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

(B) Since $A.M. > G.M.$ 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,we get:
$\frac{(a + b)(b + c)(c + a)}{8} > \sqrt{ab \cdot bc \cdot ca}$
$\frac{(a + b)(b + c)(c + a)}{8} > \sqrt{a^2 b^2 c^2}$
$\frac{(a + b)(b + c)(c + a)}{8} > abc$
$(a + b)(b + c)(c + a) > 8abc$.

(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 $(i)$:
$4(\frac{1}{r} + 1 + r) = 14$
$\frac{1}{r} + 1 + r = 3.5$
$r + \frac{1}{r} = 2.5$
$2r^2 - 5r + 2 = 0$
$(2r - 1)(r - 2) = 0$
So,$r = 2$ or $r = 0.5$.
If $r = 2$,the numbers are $2, 4, 8$.
If $r = 0.5$,the numbers are $8, 4, 2$.
In both cases,the greatest number is $8$.

(A) Let the two quantities be $a$ and $b$.
Since $a, {A_1}, {A_2}, b$ are in $A.P.$,we have ${A_1} - a = b - {A_2}$,which implies ${A_1} + {A_2} = a + b$ ......$(i)$
Since $a, {G_1}, {G_2}, b$ are in $G.P.$,we have $\frac{{{G_1}}}{a} = \frac{b}{{{G_2}}}$,which implies ${G_1}{G_2} = ab$ ......$(ii)$
Since $a, {H_1}, {H_2}, b$ are in $H.P.$,we have $\frac{1}{{{H_1}}} - \frac{1}{a} = \frac{1}{b} - \frac{1}{{{H_2}}}$,which implies $\frac{1}{{{H_1}}} + \frac{1}{{{H_2}}} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$.
Substituting the values from $(i)$ and $(ii)$,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 $H = \frac{2xy}{x+y} = 4$.
Since $A = \frac{x+y}{2}$,we have $x+y = 2A$. Substituting this into the harmonic mean formula: $\frac{2xy}{2A} = 4$ $\Rightarrow \frac{xy}{A} = 4$ $\Rightarrow xy = 4A$.
Since $G^2 = xy$,we have $G^2 = 4A$.
Given the relation $2A + G^2 = 27$,substitute $G^2 = 4A$:
$2A + 4A = 27$ $\Rightarrow 6A = 27$ $\Rightarrow A = \frac{27}{6} = 4.5$.
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 $b$.
Given that the ratio of the numbers is $4:1$,we have $a = 4b$.
We know that $A.M. = \frac{a+b}{2}$ and $G.M. = \sqrt{ab}$.
According to the problem,$A.M. = G.M. + 2$.
Substituting the values,we get $\frac{4b+b}{2} = \sqrt{4b \times b} + 2$.
$\frac{5b}{2} = \sqrt{4b^2} + 2$.
$\frac{5b}{2} = 2b + 2$.
Multiplying by $2$,we get $5b = 4b + 4$.
$b = 4$.
Since $a = 4b$,$a = 4(4) = 16$.
Thus,the numbers are $16$ and $4$.

(B) We know the fundamental inequality for any two positive numbers: $A.M. > G.M. > H.M.$
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 $H.M. = \frac{144}{15}$.
The question asks for the values in the order $H.M., G.M., A.M.$
Thus,the required order is $\frac{144}{15}, 12, 15$.

(C) Given that $a = \frac{x + y}{2}$,$g = \sqrt{xy}$,and $h = \frac{2xy}{x + y}$.
Calculate $g^2$:
$g^2 = (\sqrt{xy})^2 = xy$ ... $(i)$
Calculate the product $ah$:
$ah = \left(\frac{x + y}{2}\right) \times \left(\frac{2xy}{x + y}\right) = xy$ ... $(ii)$
From $(i)$ and $(ii)$,we observe that $g^2 = ah$,which implies $g = \sqrt{ah}$.
Therefore,$g$ is the geometric mean between $a$ and $h$.

(D) Given $a, b, c$ are in $A.P.$,let $b = a + d$ and $c = a + 2d$,where $d > 0$ since $a < b < c$.
Given $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^2 = a^2 c^2$,which implies $b^4 = a^2 c^2$.
Taking the square root,$b^2 = \pm ac$.
If $b^2 = ac$,then $a, b, c$ are in $G.P.$ Since they are also in $A.P.$,$a = b = c$,which contradicts $a < b < c$.
Thus,$b^2 = -ac$.
Substituting $b = a + d$ and $c = a + 2d$:
$(a + d)^2 = -a(a + 2d)$
$a^2 + d^2 + 2ad = -a^2 - 2ad$
$2a^2 + 4ad + d^2 = 0$.
Given $a + b + c = \frac{3}{2}$,we have $a + (a + d) + (a + 2d) = 3a + 3d = \frac{3}{2}$,so $a + d = \frac{1}{2}$,which means $d = \frac{1}{2} - a$.
Substituting $d$ into the equation $2a^2 + 4ad + d^2 = 0$:
$2a^2 + 4a(\frac{1}{2} - a) + (\frac{1}{2} - a)^2 = 0$
$2a^2 + 2a - 4a^2 + \frac{1}{4} - a + a^2 = 0$
$-a^2 + a + \frac{1}{4} = 0$
$4a^2 - 4a - 1 = 0$.
Using the quadratic formula $a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-1)}}{2(4)} = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1}{2} \pm \frac{1}{\sqrt{2}}$.
Since $d = \frac{1}{2} - a > 0$,we must have $a < \frac{1}{2}$.
Therefore,$a = \frac{1}{2} - \frac{1}{\sqrt{2}}$.

(A) Given that $a^2(b + c), b^2(c + a), c^2(a + b)$ are in $AP$.
Dividing each term by $abc$,we get:
$\frac{a^2(b + c)}{abc}, \frac{b^2(c + a)}{abc}, \frac{c^2(a + b)}{abc}$ are in $AP$.
This simplifies to:
$\frac{a(b + c)}{bc}, \frac{b(c + a)}{ac}, \frac{c(a + b)}{ab}$ are in $AP$.
$\frac{ab + ac}{bc}, \frac{bc + ab}{ac}, \frac{ca + bc}{ab}$ are in $AP$.
Adding $1$ to each term:
$\frac{ab + ac}{bc} + 1, \frac{bc + ab}{ac} + 1, \frac{ca + bc}{ab} + 1$ are in $AP$.
$\frac{ab + ac + bc}{bc}, \frac{bc + ab + ac}{ac}, \frac{ca + bc + ab}{ab}$ are in $AP$.
Dividing by $(ab + bc + ca)$,we get:
$\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}$ are in $AP$.
Multiplying by $abc$,we get:
$a, b, c$ are in $AP$.

(A) Given that $x, y, z$ are in $AP$,we have $2y = x + z$.
This implies $z = 2y - x$.
Given that $x, y, t$ are in $GP$,we have $y^2 = xt$.
This implies $t = \frac{y^2}{x}$.
Now,consider the sequence $x, x - y, t - z$.
Substitute $t$ and $z$:
$t - z = \frac{y^2}{x} - (2y - x) = \frac{y^2 - 2xy + x^2}{x} = \frac{(x - y)^2}{x}$.
For the sequence $x, x - y, t - z$ to be in $GP$,the square of the middle term must equal the product of the first and third terms:
$(x - y)^2 = x \times (t - z)$.
Substituting the expression for $(t - z)$:
$x \times \frac{(x - y)^2}{x} = (x - y)^2$.
Since the condition holds,$x, x - y, t - z$ are in $GP$.

(B) The geometric mean between $a$ and $b$ is given by $\sqrt{ab} = (ab)^{1/2}$.
Given that $\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = (ab)^{1/2}$.
Cross-multiplying,we get $a^{n+1} + b^{n+1} = (ab)^{1/2} (a^n + b^n)$.
$a^{n+1} + b^{n+1} = a^{n+1/2} b^{1/2} + a^{1/2} b^{n+1/2}$.
Rearranging the terms: $a^{n+1} - a^{n+1/2} b^{1/2} = a^{1/2} b^{n+1/2} - b^{n+1}$.
$a^{n+1/2} (a^{1/2} - b^{1/2}) = b^{n+1/2} (a^{1/2} - b^{1/2})$.
Assuming $a \neq b$,we can divide by $(a^{1/2} - b^{1/2})$:
$a^{n+1/2} = b^{n+1/2}$.
$(a/b)^{n+1/2} = 1 = (a/b)^0$.
Equating the exponents,$n + 1/2 = 0$,which gives $n = -1/2$.

(C) Let the two numbers be $a$ and $b$.
Given that $A = \frac{a+b}{2}$ and $G = \sqrt{ab}$.
From these,we have $a+b = 2A$ and $ab = G^2$.
We know that $(a-b)^2 = (a+b)^2 - 4ab$.
Substituting the values,$(a-b)^2 = (2A)^2 - 4G^2 = 4(A^2 - G^2)$.
Therefore,$a-b = 2\sqrt{A^2 - G^2}$ (assuming $a > b$).
Adding the equations $a+b = 2A$ and $a-b = 2\sqrt{A^2 - G^2}$,we get $2a = 2A + 2\sqrt{A^2 - G^2}$,which implies $a = A + \sqrt{A^2 - G^2}$.
Subtracting the equations,we get $2b = 2A - 2\sqrt{A^2 - G^2}$,which implies $b = A - \sqrt{A^2 - G^2}$.
Thus,the numbers are $A \pm \sqrt{A^2 - G^2}$.

(D) Given the expression $S = a^{-5} + a^{-4} + 3a^{-3} + 1 + a^8 + a^{10}$ for $a > 0$.
Since all terms are positive,we apply the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.
We have $7$ terms: $a^{-5}, a^{-4}, a^{-3}, a^{-3}, a^{-3}, a^8, a^{10}$.
$\frac{a^{-5} + a^{-4} + a^{-3} + a^{-3} + a^{-3} + a^8 + a^{10}}{7} \geq (a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot a^8 \cdot a^{10})^{1/7}$
$\frac{a^{-5} + a^{-4} + 3a^{-3} + a^8 + a^{10}}{7} \geq (a^{-5-4-3-3-3+8+10})^{1/7}$
$\frac{a^{-5} + a^{-4} + 3a^{-3} + a^8 + a^{10}}{7} \geq (a^0)^{1/7} = 1$
$a^{-5} + a^{-4} + 3a^{-3} + a^8 + a^{10} \geq 7$
Adding the constant $1$ from the original expression:
$S = (a^{-5} + a^{-4} + 3a^{-3} + a^8 + a^{10}) + 1 \geq 7 + 1 = 8$
The minimum value is $8$.

(B) Let the first term $a = 1$ and the common difference be $d$.
The terms of the arithmetic progression are $t_n = a + (n-1)d$.
Thus,$t_2 = 1 + d$,$t_{10} = 1 + 9d$,and $t_{34} = 1 + 33d$.
Since $t_2, t_{10}, t_{34}$ are in geometric progression,we have $(t_{10})^2 = t_2 \times t_{34}$.
$(1 + 9d)^2 = (1 + d)(1 + 33d)$.
$1 + 18d + 81d^2 = 1 + 33d + d + 33d^2$.
$1 + 18d + 81d^2 = 1 + 34d + 33d^2$.
$81d^2 - 33d^2 + 18d - 34d = 0$.
$48d^2 - 16d = 0$.
$16d(3d - 1) = 0$.
Since $d \neq 0$ for a non-trivial progression,$3d - 1 = 0$,which gives $d = 1/3$.

(D) Given that $a, b,$ and $c$ are in arithmetic progression $(AP)$.
Therefore,$2b = a + c$.
Consider the terms $2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1}$.
For these terms to be in geometric progression $(GP)$,the ratio of consecutive terms must be equal:
$\frac{2^{bx + 1}}{2^{ax + 1}} = \frac{2^{cx + 1}}{2^{bx + 1}}$
$2^{(b - a)x} = 2^{(c - b)x}$
Since $a, b, c$ are in $AP$,we have $b - a = c - b = d$ (common difference).
Thus,$2^{dx} = 2^{dx}$,which is always true for any $x \neq 0$.
Therefore,$2^{ax + 1}, 2^{bx + 1},$ and $2^{cx + 1}$ are in geometric progression for every $x \neq 0$.

(B) Let the two numbers be $a$ and $b$.
Given that the geometric mean $G = \sqrt{ab} = 4$,so $ab = 16$.
Given that the arithmetic mean $A = \frac{a+b}{2} = 5$,so $a+b = 10$.
The relationship between arithmetic mean $(A)$,geometric mean $(G)$,and harmonic mean $(H)$ is $G^2 = AH$.
Substituting the values,$4^2 = 5 \times H$.
$16 = 5H$.
Therefore,$H = \frac{16}{5}$.

(D) The arithmetic mean $(AM)$ of $p$ and $q$ is $\frac{p+q}{2}$.
The geometric mean $(GM)$ of $p$ and $q$ is $\sqrt{pq}$.
Given that $AM = 2 \times GM$,we have $\frac{p+q}{2} = 2\sqrt{pq}$.
This implies $p+q = 4\sqrt{pq}$.
Squaring both sides,we get $(p+q)^2 = 16pq$,which simplifies to $p^2 + 2pq + q^2 = 16pq$,or $p^2 - 14pq + q^2 = 0$.
Dividing by $q^2$,we get $(\frac{p}{q})^2 - 14(\frac{p}{q}) + 1 = 0$.
Let $x = \frac{p}{q}$. Then $x^2 - 14x + 1 = 0$.
Using the quadratic formula,$x = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2} = \frac{14 \pm 8\sqrt{3}}{2} = 7 \pm 4\sqrt{3}$.
Since $p > q$,$x > 1$,so $x = 7 + 4\sqrt{3} = \frac{7 + 4\sqrt{3}}{1}$.
Note that $(2 + \sqrt{3})^2 = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}$.
Thus,the ratio $p:q$ is $(7 + 4\sqrt{3}) : 1$.

(A) We know the relationship between Arithmetic Mean $(AM)$,Geometric Mean $(GM)$,and Harmonic Mean $(HM)$ for two numbers is given by the formula:
$GM^2 = AM \times HM$
Given:
$AM = 16$
$HM = \frac{63}{4}$
Substituting the values:
$GM^2 = 16 \times \frac{63}{4}$
$GM^2 = 4 \times 63$
$GM^2 = 252$
$GM = \sqrt{252}$
$GM = \sqrt{36 \times 7}$
$GM = 6\sqrt{7}$
Therefore,the correct option is $A$.

(B) Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$. Let the first term of the Geometric Progression be $A$ and the common ratio be $R$.
$a + (p - 1)d = A R^{p - 1} = x$
$a + (q - 1)d = A R^{q - 1} = y$
$a + (r - 1)d = A R^{r - 1} = z$
Taking logarithms on both sides for each equation:
$\ln(x) = \ln(A) + (p - 1)\ln(R)$
$\ln(y) = \ln(A) + (q - 1)\ln(R)$
$\ln(z) = \ln(A) + (r - 1)\ln(R)$
Consider the expression $E = x^{y - z} \cdot y^{z - x} \cdot z^{x - y}$.
Taking $\ln(E) = (y - z)\ln(x) + (z - x)\ln(y) + (x - y)\ln(z)$.
Substituting the values of $\ln(x), \ln(y), \ln(z)$:
$\ln(E) = (y - z)(\ln(A) + (p - 1)\ln(R)) + (z - x)(\ln(A) + (q - 1)\ln(R)) + (x - y)(\ln(A) + (r - 1)\ln(R))$
$\ln(E) = \ln(A)(y - z + z - x + x - y) + \ln(R)((p - 1)(y - z) + (q - 1)(z - x) + (r - 1)(x - y))$
Since $(y - z + z - x + x - y) = 0$ and the second term simplifies to $0$ due to the cyclic nature of the terms in an $A$.$P$. and $G$.$P$. relationship:
$\ln(E) = 0 \implies E = e^0 = 1$.

(B) Given that the arithmetic mean $A = \frac{a+b}{2}$ and geometric mean $G = \sqrt{ab}$.
According to the problem,$A = 2G$.
$\frac{a+b}{2} = 2\sqrt{ab} \implies \frac{a+b}{\sqrt{ab}} = 4$.
Dividing by $\sqrt{ab}$,we get $\frac{a}{\sqrt{ab}} + \frac{b}{\sqrt{ab}} = 4 \implies \sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} = 4$.
Let $x = \sqrt{\frac{a}{b}}$. Then $x + \frac{1}{x} = 4 \implies x^2 - 4x + 1 = 0$.
Using the quadratic formula,$x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$.
Since $x = \sqrt{\frac{a}{b}}$,we have $\frac{a}{b} = x^2 = (2 \pm \sqrt{3})^2 = 4 + 3 \pm 4\sqrt{3} = 7 \pm 4\sqrt{3}$.
Alternatively,using the componendo and dividendo method on $\frac{a+b}{2\sqrt{ab}} = \frac{2}{1}$:
$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{2+1}{2-1} = 3$.
$\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = 3 \implies \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{3}$.
Applying componendo and dividendo again: $\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.
$\frac{a}{b} = \left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2 = \frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}} = \frac{4+2\sqrt{3}}{4-2\sqrt{3}} = \frac{2+\sqrt{3}}{2-\sqrt{3}}$.

(C) The Arithmetic Mean of $a$ and $b$ is given by $\frac{a+b}{2}$.
Given,$\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a+b}{2}$.
Cross-multiplying,we get $2(a^{n+1} + b^{n+1}) = (a+b)(a^n + b^n)$.
$2a^{n+1} + 2b^{n+1} = a^{n+1} + ab^n + ba^n + b^{n+1}$.
Rearranging the terms: $a^{n+1} - ab^n - ba^n + b^{n+1} = 0$.
$a^n(a - b) - b^n(a - b) = 0$.
$(a^n - b^n)(a - b) = 0$.
Since $a \neq b$,we must have $a^n - b^n = 0$,which implies $a^n = b^n$.
Thus,$(\frac{a}{b})^n = 1 = (\frac{a}{b})^0$.
Therefore,$n = 0$.

(C) For any two positive real numbers $a$ and $b$,the arithmetic mean $A = \frac{a+b}{2}$,the geometric mean $G = \sqrt{ab}$,and the harmonic mean $H = \frac{2ab}{a+b}$.
We know that for any two positive numbers,$A \ge G \ge H$.
Specifically,$A \times H = \frac{a+b}{2} \times \frac{2ab}{a+b} = ab = G^2$.
Since $G^2 = AH$,it implies $G$ is the geometric mean of $A$ and $H$,which confirms $A > G > H$ for distinct numbers.

(A) For any two distinct positive numbers $a$ and $b$,we know that $A_1 > G_1 > H_1$.
By definition,$A_2 = \frac{A_1 + H_1}{2}$.
Since $A_1 > H_1$,it follows that $A_1 > A_2 > H_1$.
Similarly,$A_3 = \frac{A_2 + H_2}{2}$.
Since $A_2 > H_2$,it follows that $A_2 > A_3 > H_2$.
Continuing this process,we get the sequence $A_1 > A_2 > A_3 > \dots > A_n > \dots$.

(C) For positive real numbers $a_1, a_2, \dots, a_n$,the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality states that $AM \ge GM$.
$\therefore \frac{a_1 + a_2 + \dots + a_n}{n} \ge (a_1 \cdot a_2 \cdot \dots \cdot a_n)^{\frac{1}{n}}$
Given $a_1 \cdot a_2 \cdot \dots \cdot a_n = 1$,we substitute this into the inequality:
$\therefore \frac{a_1 + a_2 + \dots + a_n}{n} \ge (1)^{\frac{1}{n}}$
$\therefore \frac{a_1 + a_2 + \dots + a_n}{n} \ge 1$
$\therefore a_1 + a_2 + \dots + a_n \ge n$
Thus,the sum of the numbers is not less than $n$.

(D) The Arithmetic Mean $(AM)$ is $\frac{a+b}{2}$ and the Harmonic Mean $(HM)$ is $\frac{2ab}{a+b}$.
Given $\frac{AM}{HM} = \frac{m}{n}$,so $\frac{(a+b)^2}{4ab} = \frac{m}{n}$.
This implies $\frac{(a+b)^2}{ab} = \frac{4m}{n}$.
Dividing by $4$,we get $\frac{(a+b)^2}{4ab} = \frac{m}{n}$.
Using the property $\frac{a}{b} = \frac{AM + \sqrt{AM^2 - GM^2}}{AM - \sqrt{AM^2 - GM^2}}$,where $GM^2 = AM \times HM$.
Since $GM^2 = \frac{a+b}{2} \times \frac{2ab}{a+b} = ab$,we have $\frac{a}{b} = \frac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}}$.

(D) If $a, b, c$ are in $A.P.$,then $2b = a + c$.
If $a, b, c$ are in $G.P.$,then $b^2 = ac$.
From the first equation,$c = 2b - a$.
Substituting this into the second equation: $b^2 = a(2b - a)$.
$b^2 = 2ab - a^2$.
$a^2 - 2ab + b^2 = 0$.
$(a - b)^2 = 0$.
This implies $a = b$.
Since $a = b$,substituting into $2b = a + c$ gives $2a = a + c$,which implies $a = c$.
Therefore,$a = b = c$.

(A) Let $a^{1/x} = b^{1/y} = c^{1/z} = k$ (where $k \neq 1$).
Then $a = k^x, b = k^y, c = k^z$.
Since $a, b, c$ are in geometric progression,we have $b^2 = ac$.
Substituting the values,we get $(k^y)^2 = k^x \cdot k^z$.
$k^{2y} = k^{x+z}$.
Equating the exponents,$2y = x + z$.
This implies that $x, y, z$ are in arithmetic progression.

(C) Let the two positive numbers be $a$ and $b$.
The geometric mean $(GM)$ is $\sqrt{ab}$ and the harmonic mean $(HM)$ is $\frac{2ab}{a+b}$.
Given the ratio $\frac{HM}{GM} = \frac{4}{5}$.
Substituting the formulas: $\frac{\frac{2ab}{a+b}}{\sqrt{ab}} = \frac{4}{5}$.
This simplifies to $\frac{2\sqrt{ab}}{a+b} = \frac{4}{5}$,which implies $\frac{\sqrt{ab}}{a+b} = \frac{2}{5}$.
Squaring both sides: $\frac{ab}{(a+b)^2} = \frac{4}{25}$.
Let $x = \frac{a}{b}$. Then $\frac{b^2 x}{b^2(x+1)^2} = \frac{4}{25}$,so $\frac{x}{(x+1)^2} = \frac{4}{25}$.
$25x = 4(x^2 + 2x + 1) \implies 4x^2 - 17x + 4 = 0$.
Factoring the quadratic: $(4x - 1)(x - 4) = 0$.
Since $a > b$,$x = \frac{a}{b} > 1$,so $x = 4$.
Therefore,$a : b = 4 : 1$.

(A) Let the two numbers be $a$ and $b$.
Since $p$ and $q$ are two arithmetic means between $a$ and $b$,the sequence $a, p, q, b$ is in Arithmetic Progression ($A$.$P$.).
Let the common difference be $d$.
Then $p = a + d$,$q = a + 2d$,and $b = a + 3d$.
From these,$d = q - p$,so $a = p - d = p - (q - p) = 2p - q$.
Also,$b = a + 3d = (2p - q) + 3(q - p) = 2p - q + 3q - 3p = 2q - p$.
Since $G$ is the geometric mean between $a$ and $b$,$G^2 = ab$.
Substituting the values of $a$ and $b$,we get $G^2 = (2p - q)(2q - p)$.

(D) Let the two positive real numbers be $a$ and $b$.
Then,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,consider the product $AH = \left( \frac{a+b}{2} \right) \times \left( \frac{2ab}{a+b} \right) = ab$.
Also,$G^2 = (\sqrt{ab})^2 = ab$.
Therefore,$G^2 = AH$.

(C) The arithmetic mean of $a$ and $b$ is $A = \frac{a + b}{2}$.
The geometric mean of $a$ and $b$ is $G = \sqrt{ab}$.
Therefore,$A - G = \frac{a + b}{2} - \sqrt{ab}$.
$A - G = \frac{a + b - 2\sqrt{ab}}{2}$.
Since $a = (\sqrt{a})^2$ and $b = (\sqrt{b})^2$,we have $A - G = \frac{(\sqrt{a})^2 + (\sqrt{b})^2 - 2\sqrt{a}\sqrt{b}}{2}$.
$A - G = \frac{(\sqrt{a} - \sqrt{b})^2}{2} = \left[ \frac{\sqrt{a} - \sqrt{b}}{\sqrt{2}} \right]^2$.

(A) Since $a, b, c$ are in an arithmetic progression,we have $a + c = 2b$ or $c - b = b - a$. Let $d = b - a = c - b$.
Given that $b - a, c - b, a$ are in a geometric progression,we have $(c - b)^2 = (b - a)a$.
Substituting $d$ into the equation,we get $d^2 = da$. Since the numbers are distinct,$d \neq 0$,so $d = a$.
Now,$b - a = a \implies b = 2a$.
Also,$c - b = a \implies c = b + a = 2a + a = 3a$.
Therefore,$a : b : c = a : 2a : 3a = 1 : 2 : 3$.

(C) Let the two positive numbers be $a$ and $b$.
Arithmetic Mean $A = \frac{a+b}{2}$,Geometric Mean $G = \sqrt{ab} = 12$,Harmonic Mean $H = \frac{2ab}{a+b}$.
Given $A + H = 25$ and $G = 12$.
Since $G^2 = AH$,we have $AH = 12^2 = 144$.
Substituting $H = \frac{144}{A}$ into $A + H = 25$:
$A + \frac{144}{A} = 25 \implies A^2 - 25A + 144 = 0$.
$(A - 16)(A - 9) = 0$,so $A = 16$ or $A = 9$.
Since $A \geq G$,we must have $A = 16$.
Thus,$\frac{a+b}{2} = 16 \implies a+b = 32$.

(C) For the $A$.$P$. $a, x, y, z, b$,the sum is $x + y + z = 15$. Since $x, y, z$ are terms between $a$ and $b$,we have $x+z = a+b$ and $y = \frac{a+b}{2}$.
Thus,$x+y+z = \frac{3}{2}(a+b) = 15$,which implies $a+b = 10$ (Equation $1$).
For the $H$.$P$. $a, x, y, z, b$,the reciprocals $\frac{1}{a}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{b}$ are in $A$.$P$.
Similarly,$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{2}(\frac{1}{a} + \frac{1}{b}) = \frac{5}{3}$.
This implies $\frac{1}{a} + \frac{1}{b} = \frac{10}{9}$,so $\frac{a+b}{ab} = \frac{10}{9}$.
Substituting $a+b=10$,we get $\frac{10}{ab} = \frac{10}{9}$,so $ab = 9$ (Equation $2$).
Solving $a+b=10$ and $ab=9$,the quadratic equation $t^2 - 10t + 9 = 0$ gives $(t-9)(t-1) = 0$.
Thus,the values are $a, b = 9, 1$.

(B) Let the two positive numbers be $a$ and $b$.
Given,$\sqrt{ab} = 6$ and $\frac{a+b}{2} = 6.5$.
From the first equation,$ab = 36$.
From the second equation,$a+b = 13$.
We know that $(a-b)^2 = (a+b)^2 - 4ab$.
Substituting the values,$(a-b)^2 = (13)^2 - 4(36) = 169 - 144 = 25$.
Thus,$a-b = 5$ (assuming $a > b$).
Solving the system $a+b = 13$ and $a-b = 5$:
Adding the equations: $2a = 18 \implies a = 9$.
Subtracting the equations: $2b = 8 \implies b = 4$.
Therefore,the numbers are $4$ and $9$.

(D) Given that $a, b, c$ are in $AP$,we have $\frac{a + c}{2} = b$,which implies $a + c = 2b$.
Given that $a, b, c$ are in $GP$,we have $b^2 = ac$.
Squaring the first equation: $(a + c)^2 = (2b)^2 = 4b^2$.
Substituting $b^2 = ac$ into the equation: $(a + c)^2 = 4ac$.
This simplifies to $(a + c)^2 - 4ac = 0$,which is $(a - c)^2 = 0$.
Therefore,$a = c$.
Substituting $a = c$ into $a + c = 2b$,we get $c + c = 2b$,so $2c = 2b$,which means $b = c$.
Thus,$a = b = c$.

(A) Let the two numbers be $x$ and $y$.
Given that the harmonic mean $H = 4$,we have $H = \frac{2xy}{x+y} = 4$,which implies $xy = 2(x+y)$.
Since $A = \frac{x+y}{2}$,we have $x+y = 2A$,so $xy = 2(2A) = 4A$.
We know that $G^2 = xy$,therefore $G^2 = 4A$.
Substitute $G^2 = 4A$ into the given equation $2A + G^2 = 27$:
$2A + 4A = 27$ $\Rightarrow 6A = 27$ $\Rightarrow A = 4.5$.
Since $A = \frac{x+y}{2} = 4.5$,we get $x+y = 9$.
Since $xy = 4A = 4(4.5) = 18$,we have the system $x+y = 9$ and $xy = 18$.
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: $(t-6)(t-3) = 0$,so $t = 6$ or $t = 3$.
Thus,the two numbers are $6$ and $3$.

(D) Let the two numbers be $x$ and $y$,where $x > y$.
Given $x - y = 48$,so $x = 48 + y$ $(1)$.
The difference between their arithmetic mean and geometric mean is $18$:
$\frac{x + y}{2} - \sqrt{xy} = 18$
$x + y - 2\sqrt{xy} = 36$ $(2)$.
Substitute $x = 48 + y$ into equation $(2)$:
$(48 + y) + y - 2\sqrt{(48 + y)y} = 36$
$48 + 2y - 36 = 2\sqrt{y^2 + 48y}$
$12 + 2y = 2\sqrt{y^2 + 48y}$
Divide by $2$:
$6 + y = \sqrt{y^2 + 48y}$
Square both sides:
$(6 + y)^2 = y^2 + 48y$
$36 + 12y + y^2 = y^2 + 48y$
$36 = 36y$
$y = 1$.
Since $x = 48 + y$,we have $x = 48 + 1 = 49$.
The larger number is $49$.

(A) Since $a, b, c$ are in geometric progression,we have $b^2 = ac$.
The equation $ax^2 + 2bx + c = 0$ becomes $ax^2 + 2\sqrt{ac}x + c = 0$,which is $(\sqrt{a}x + \sqrt{c})^2 = 0$.
Thus,the roots are $x = -\sqrt{\frac{c}{a}}, -\sqrt{\frac{c}{a}}$.
For the equation $dx^2 + 2ex + f = 0$ to have the same roots,the root $-\sqrt{\frac{c}{a}}$ must satisfy it.
Substituting $x = -\sqrt{\frac{c}{a}}$ into $dx^2 + 2ex + f = 0$,we get $d(\frac{c}{a}) - 2e\sqrt{\frac{c}{a}} + f = 0$.
Dividing by $c$,we get $\frac{d}{a} - \frac{2e}{c}\sqrt{\frac{c}{a}} + \frac{f}{c} = 0$.
Since $\sqrt{\frac{c}{a}} = \frac{c}{\sqrt{ac}} = \frac{c}{b}$,we have $\frac{d}{a} + \frac{f}{c} = \frac{2e}{c} \cdot \frac{c}{b} = \frac{2e}{b}$.
This is the condition for $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ to be in Arithmetic Progression.

(C) We use the Arithmetic Mean-Harmonic Mean inequality $(AM \geq HM)$.
For three positive numbers $a, b, c$,the Arithmetic Mean is $\frac{a+b+c}{3}$ and the Harmonic Mean is $\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$.
According to the inequality,$\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$.
Multiplying both sides by $3 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$,we get $(a+b+c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq 3 \times 3 = 9$.
Thus,the minimum value is $9$.

(B) Let $x_1, x_2, \dots, x_n$ be $n$ positive numbers such that $x_1 \times x_2 \times \dots \times x_n = 1$ $(1)$.
By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:
$\frac{x_1 + x_2 + \dots + x_n}{n} \ge (x_1 \times x_2 \times \dots \times x_n)^{1/n}$.
Substituting $(1)$ into the inequality:
$\frac{x_1 + x_2 + \dots + x_n}{n} \ge (1)^{1/n} = 1$.
Therefore,$x_1 + x_2 + \dots + x_n \ge n$.

(B) Let the three positive numbers in $G.P.$ be $a, ar, ar^2$ where $r > 1$ for an increasing $G.P.$
If the middle term is doubled,the new numbers are $a, 2ar, ar^2$.
Since these numbers are in $A.P.$,the middle term is the arithmetic mean of the other two:
$2(2ar) = a + ar^2$
$4ar = a(1 + r^2)$
Since $a > 0$,we can divide by $a$:
$r^2 - 4r + 1 = 0$
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$
Since the $G.P.$ is increasing,$r > 1$. Thus,$r = 2 + \sqrt{3}$.

(D) Let the first term of the $A.P.$ be $a$ and the common difference be $d$. The terms are given by $T_n = a + (n-1)d$.
The $2^{nd}, 5^{th},$ and $9^{th}$ terms are $a+d, a+4d,$ and $a+8d$ respectively.
Since these terms are in $G.P.$,the square of the middle term equals the product of the first and third terms:
$(a+4d)^2 = (a+d)(a+8d)$
Expanding both sides:
$a^2 + 8ad + 16d^2 = a^2 + 9ad + 8d^2$
Subtracting $a^2$ and rearranging terms:
$8d^2 = ad$
Since the $A.P.$ is non-constant,$d \neq 0$,so we can divide by $d$:
$a = 8d$
The terms of the $G.P.$ are:
$T_2 = a+d = 8d+d = 9d$
$T_5 = a+4d = 8d+4d = 12d$
$T_9 = a+8d = 8d+8d = 16d$
The common ratio $r$ is given by:
$r = \frac{T_5}{T_2} = \frac{12d}{9d} = \frac{4}{3}$

(A) Given that $\alpha, \beta, \gamma$ are geometric means between $(ca, ab), (ab, bc), (bc, ca)$ respectively.
Therefore,$\alpha^2 = (ca)(ab) = a^2bc$,$\beta^2 = (ab)(bc) = ab^2c$,and $\gamma^2 = (bc)(ca) = abc^2$.
Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.
We want to check the progression of $\alpha^2, \beta^2, \gamma^2$,which are $a^2bc, ab^2c, abc^2$.
Dividing each term by $abc$,we get $a, b, c$.
Since $a, b, c$ are in $A.P.$,it follows that $a^2bc, ab^2c, abc^2$ are also in $A.P.$ because they are obtained by multiplying each term of the $A.P.$ by the constant $abc$.

(D) Given ${a_1 = h_1 = 2}$ and ${a_{10} = h_{10} = 3}$.
For the $A.P.$,${a_{10} = a_1 + 9d = 3}$,so ${2 + 9d = 3}$,which gives ${d = \frac{1}{9}}$.
Thus,${a_4 = a_1 + 3d = 2 + 3(\frac{1}{9}) = 2 + \frac{1}{3} = \frac{7}{3}}$.
For the $H.P.$,the reciprocals $\frac{1}{h_n}$ are in $A.P.$ Let ${H_n = \frac{1}{h_n}}$.
Then ${H_1 = \frac{1}{2}}$ and ${H_{10} = \frac{1}{3}}$.
${H_{10} = H_1 + 9D = \frac{1}{3}}$,so ${9D = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}}$,which gives ${D = -\frac{1}{54}}$.
Then ${H_7 = H_1 + 6D = \frac{1}{2} + 6(-\frac{1}{54}) = \frac{1}{2} - \frac{1}{9} = \frac{9-2}{18} = \frac{7}{18}}$.
Since ${h_7 = \frac{1}{H_7}}$,we have ${h_7 = \frac{18}{7}}$.
Therefore,${a_4 h_7 = \frac{7}{3} \times \frac{18}{7} = 6}$.

(D) We use the $AM \ge GM$ inequality for positive real numbers.
For option $A$: By $AM \ge GM$,$\frac{a_1^3 + a_2^3 + a_3^3}{3} \ge \sqrt[3]{a_1^3 a_2^3 a_3^3} = a_1 a_2 a_3$,so $a_1^3 + a_2^3 + a_3^3 \ge 3a_1 a_2 a_3$. This is true.
For option $B$: By $AM \ge GM$,$\frac{a_1/a_2 + a_2/a_3 + a_3/a_1}{3} \ge \sqrt[3]{\frac{a_1}{a_2} \cdot \frac{a_2}{a_3} \cdot \frac{a_3}{a_1}} = 1$,so $\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1} \ge 3$. This is true.
For option $C$: By $AM \ge HM$,$(a_1 + a_2 + a_3) \ge \frac{9}{1/a_1 + 1/a_2 + 1/a_3}$,which implies $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) \ge 9$. This is true.
For option $D$: The inequality $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)^3 \le 27$ is generally false. For example,if $a_1=a_2=a_3=1$,then $(3)(3^3) = 81 \not\le 27$. Thus,option $D$ is not true.

(C) Let the total distance of the journey be $2d$.
Time taken for the first half of the journey is $t_1 = \frac{d}{v_1}$.
Time taken for the second half of the journey is $t_2 = \frac{d}{v_2}$.
Average velocity $V_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{t_1 + t_2}$.
Substituting the values,$V_{av} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}} = \frac{2d}{d(\frac{v_2 + v_1}{v_1 v_2})} = \frac{2v_1 v_2}{v_1 + v_2}$.

(B) By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for non-negative real numbers $x, y, z$,we have:
$\frac{x^{2017} + y^{2017} + z^{2017}}{3} \geq \sqrt[3]{x^{2017} y^{2017} z^{2017}}$
$x^{2017} + y^{2017} + z^{2017} \geq 3(xyz)^{2017/3}$.
However,this does not directly lead to the expression $2017xyz$.
Consider the case where $x = y = z = k$. Then $E = 3k^{2017} - 2017k^3$.
If $x=y=z=1$,then $E = 1 + 1 + 1 - 2017 = -2014$.
If $x=y=z=0$,then $E = 0$.
Since $x, y, z \geq 0$,we can test values. If we set $y=z=0$,then $E = x^{2017} \geq 0$.
If we set $x=y=z=1$,$E = -2014$.
If we consider the function $f(x, y, z) = x^{2017} + y^{2017} + z^{2017} - 2017xyz$,for $x, y, z geq 0$,the minimum value occurs at $x=y=z=1$,which gives $1+1+1-2017 = -2014$.

(C) Given,geometric mean $G = \sqrt{x_1 x_2} = 18$,so $x_1 x_2 = 18^2 = 324$.
Given,harmonic mean $H = \frac{2 x_1 x_2}{x_1 + x_2} = 16\frac{8}{13} = \frac{216}{13}$.
Substituting $x_1 x_2 = 324$ into the harmonic mean formula:
$\frac{2(324)}{x_1 + x_2} = \frac{216}{13} \Rightarrow x_1 + x_2 = \frac{648 \times 13}{216} = 3 \times 13 = 39$.
We know that $(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1 x_2$.
$(x_1 - x_2)^2 = (39)^2 - 4(324) = 1521 - 1296 = 225$.
Therefore,$|x_1 - x_2| = \sqrt{225} = 15$.