:
$R = \frac{T_q}{T_p} = \frac{T_r}{T_q} = \frac{T_s}{T_r} = \frac{T_q - T_r}{T_p - T_q} = \frac{T_r - T_s}{T_q - T_r}$.
Substituting the values:
$T_q - T_r = (q - r)d$ and $T_p - T_q = (p - q)d$.
Thus,$R = \frac{(q - r)d}{(p - q)d} = \frac{q - r}{p - q}$.
Similarly,$R = \frac{r - s}{q - r}$.
Since $\frac{q - r}{p - q} = \frac{r - s}{q - r}$,the terms $(p - q), (q - r), (r - s)$ are in $G.P.$

(C) The arithmetic mean of $a$ and $b$ is $A = \frac{a + b}{2}$.
The geometric mean of $a$ and $b$ is $G = \sqrt{ab}$.
Then,$A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2}$.
Since $a = (\sqrt{a})^2$ and $b = (\sqrt{b})^2$,we have $A - G = \frac{(\sqrt{a})^2 + (\sqrt{b})^2 - 2\sqrt{a}\sqrt{b}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2}$.
This can be rewritten as $[\frac{\sqrt{a} - \sqrt{b}}{\sqrt{2}}]^2$.

(A) Let $a^{1/x} = b^{1/y} = c^{1/z} = k$.
Then $a = k^x, b = k^y, c = k^z$.
Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Substituting the values of $a, b, c$ in terms of $k$,we get $(k^y)^2 = k^x \cdot k^z$.
This simplifies to $k^{2y} = k^{x+z}$.
Equating the exponents,we get $2y = x + z$.
This condition implies that $x, y, z$ are in $A.P.$

(C) Let the two numbers be $a$ and $b$.
Given that the arithmetic mean $A = \frac{a + b}{2}$,so $a + b = 2A$.
Given that the geometric mean $G = \sqrt{ab}$,so $ab = G^2$.
The quadratic equation whose roots are $a$ and $b$ is given by $x^2 - (a + b)x + ab = 0$.
Substituting the values,we get $x^2 - 2Ax + G^2 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have:
$x = \frac{2A \pm \sqrt{(2A)^2 - 4G^2}}{2} = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} = \frac{2A \pm 2\sqrt{A^2 - G^2}}{2} = A \pm \sqrt{A^2 - G^2}$.
Since $A^2 - G^2 = (A + G)(A - G)$,the numbers are $A \pm \sqrt{(A + G)(A - G)}$.

(B) We know that for any two distinct positive real numbers $a$ and $b$,the Arithmetic Mean $(A)$ is strictly greater than the Geometric Mean $(G)$.
$A = \frac{a + b}{2}$ and $G = \sqrt{ab}$.
Since $A > G$,we have $\frac{a + b}{2} > \sqrt{ab}$.
Multiplying both sides by $2$,we get $(a + b) > 2\sqrt{ab}$ or $2\sqrt{ab} < (a + b)$.

(C) Given that ${b^2}, {a^2}, {c^2}$ are in $A.P.$
Therefore,${a^2} - {b^2} = {c^2} - {a^2}$.
This implies $(a - b)(a + b) = (c - a)(c + a)$.
Dividing both sides by $(a + b)(b + c)(c + a)$,we get:
$\frac{a - b}{(b + c)(c + a)} = \frac{c - a}{(a + b)(b + c)}$.
Using the property of $A.P.$,we can rewrite the terms to show that $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in $A.P.$ is not the direct path,but rather:
Since ${a^2}, {b^2}, {c^2}$ are in $A.P.$,then $a+b, b+c, c+a$ are in $H.P.$ if their reciprocals are in $A.P.$
Thus,$a + b, b + c, c + a$ are in $H.P.$

(D) Given that $a, b, c$ are in $A.P.$,we have $b = \frac{a + c}{2} \implies 2b = a + c$ .....$(i)$
Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$ .....(ii)
From $(i)$,$a + c = 2b$. Squaring both sides,we get $(a + c)^2 = 4b^2$.
Substituting $b^2 = ac$ from (ii) into the equation,we get $(a + c)^2 = 4ac$.
This simplifies to $a^2 + 2ac + c^2 = 4ac$,which is $a^2 - 2ac + c^2 = 0$.
Thus,$(a - c)^2 = 0$,which implies $a = c$.
Substituting $a = c$ into $(i)$,we get $b = \frac{a + a}{2} = a$.
Therefore,$a = b = c$.

(B) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $c = 2b - a$.
Since $a, b, d$ are in $G.P.$,we have $b^2 = ad$,which implies $d = \frac{b^2}{a}$.
We need to check the sequence $a, a - b, d - c$.
Let the terms be $T_1 = a$,$T_2 = a - b$,and $T_3 = d - c$.
Substituting $c$ and $d$:
$T_3 = \frac{b^2}{a} - (2b - a) = \frac{b^2 - 2ab + a^2}{a} = \frac{(a - b)^2}{a}$.
Now,check the ratio $\frac{T_2}{T_1} = \frac{a - b}{a}$ and $\frac{T_3}{T_2} = \frac{(a - b)^2 / a}{a - b} = \frac{a - b}{a}$.
Since the ratios are equal,the sequence $a, a - b, d - c$ is in $G.P.$

(C) Given that $x, 1, z$ are in $A.P.$,the middle term is the arithmetic mean: $1 = \frac{x + z}{2}$,which implies $x + z = 2$......$(i)$
Given that $x, 2, z$ are in $G.P.$,the middle term is the geometric mean: $2^2 = xz$,which implies $xz = 4$......$(ii)$
For $x, 4, z$ to be in $H.P.$,the middle term must be the harmonic mean of $x$ and $z$,which is $\frac{2xz}{x + z}$.
Substituting the values from $(i)$ and $(ii)$ into the expression for the harmonic mean:
$\text{Harmonic Mean} = \frac{2(4)}{2} = \frac{8}{2} = 4$.
Since the middle term is $4$,$x, 4, z$ are in $H.P.$

(C) Given that $x = 1 + a + a^2 + \dots = \frac{1}{1-a}$,$y = 1 + b + b^2 + \dots = \frac{1}{1-b}$,and $z = 1 + c + c^2 + \dots = \frac{1}{1-c}$ for $|a|, |b|, |c| < 1$.
Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Subtracting each term from $1$,we get $1-a, 1-b, 1-c$ are also in $A.P.$ because $(1-a) + (1-c) = 2 - (a+c) = 2 - 2b = 2(1-b)$.
Since $1-a, 1-b, 1-c$ are in $A.P.$,their reciprocals $\frac{1}{1-a}, \frac{1}{1-b}, \frac{1}{1-c}$ must be in $H.P.$
Therefore,$x, y, z$ are in $H.P.$

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$ ..... $(i)$
Given that $b, c, d$ are in $G.P.$,we have $c^2 = bd$ ..... $(ii)$
Given that $c, d, e$ are in $H.P.$,we have $d = \frac{2ce}{c + e}$ ..... $(iii)$
From $(ii)$,substitute $b = \frac{a + c}{2}$ and $d = \frac{2ce}{c + e}$:
$c^2 = \left(\frac{a + c}{2}\right) \left(\frac{2ce}{c + e}\right)$
$c^2 = \frac{(a + c)ce}{c + e}$
$c^2(c + e) = (a + c)ce$
$c^3 + c^2e = ace + c^2e$
$c^3 = ace$
Since $c \neq 0$,we divide by $c$ to get $c^2 = ae$.
Therefore,$a, c, e$ are in $G.P.$

(B) Let the two numbers be $a$ and $b$.
We know that for any two positive numbers,the relationship between their Arithmetic Mean $(A.M.)$,Geometric Mean $(G.M.)$,and Harmonic Mean $(H.M.)$ is given by the formula:
$(G.M.)^2 = A.M. \times H.M.$
Given $A.M. = 27$ and $H.M. = 12$.
Substituting these values into the formula:
$(G.M.)^2 = 27 \times 12$
$(G.M.)^2 = 324$
$G.M. = \sqrt{324} = 18$.

(C) The relationship between Arithmetic Mean $(A.M.)$,Geometric Mean $(G.M.)$,and Harmonic Mean $(H.M.)$ for two numbers is given by the formula:
$G.M.^2 = A.M. \times H.M.$
Given that $G.M. = 18$ and $A.M. = 27$,we substitute these values into the formula:
$(18)^2 = 27 \times H.M.$
$324 = 27 \times H.M.$
$H.M. = \frac{324}{27}$
$H.M. = 12$

(B) Given $A.M. = 2(G.M.)$.
$\frac{a + b}{2} = 2\sqrt{ab}$.
$\frac{a + b}{\sqrt{ab}} = 4$.
Let $x = \sqrt{\frac{a}{b}}$. Then $\frac{a}{b} = x^2$ and $\frac{b}{a} = \frac{1}{x^2}$.
Dividing by $\sqrt{ab}$,we get $\sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} = 4$.
$x + \frac{1}{x} = 4$.
$x^2 - 4x + 1 = 0$.
Using the quadratic formula,$x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.
Since $\frac{a}{b} = x^2$,we have $\frac{a}{b} = (2 \pm \sqrt{3})^2 = 4 + 3 \pm 4\sqrt{3} = 7 \pm 4\sqrt{3}$.
Alternatively,using the componendo and dividendo rule on $\frac{a+b}{2\sqrt{ab}} = \frac{2}{1}$:
$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{2+1}{2-1} = 3$.
$\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = 3$.
$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \sqrt{3}$.
Applying componendo and dividendo again:
$\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.
$\sqrt{\frac{a}{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.
$\frac{a}{b} = \frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}} = \frac{4+2\sqrt{3}}{4-2\sqrt{3}} = \frac{2+\sqrt{3}}{2-\sqrt{3}}$.

(C) Let $A_j$ and $H_j$ denote the $j^{th}$ $A.M.$ and $H.M.$ respectively,where $j = 1, 2, \dots, 9$,inserted between $2$ and $3$.
For the $A.M.s$,the sequence $2, A_1, A_2, \dots, A_9, 3$ is in $A.P.$ with $11$ terms.
Common difference $d = \frac{3 - 2}{9 + 1} = \frac{1}{10}$.
Thus,$A_j = 2 + j \times d = 2 + \frac{j}{10}$.
For the $H.M.s$,the sequence $2, H_1, H_2, \dots, H_9, 3$ is in $H.P.$
Therefore,$\frac{1}{2}, \frac{1}{H_1}, \frac{1}{H_2}, \dots, \frac{1}{H_9}, \frac{1}{3}$ is in $A.P.$
Common difference $D = \frac{\frac{1}{3} - \frac{1}{2}}{9 + 1} = \frac{-\frac{1}{6}}{10} = -\frac{1}{60}$.
Thus,$\frac{1}{H_j} = \frac{1}{2} + j \times D = \frac{1}{2} - \frac{j}{60}$.
Now,calculate $A_j + \frac{6}{H_j}$:
$A_j + 6 \times \frac{1}{H_j} = (2 + \frac{j}{10}) + 6 \times (\frac{1}{2} - \frac{j}{60})$
$= 2 + \frac{j}{10} + 3 - \frac{6j}{60}$
$= 5 + \frac{j}{10} - \frac{j}{10} = 5$.

(A) Given that $a, b, c$ are in $A.P.$
$2b = a + c$ ......$(i)$
Given that $a^2, b^2, c^2$ are in $H.P.$
$b^2 = \frac{2a^2c^2}{a^2 + c^2}$
$b^2(a^2 + c^2) = 2a^2c^2$
$b^2((a+c)^2 - 2ac) = 2a^2c^2$
Substituting $a+c = 2b$ from $(i)$:
$b^2(4b^2 - 2ac) = 2a^2c^2$
$4b^4 - 2acb^2 - 2a^2c^2 = 0$
$2b^4 - acb^2 - a^2c^2 = 0$
$(2b^2 + ac)(b^2 - ac) = 0$
Since $a, b, c$ are real,$2b^2 + ac = 0$ implies $a, b, c$ are $0$ or complex,so we consider $b^2 = ac$.
If $b^2 = ac$,then $a, b, c$ are in $G.P.$
Since $a, b, c$ are in both $A.P.$ and $G.P.$,we must have $a = b = c$.

(D) Let the four numbers be $\frac{a}{r}, a, ar, x$.
Since the last three numbers $a, ar, x$ are in $A.P.$ with common difference $d = 6$,we have $ar - a = 6$ and $x - ar = 6$.
Thus,$x = ar + 6$.
Since $ar = a + 6$,we have $x = (a + 6) + 6 = a + 12$.
Given that the first and last numbers are equal,$\frac{a}{r} = x = a + 12$.
From $ar - a = 6$,we get $a(r - 1) = 6$,so $r - 1 = \frac{6}{a}$,which means $r = 1 + \frac{6}{a} = \frac{a + 6}{a}$.
Substituting $r$ into $\frac{a}{r} = a + 12$:
$\frac{a}{(a + 6)/a} = a + 12$
$\frac{a^2}{a + 6} = a + 12$
$a^2 = (a + 12)(a + 6)$
$a^2 = a^2 + 6a + 12a + 72$
$0 = 18a + 72$
$18a = -72$
$a = -4$.
The first number is $\frac{a}{r} = a + 12 = -4 + 12 = 8$.

(D) Given that $\frac{H.M.}{G.M.} = \frac{12}{13}$.
We know that $H.M. = \frac{2ab}{a+b}$ and $G.M. = \sqrt{ab}$.
So,$\frac{2ab}{(a+b)\sqrt{ab}} = \frac{12}{13} \Rightarrow \frac{2\sqrt{ab}}{a+b} = \frac{12}{13}$.
This implies $\frac{a+b}{2\sqrt{ab}} = \frac{13}{12}$.
Using componendo and dividendo,$\frac{(a+b)+2\sqrt{ab}}{(a+b)-2\sqrt{ab}} = \frac{13+12}{13-12} = \frac{25}{1}$.
$\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = 25$.
Taking the square root,$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = 5$.
Again,using componendo and dividendo,$\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{5+1}{5-1} = \frac{6}{4} = \frac{3}{2}$.
Squaring both sides,$\frac{a}{b} = \frac{9}{4}$.
Thus,the ratio is $9:4$,which is not among the given options.

(B) Let the two numbers be $a$ and $b$. Given $\frac{a}{b} = \frac{9}{1}$,so $a = 9b$.
The geometric mean $G$ is given by $G = \sqrt{ab}$.
The harmonic mean $H$ is given by $H = \frac{2ab}{a+b}$.
Substituting $a = 9b$ into the expressions:
$G = \sqrt{9b \cdot b} = \sqrt{9b^2} = 3b$.
$H = \frac{2(9b)(b)}{9b + b} = \frac{18b^2}{10b} = \frac{9b}{5}$.
The ratio of the geometric mean to the harmonic mean is $\frac{G}{H} = \frac{3b}{\frac{9b}{5}} = 3b \cdot \frac{5}{9b} = \frac{15}{9} = \frac{5}{3}$.
Thus,the ratio is $5:3$.

(D) Let $\alpha$ and $\beta$ be the first and $(2n - 1)^{th}$ terms of the $A.P.$,$G.P.$,and $H.P.$ respectively.
For $A.P.$: The $n^{th}$ term is $a = \frac{\alpha + \beta}{2}$ $(i)$
For $G.P.$: The $n^{th}$ term is $b = \sqrt{\alpha \beta}$ (ii)
For $H.P.$: The $n^{th}$ term is $c = \frac{2\alpha \beta}{\alpha + \beta}$ (iii)
From $(i)$,(ii),and (iii),we observe that $a, b, c$ are the Arithmetic Mean,Geometric Mean,and Harmonic Mean of $\alpha$ and $\beta$ respectively.
We know that for any two positive numbers $\alpha$ and $\beta$,$A.M. \ge G.M. \ge H.M.$
Thus,$a \ge b \ge c$,which matches option $(a)$.
Also,$ac = \left(\frac{\alpha + \beta}{2}\right) \left(\frac{2\alpha \beta}{\alpha + \beta}\right) = \alpha \beta = b^2$.
Therefore,$ac - b^2 = 0$,which matches option $(c)$.
Hence,both $(a)$ and $(c)$ are correct.

(B) Let $a$ and $b$ be the first and last terms of the three progressions,each having $(2n + 1)$ terms.
The middle term of the $A.P.$ is $A = \frac{a + b}{2}$.
The middle term of the $G.P.$ is $G = \sqrt{ab}$.
The middle term of the $H.P.$ is $H = \frac{2ab}{a + b}$.
We observe that $G^2 = (\sqrt{ab})^2 = ab$ and $A \times H = \left(\frac{a + b}{2}\right) \times \left(\frac{2ab}{a + b}\right) = ab$.
Since $G^2 = A \times H$,the middle terms $A, G, H$ are in $G.P.$

(B) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Since $a + x, b + x, c + x$ are in $H.P.$,their reciprocals are in $A.P.$,or more directly,the middle term is the harmonic mean:
$b + x = \frac{2(a + x)(c + x)}{(a + x) + (c + x)}$
$(b + x)(a + c + 2x) = 2(ac + x(a + c) + x^2)$
$ab + bc + 2bx + ax + cx + 2x^2 = 2ac + 2x(a + c) + 2x^2$
$ab + bc + 2bx = 2ac + x(a + c)$
$x(2b - a - c) = 2ac - ab - bc$
Since $ac = b^2$,we substitute:
$x(2b - a - c) = 2b^2 - ab - bc = b(2b - a - c)$
If $2b - a - c \neq 0$,then $x = b$.

(C) Given that $\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}$ are in $A.P.$
This implies $b - \frac{a + b}{1 - ab} = \frac{b + c}{1 - bc} - b$
$\Rightarrow \frac{b(1 - ab) - (a + b)}{1 - ab} = \frac{(b + c) - b(1 - bc)}{1 - bc}$
$\Rightarrow \frac{b - ab^2 - a - b}{1 - ab} = \frac{b + c - b + bc^2}{1 - bc}$
$\Rightarrow \frac{-a(1 + b^2)}{1 - ab} = \frac{c(1 + b^2)}{1 - bc}$
Dividing both sides by $(1 + b^2)$,we get $\frac{-a}{1 - ab} = \frac{c}{1 - bc}$
$\Rightarrow -a(1 - bc) = c(1 - ab)$
$\Rightarrow -a + abc = c - abc$
$\Rightarrow 2abc = a + c$
Dividing both sides by $abc$,we get $2b = \frac{1}{c} + \frac{1}{a}$
This is the condition for $\frac{1}{a}, b, \frac{1}{c}$ to be in $A.P.$,which means $a, \frac{1}{b}, c$ are in $H.P.$

(B) Given that $2(y - a)$ is the $H.M.$ between $y - x$ and $y - z$,therefore $y - x, 2(y - a), y - z$ are in $H.P.$
This implies that $\frac{1}{y - x}, \frac{1}{2(y - a)}, \frac{1}{y - z}$ are in $A.P.$
Thus,$\frac{1}{2(y - a)} - \frac{1}{y - x} = \frac{1}{y - z} - \frac{1}{2(y - a)}$.
$\frac{2(y - a) - (y - x)}{2(y - a)(y - x)} = \frac{2(y - a) - (y - z)}{2(y - a)(y - z)}$
$\frac{y - 2a + x}{y - x} = \frac{y - 2a + z}{y - z}$
Rearranging terms: $\frac{y - 2a + x}{y - x} = \frac{y - 2a + z}{y - z}$
By simplifying,we get $\frac{x - a + y - a}{-(x - a) + (y - a)} = \frac{y - a + z - a}{-(y - a) + (z - a)}$.
Applying componendo and dividendo,we obtain $\frac{x - a}{y - a} = \frac{y - a}{z - a}$.
Therefore,$x - a, y - a, z - a$ are in $G.P.$

(C) Given that the ratio of $A.M.$ to $H.M.$ is $m:n$,we have $\frac{(a+b)/2}{2ab/(a+b)} = \frac{m}{n}$.
This simplifies to $\frac{(a+b)^2}{4ab} = \frac{m}{n}$,which implies $\frac{(a+b)^2}{ab} = \frac{4m}{n}$.
Using the property of componendo and dividendo on $\frac{(a+b)^2}{4ab} = \frac{m}{n}$,we get $\frac{(a+b)^2 + 4ab}{(a+b)^2 - 4ab} = \frac{m+n}{m-n}$.
This simplifies to $\frac{(a+b)^2}{(a-b)^2} = \frac{m+n}{m-n}$ is incorrect. Let us re-evaluate: $\frac{(a+b)^2}{4ab} = \frac{m}{n} \Rightarrow \frac{4ab}{(a+b)^2} = \frac{n}{m}$.
Applying componendo and dividendo: $\frac{(a+b)^2 + 4ab}{(a+b)^2 - 4ab} = \frac{m+n}{m-n} \Rightarrow \frac{a^2+b^2+6ab}{a^2+b^2-2ab} = \frac{m+n}{m-n}$ (not helpful).
Correct approach: $\frac{(a+b)^2}{4ab} = \frac{m}{n} \Rightarrow \frac{(a+b)^2}{ab} = \frac{4m}{n}$.
We know $(a+b)^2 - (a-b)^2 = 4ab$,so $\frac{(a+b)^2}{(a-b)^2} = \frac{m}{m-n}$.
Taking square root,$\frac{a+b}{a-b} = \frac{\sqrt{m}}{\sqrt{m-n}}$.
Applying componendo and dividendo again: $\frac{(a+b)+(a-b)}{(a+b)-(a-b)} = \frac{\sqrt{m} + \sqrt{m-n}}{\sqrt{m} - \sqrt{m-n}}$.
Thus,$\frac{2a}{2b} = \frac{\sqrt{m} + \sqrt{m-n}}{\sqrt{m} - \sqrt{m-n}}$,so $\frac{a}{b} = \frac{\sqrt{m} + \sqrt{m-n}}{\sqrt{m} - \sqrt{m-n}}$.

(B) Let the two numbers be $a$ and $b$.
Given that the geometric mean is $6$,we have $\sqrt{ab} = 6$,which implies $ab = 36$.
Given that the arithmetic mean is $6.5$,we have $\frac{a+b}{2} = 6.5$,which implies $a+b = 13$.
We can form a quadratic equation $x^2 - (a+b)x + ab = 0$ to find the numbers.
Substituting the values,we get $x^2 - 13x + 36 = 0$.
Factoring the quadratic equation: $x^2 - 9x - 4x + 36 = 0$.
$x(x - 9) - 4(x - 9) = 0$.
$(x - 4)(x - 9) = 0$.
Thus,the numbers are $4$ and $9$.

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Also,$a, c - b, b - a$ are in $G.P.$,so $(c - b)^2 = a(b - a)$.
Since $a, b, c$ are in $A.P.$,let $a = x - d, b = x, c = x + d$.
Then $c - b = d$ and $b - a = d$.
The $G.P.$ condition becomes $d^2 = (x - d)(d)$.
Since $a \ne b \ne c$,$d \ne 0$,so $d = x - d$,which implies $x = 2d$.
Thus,$a = 2d - d = d$,$b = 2d$,and $c = 2d + d = 3d$.
Therefore,$a:b:c = d:2d:3d = 1:2:3$.

(A) We know that for any two positive numbers $a$ and $b$,the relationship between Arithmetic Mean $(A.M.)$,Geometric Mean $(G.M.)$,and Harmonic Mean $(H.M.)$ is given by:
$(G.M.)^2 = (A.M.) \times (H.M.)$
Given $A.M. = 9$ and $H.M. = 36$,we substitute these values into the formula:
$(G.M.)^2 = 9 \times 36$
$(G.M.)^2 = 324$
Taking the square root on both sides:
$G.M. = \sqrt{324} = 18$
Therefore,the correct option is $A$.

(D) We have $H.M. = \frac{2ab}{a + b}$ and $G.M. = \sqrt{ab}$.
Given $\frac{H.M.}{G.M.} = \frac{4}{5}$.
$\Rightarrow \frac{2ab/(a + b)}{\sqrt{ab}} = \frac{4}{5}$ $\Rightarrow \frac{2\sqrt{ab}}{a + b} = \frac{4}{5}$.
$\Rightarrow \frac{a + b}{2\sqrt{ab}} = \frac{5}{4}$.
Applying componendo and dividendo:
$\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{5 + 4}{5 - 4}$ $\Rightarrow \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{9}{1}$.
Taking square root on both sides:
$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{3}{1}$.
Applying componendo and dividendo again:
$\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{3 + 1}{3 - 1}$ $\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}} = \frac{4}{2} = 2$.
Squaring both sides:
$\frac{a}{b} = 2^2 = 4$.
Thus,$a:b = 4:1$ or $b:a = 1:4$.

(A) Given that $A.M. = G.M. = H.M.$
We know that for any two positive numbers $a$ and $b$,the relationship between Arithmetic Mean $(A.M.)$,Geometric Mean $(G.M.)$,and Harmonic Mean $(H.M.)$ is given by $A.M. \ge G.M. \ge H.M.$
Equality holds if and only if $a = b$.
Alternatively,from $A.M. = G.M.$,we have $\frac{a+b}{2} = \sqrt{ab}$.
Squaring both sides,we get $\frac{(a+b)^2}{4} = ab$,which implies $(a+b)^2 = 4ab$.
$(a+b)^2 - 4ab = 0$,which simplifies to $(a-b)^2 = 0$.
Therefore,$a - b = 0$,or $a = b$.

(C) Given that $a, b, c$ are in $A.P.$
$ \Rightarrow 2b = a + c $
Consider the terms $10^{ax + 10}, 10^{bx + 10}, 10^{cx + 10}$.
For these terms to be in $G.P.$,the square of the middle term must equal the product of the first and third terms:
$(10^{bx + 10})^2 = 10^{2(bx + 10)} = 10^{2bx + 20}$
Product of first and third terms: $10^{ax + 10} \times 10^{cx + 10} = 10^{ax + cx + 20} = 10^{(a+c)x + 20}$
Since $a + c = 2b$,we have $10^{(a+c)x + 20} = 10^{2bx + 20}$.
Thus,$(10^{bx + 10})^2 = 10^{ax + 10} \times 10^{cx + 10}$ holds true for all values of $x$.
Therefore,the terms are in $G.P.$ for all values of $x$.

(C) Let the three non-zero real numbers in $A.P.$ be $(a - d), a, (a + d)$.
Since their squares form a $G.P.$,we have $(a - d)^2, a^2, (a + d)^2$ in $G.P.$
Therefore,$(a^2)^2 = (a - d)^2(a + d)^2$.
$a^4 = ((a - d)(a + d))^2 = (a^2 - d^2)^2$.
$a^4 = a^4 - 2a^2d^2 + d^4$.
$d^4 - 2a^2d^2 = 0$.
$d^2(d^2 - 2a^2) = 0$.
Since the numbers are non-zero,$d \neq 0$ is not necessarily true,but if $d=0$,the numbers are $a, a, a$,which form a $G.P.$ with common ratio $r = 1$.
If $d^2 = 2a^2$,then $d = \pm \sqrt{2}a$.
Case $1$: $d = 0$. The numbers are $a, a, a$. The squares are $a^2, a^2, a^2$. Common ratio $r = 1$.
Case $2$: $d = \sqrt{2}a$. The numbers are $a(1-\sqrt{2}), a, a(1+\sqrt{2})$. The squares are $a^2(3-2\sqrt{2}), a^2, a^2(3+2\sqrt{2})$. The common ratio $r = \frac{a^2}{a^2(3-2\sqrt{2})} = 3+2\sqrt{2}$.
Case $3$: $d = -\sqrt{2}a$. The numbers are $a(1+\sqrt{2}), a, a(1-\sqrt{2})$. The squares are $a^2(3+2\sqrt{2}), a^2, a^2(3-2\sqrt{2})$. The common ratio $r = \frac{a^2}{a^2(3+2\sqrt{2})} = 3-2\sqrt{2}$.
Thus,there are $3$ possible common ratios.

(C) Let the two numbers be $p$ and $q$.
Since ${G_1}$ and ${G_2}$ are two geometric means between $p$ and $q$,the sequence $p, G_1, G_2, q$ is in geometric progression.
Let the common ratio be $r$. Then $G_1 = pr$,$G_2 = pr^2$,and $q = pr^3$,so $r = (q/p)^{1/3}$.
Thus,${G_1} = p(q/p)^{1/3} = p^{2/3}q^{1/3}$ and ${G_2} = p(q/p)^{2/3} = p^{1/3}q^{2/3}$.
The arithmetic mean $A$ is given by $A = \frac{p+q}{2}$,so $p+q = 2A$.
Now,$\frac{{G_1^2}}{{{G_2}}} + \frac{{G_2^2}}{{{G_1}}} = \frac{(p^{2/3}q^{1/3})^2}{p^{1/3}q^{2/3}} + \frac{(p^{1/3}q^{2/3})^2}{p^{2/3}q^{1/3}} = \frac{p^{4/3}q^{2/3}}{p^{1/3}q^{2/3}} + \frac{p^{2/3}q^{4/3}}{p^{2/3}q^{1/3}} = p + q$.
Substituting $p+q = 2A$,we get the value as $2A$.

(D) Let the three terms of the decreasing $A.P.$ be $a+d, a, a-d$,where $d > 0$.
Sum: $(a+d) + a + (a-d) = 27$ $\Rightarrow 3a = 27$ $\Rightarrow a = 9$.
The terms are $9+d, 9, 9-d$.
Adding $-1, -1, 3$ respectively gives: $(9+d-1), (9-1), (9-d+3) = (8+d), 8, (12-d)$.
Since these are in $G.P.$,the square of the middle term equals the product of the extremes:
$8^2 = (8+d)(12-d)$
$64 = 96 - 8d + 12d - d^2$
$d^2 - 4d - 32 = 0$
$(d-8)(d+4) = 0$.
Since the $A.P.$ is decreasing,$d$ must be positive,so $d = 8$.
The numbers are $9+8, 9, 9-8$,which are $17, 9, 1$.

(A) Let $a$ and $b$ be two numbers.
The sum of $n$ $A.M.s$ is given by $n \times (\text{single } A.M.)$,so ${A_1} + {A_2} = 2 \times \left( \frac{a + b}{2} \right) = a + b$.
The product of $n$ $G.M.s$ is given by $(\text{single } G.M.)^n$,so ${G_1}{G_2} = (\sqrt{ab})^2 = ab$.
Since ${H_1}, {H_2}$ are $H.M.s$ between $a$ and $b$,the sequence $\frac{1}{a}, \frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{b}$ is in $A.P$.
Thus,$\frac{1}{H_1} + \frac{1}{H_2} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$.
This simplifies to $\frac{H_1 + H_2}{H_1 H_2} = \frac{a + b}{ab}$.
Substituting the values,we get $\frac{H_1 + H_2}{H_1 H_2} = \frac{A_1 + A_2}{G_1 G_2}$.
Rearranging the terms,we have $\frac{G_1 G_2}{H_1 H_2} \times \frac{H_1 + H_2}{A_1 + A_2} = 1$.

(B) Let the first term of the $A.P.$ be $a = 1$ and the common difference be $d$.
The terms are given by $T_n = a + (n-1)d$.
Thus,$T_2 = 1 + d$,$T_{10} = 1 + 9d$,and $T_{34} = 1 + 33d$.
Since $T_2, T_{10}, T_{34}$ are in $G.P.$,we have $(T_{10})^2 = T_2 \times T_{34}$.
$(1 + 9d)^2 = (1 + d)(1 + 33d)$.
$1 + 81d^2 + 18d = 1 + 33d + d + 33d^2$.
$1 + 81d^2 + 18d = 1 + 34d + 33d^2$.
$48d^2 - 16d = 0$.
$16d(3d - 1) = 0$.
This gives $d = 0$ or $d = \frac{1}{3}$.
Since the common difference is typically non-zero for such problems,the value is $\frac{1}{3}$.

(D) Given that $a, b, c$ are in $A.P.$,we have $b - a = c - b$.
Let the terms be $T_1 = 2^{ax + 1}$,$T_2 = 2^{bx + 1}$,and $T_3 = 2^{cx + 1}$.
For these terms to be in $G.P.$,the ratio $\frac{T_2}{T_1}$ must equal $\frac{T_3}{T_2}$.
$\frac{T_2}{T_1} = \frac{2^{bx + 1}}{2^{ax + 1}} = 2^{(b - a)x}$.
$\frac{T_3}{T_2} = \frac{2^{cx + 1}}{2^{bx + 1}} = 2^{(c - b)x}$.
Since $b - a = c - b$,it follows that $2^{(b - a)x} = 2^{(c - b)x}$ for all $x \ne 0$.
Thus,the terms form a $G.P.$ for all $x \ne 0$.

(B) Let $h_1, h_2, h_3$ be the altitudes from vertices $P, Q, R$ to the opposite sides $a, b, c$ respectively.
We know that the area of the triangle $\Delta = \frac{1}{2} a h_1 = \frac{1}{2} b h_2 = \frac{1}{2} c h_3$.
Thus,$h_1 = \frac{2\Delta}{a}$,$h_2 = \frac{2\Delta}{b}$,and $h_3 = \frac{2\Delta}{c}$.
Given that $h_1, h_2, h_3$ are in $A.P.$,we have $2h_2 = h_1 + h_3$.
Substituting the values,we get $2(\frac{2\Delta}{b}) = \frac{2\Delta}{a} + \frac{2\Delta}{c}$.
Dividing by $2\Delta$,we get $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.
This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$
Therefore,$a, b, c$ are in $H.P.$

(A) Since $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $c - b = b - a$.
Given that $b - a, c - b, a$ are in $G.P.$,we have $(c - b)^2 = (b - a)a$.
Substituting $c - b = b - a$ into the $G.P.$ condition,we get $(b - a)^2 = (b - a)a$.
Since $a, b, c$ are unequal,$b - a \neq 0$,so we can divide by $(b - a)$ to get $b - a = a$,which means $b = 2a$.
Substituting $b = 2a$ into $2b = a + c$,we get $2(2a) = a + c$,so $4a = a + c$,which gives $c = 3a$.
Thus,$a : b : c = a : 2a : 3a = 1 : 2 : 3$.

(D) Given $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Since $a^2, b^2, c^2$ are in $H.P.$,we have $\frac{1}{b^2} - \frac{1}{a^2} = \frac{1}{c^2} - \frac{1}{b^2}$.
This simplifies to $\frac{a^2 - b^2}{a^2 b^2} = \frac{b^2 - c^2}{b^2 c^2}$,which implies $\frac{(a-b)(a+b)}{a^2} = \frac{(b-c)(b+c)}{c^2}$.
Since $a-b = b-c$,if $a \neq b$,we can divide by $(a-b)$ to get $\frac{a+b}{a^2} = \frac{b+c}{c^2}$.
$c^2(a+b) = a^2(b+c) \Rightarrow c^2 a + c^2 b = a^2 b + a^2 c$.
$ac(c-a) = b(a^2-c^2) = b(a-c)(a+c)$.
Since $a \neq c$,we divide by $(c-a)$ to get $ac = -b(a+c)$.
Substituting $a+c = 2b$,we get $ac = -b(2b) = -2b^2$,or $b^2 = \frac{-ac}{2} = (\frac{-a}{2})c$.
This confirms that $\frac{-a}{2}, b, c$ are in $G.P.$

(A) Let the two positive numbers be $a_1$ and $a_2$.
The arithmetic mean is $A = \frac{a_1 + a_2}{2}$,which implies $a_1 + a_2 = 2A$.
The geometric mean is $G = \sqrt{a_1 a_2}$,which implies $G^2 = a_1 a_2$.
The harmonic mean is $H = \frac{2a_1 a_2}{a_1 + a_2}$.
Substituting the values of $a_1 + a_2$ and $a_1 a_2$ into the formula for $H$:
$H = \frac{2(G^2)}{2A} = \frac{G^2}{A}$.

(A) Let the two numbers be $a$ and $b$.
We know that the relationship between Arithmetic Mean $(A)$,Geometric Mean $(G)$,and Harmonic Mean $(H)$ is $G^2 = AH$.
Given $H = 14\frac{2}{5} = \frac{72}{5}$ and $G = 24$.
Substituting the values: $(24)^2 = A \times \frac{72}{5}$.
$576 = A \times \frac{72}{5} \Rightarrow A = \frac{576 \times 5}{72} = 8 \times 5 = 40$.
Since $A = \frac{a+b}{2} = 40$,we have $a+b = 80$ $(i)$.
Since $G = \sqrt{ab} = 24$,we have $ab = 576$ $(ii)$.
Using the identity $(a-b)^2 = (a+b)^2 - 4ab$,we get $(a-b)^2 = (80)^2 - 4(576) = 6400 - 2304 = 4096$.
Thus,$a-b = \sqrt{4096} = 64$ $(iii)$.
Solving equations $(i)$ and $(iii)$ by adding them: $2a = 144 \Rightarrow a = 72$.
Substituting $a=72$ into $(i)$: $72 + b = 80 \Rightarrow b = 8$.
The greater number is $72$.

(B) Given that $\frac{A.M.}{G.M.} = \frac{p}{q}$,where $A.M. = \frac{x+y}{2}$ and $G.M. = \sqrt{xy}$.
So,$\frac{x+y}{2\sqrt{xy}} = \frac{p}{q} \dots (i)$.
Using the property of componendo and dividendo,let $x = u^2$ and $y = v^2$. Then $\frac{u^2+v^2}{2uv} = \frac{p}{q}$.
$\frac{u^2+v^2+2uv}{u^2+v^2-2uv} = \frac{p+q}{p-q} \implies \frac{(u+v)^2}{(u-v)^2} = \frac{p+q}{p-q}$.
Alternatively,from $(i)$,$\frac{x+y}{2\sqrt{xy}} = \frac{p}{q} \implies \frac{x+y}{2\sqrt{xy}} + 1 = \frac{p}{q} + 1$ and $\frac{x+y}{2\sqrt{xy}} - 1 = \frac{p}{q} - 1$.
This leads to $\frac{(\sqrt{x}+\sqrt{y})^2}{2\sqrt{xy}} = \frac{p+q}{q}$ and $\frac{(\sqrt{x}-\sqrt{y})^2}{2\sqrt{xy}} = \frac{p-q}{q}$.
Dividing these two,we get $\frac{(\sqrt{x}+\sqrt{y})^2}{(\sqrt{x}-\sqrt{y})^2} = \frac{p+q}{p-q}$.
Taking the square root,$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \sqrt{\frac{p+q}{p-q}}$.
Applying componendo and dividendo again,$\frac{\sqrt{x}}{\sqrt{y}} = \frac{\sqrt{p+q} + \sqrt{p-q}}{\sqrt{p+q} - \sqrt{p-q}}$.
Squaring both sides,$\frac{x}{y} = \frac{(p+q) + (p-q) + 2\sqrt{p^2-q^2}}{(p+q) + (p-q) - 2\sqrt{p^2-q^2}} = \frac{2p + 2\sqrt{p^2-q^2}}{2p - 2\sqrt{p^2-q^2}} = \frac{p + \sqrt{p^2-q^2}}{p - \sqrt{p^2-q^2}}$.