Geometric progression Questions in English

(C) Let the three terms in $GP$ be $\frac{a}{r}, a, ar$.
Given that their sum is $155$:
$\frac{a}{r} + a + ar = 155$ $(1)$
Also,the first term is $120$ less than the third term:
$ar - \frac{a}{r} = 120$ $(2)$
From $(2)$,$a(r - \frac{1}{r}) = 120 \implies a(\frac{r^2 - 1}{r}) = 120$.
From $(1)$,$a(\frac{1 + r + r^2}{r}) = 155$.
Dividing the two equations:
$\frac{r^2 - 1}{r^2 + r + 1} = \frac{120}{155} = \frac{24}{31}$.
$31r^2 - 31 = 24r^2 + 24r + 24$.
$7r^2 - 24r - 55 = 0$.
Solving the quadratic equation: $7r^2 - 35r + 11r - 55 = 0 \implies 7r(r - 5) + 11(r - 5) = 0$.
So,$r = 5$ or $r = -\frac{11}{7}$.
If $r = 5$,then $a(\frac{1}{5} + 1 + 5) = 155 \implies a(\frac{31}{5}) = 155 \implies a = 25$.
The terms are $\frac{25}{5}, 25, 25 \times 5$,which are $5, 25, 125$.

(D) Let the geometric progression be $a, ar, ar^2, ar^3$.
Here,the terms are $a_1 = a$,$a_2 = b = ar$,$a_3 = c = ar^2$,and $a_4 = d = ar^3$.
We check the product of the extremes and the means:
$ad = a \times ar^3 = a^2r^3$
$bc = ar \times ar^2 = a^2r^3$
Therefore,$ad = bc$.

(C) Let the two numbers be $p$ and $q$.
The two geometric means $G_1$ and $G_2$ between $p$ and $q$ are given by $G_1 = p(q/p)^{1/3} = p^{2/3}q^{1/3}$ and $G_2 = p(q/p)^{2/3} = p^{1/3}q^{2/3}$.
Now,calculate the expression $\frac{G_1^2}{G_2} + \frac{G_2^2}{G_1}$:
$\frac{G_1^2}{G_2} = \frac{(p^{2/3}q^{1/3})^2}{p^{1/3}q^{2/3}} = \frac{p^{4/3}q^{2/3}}{p^{1/3}q^{2/3}} = p^{4/3-1/3} = p$.
Similarly,$\frac{G_2^2}{G_1} = \frac{(p^{1/3}q^{2/3})^2}{p^{2/3}q^{1/3}} = \frac{p^{2/3}q^{4/3}}{p^{2/3}q^{1/3}} = q^{4/3-1/3} = q$.
Thus,$\frac{G_1^2}{G_2} + \frac{G_2^2}{G_1} = p + q$.
Since the arithmetic mean $A = \frac{p+q}{2}$,we have $p+q = 2A$.
Therefore,the value is $2A$.

(B) Let $x = 0.125125125 \dots$ (Equation $1$).
Since the repeating block is of $3$ digits,multiply both sides by $1000$:
$1000x = 125.125125125 \dots$ (Equation $2$).
Subtract Equation $1$ from Equation $2$:
$1000x - x = 125.125125 \dots - 0.125125 \dots$
$999x = 125$
$x = \frac{125}{999}$.

(B) Given $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$.
Applying componendo and dividendo rule to the first two parts:
$\frac{(a + bx) + (a - bx)}{(a + bx) - (a - bx)} = \frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)}$
$\frac{2a}{2bx} = \frac{2b}{2cx} \implies \frac{a}{b} = \frac{b}{c} \implies b^2 = ac$.
Similarly,applying it to the last two parts:
$\frac{2b}{2cx} = \frac{2c}{2dx} \implies \frac{b}{c} = \frac{c}{d} \implies c^2 = bd$.
Since $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$,the terms $a, b, c, d$ are in Geometric Progression.

(C) Given that $G$ is the geometric mean of $x$ and $y$,we have $G = \sqrt{xy}$,which implies $G^2 = xy$.
Now,consider the expression $\frac{1}{G^2 - x^2} + \frac{1}{G^2 - y^2}$.
Substituting $G^2 = xy$ into the expression:
$= \frac{1}{xy - x^2} + \frac{1}{xy - y^2}$
$= \frac{1}{x(y - x)} + \frac{1}{y(x - y)}$
$= \frac{1}{x(y - x)} - \frac{1}{y(y - x)}$
$= \frac{1}{y - x} \left( \frac{1}{x} - \frac{1}{y} \right)$
$= \frac{1}{y - x} \left( \frac{y - x}{xy} \right)$
$= \frac{1}{xy} = \frac{1}{G^2}$.

(D) Let the geometric progression be $a, ar, ar^2, ar^3, \dots$ where $a > 0$ and $r > 0$.
According to the problem,each term is equal to the sum of the two terms following it:
$ar^{n-1} = ar^n + ar^{n+1}$
Dividing both sides by $ar^{n-1}$ (since $a \neq 0$ and $r \neq 0$):
$1 = r + r^2$
Rearranging the terms gives the quadratic equation:
$r^2 + r - 1 = 0$
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$
Since all terms are positive,the common ratio $r$ must be positive.
Therefore,we take the positive root:
$r = \frac{\sqrt{5} - 1}{2} = \frac{1}{2}(\sqrt{5} - 1)$

(C) We are given the equation: $1 + r + r^2 + \dots + r^n = (1 + r) (1 + r^2) (1 + r^4) (1 + r^8)$.
Using the formula for the sum of a geometric series,the left side is $\frac{1 - r^{n+1}}{1 - r}$.
So,$\frac{1 - r^{n+1}}{1 - r} = (1 + r) (1 + r^2) (1 + r^4) (1 + r^8)$.
Multiplying both sides by $(1 - r)$,we get:
$1 - r^{n+1} = (1 - r)(1 + r)(1 + r^2)(1 + r^4)(1 + r^8)$.
Using the identity $(1 - x)(1 + x) = 1 - x^2$ repeatedly:
$(1 - r)(1 + r) = 1 - r^2$
$(1 - r^2)(1 + r^2) = 1 - r^4$
$(1 - r^4)(1 + r^4) = 1 - r^8$
$(1 - r^8)(1 + r^8) = 1 - r^{16}$.
Thus,$1 - r^{n+1} = 1 - r^{16}$.
Comparing the exponents,$n + 1 = 16$,which gives $n = 15$.

(B) Let the first term be $a = 1$ and the common ratio be $r$.
The terms of the geometric progression are $T_1 = 1$,$T_2 = r$,and $T_3 = r^2$.
We are given the expression $f(r) = 4T_2 + 5T_3 = 4r + 5r^2$.
To find the minimum value,we differentiate $f(r)$ with respect to $r$ and set it to $0$:
$f'(r) = \frac{d}{dr}(4r + 5r^2) = 4 + 10r$.
Setting $f'(r) = 0$ gives $4 + 10r = 0$,which implies $10r = -4$.
Thus,$r = -\frac{4}{10} = -\frac{2}{5}$.
Since $f''(r) = 10 > 0$,the function has a minimum at $r = -\frac{2}{5}$.

(C) The sum of $n$ terms of a geometric progression is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
We know the last term $l = ar^{n-1} = 243$ and $r = 3$.
Substituting $ar^n = l \times r = 243 \times 3 = 729$ into the sum formula:
$S_n = \frac{ar^n - a}{r - 1} = 364$
$\frac{729 - a}{3 - 1} = 364$
$729 - a = 364 \times 2$
$729 - a = 728$
$a = 1$
Now,using $ar^{n-1} = 243$ with $a = 1$ and $r = 3$:
$1 \times 3^{n-1} = 243$
$3^{n-1} = 3^5$
$n - 1 = 5$
$n = 6$.

(C) Let the first term be $a$ and the common ratio be $r$. The terms are $a, ar, ar^2, ar^3, \dots$
Given that the third term $a_3 = a_1^2$,we have $ar^2 = a^2$.
Since $a \neq 0$,we can divide by $a$ to get $r^2 = a$.
Given the second term $a_2 = ar = 8$,we substitute $a = r^2$ into this equation:
$r^2 \cdot r = 8 \implies r^3 = 8 \implies r = 2$.
Now,find $a$: $a = r^2 = 2^2 = 4$.
The sixth term is given by $a_6 = ar^5$.
$a_6 = 4 \times (2)^5 = 4 \times 32 = 128$.

(A) Let the $n$ geometric means be $G_1, G_2, \dots, G_n$.
Then the sequence $a, G_1, G_2, \dots, G_n, b$ forms a geometric progression with $n+2$ terms.
Let $r$ be the common ratio.
The $(n+2)$-th term is given by $T_{n+2} = a \cdot r^{(n+2)-1} = a \cdot r^{n+1}$.
Since $T_{n+2} = b$,we have $b = a \cdot r^{n+1}$.
Therefore,$r^{n+1} = \frac{b}{a}$.
Thus,$r = (\frac{b}{a})^{\frac{1}{n+1}}$.

(B) Let the infinite geometric series be $a, ar, ar^2, ar^3, \dots$ where $a = 1$.
According to the problem,each term is equal to the sum of all subsequent terms.
For the first term: $a = ar + ar^2 + ar^3 + \dots$
Since $a = 1$,we have $1 = ar + ar^2 + ar^3 + \dots$
This is an infinite geometric series with first term $ar$ and common ratio $r$.
The sum of an infinite geometric series is given by $S = \frac{\text{first term}}{1 - r}$.
So,$1 = \frac{ar}{1 - r}$.
Substituting $a = 1$: $1 = \frac{r}{1 - r}$.
$1 - r = r \implies 2r = 1 \implies r = 1/2$.
The fourth term is $T_4 = ar^3$.
$T_4 = 1 \times (1/2)^3 = 1/8$.

(B) Let the three terms in geometric progression be $\frac{a}{r}, a, ar$.
The product of these terms is $(\frac{a}{r}) \times a \times (ar) = a^3 = 216 = (6)^3$,so $a = 6$.
The sum of the products of these terms taken two at a time is $(\frac{a}{r} \times a) + (a \times ar) + (ar \times \frac{a}{r}) = 156$.
Substituting $a = 6$: $\frac{36}{r} + 36r + 36 = 156$.
$\frac{36}{r} + 36r = 120$.
Dividing by $12$: $\frac{3}{r} + 3r = 10$.
$3r^2 - 10r + 3 = 0$.
$(3r - 1)(r - 3) = 0$.
So,$r = 3$ or $r = \frac{1}{3}$.
For $a = 6$ and $r = 3$,the terms are $\frac{6}{3}, 6, 6 \times 3$,which are $2, 6, 18$.

(A) Let the $n$ terms of the geometric progression be $a, ar, ar^2, \dots, ar^{n-1}$.
$S = a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$.
$P = a \cdot (ar) \cdot (ar^2) \cdot \dots \cdot (ar^{n-1}) = a^n r^{0+1+2+\dots+(n-1)} = a^n r^{\frac{n(n-1)}{2}}$.
$P^2 = a^{2n} r^{n(n-1)}$.
$R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \dots + \frac{1}{ar^{n-1}} = \frac{r^{n-1} + r^{n-2} + \dots + 1}{ar^{n-1}} = \frac{1}{ar^{n-1}} \cdot \frac{r^n - 1}{r - 1}$.
Now,$\frac{S}{R} = \frac{a(r^n - 1)}{r - 1} \cdot \frac{ar^{n-1}(r - 1)}{r^n - 1} = a^2 r^{n-1}$.
Therefore,$(\frac{S}{R})^n = (a^2 r^{n-1})^n = a^{2n} r^{n(n-1)}$.
Comparing this with $P^2$,we get $P^2 = (\frac{S}{R})^n$.

(C) The geometric mean of $n$ terms $a_1, a_2, \dots, a_n$ is given by $(a_1 \times a_2 \times \dots \times a_n)^{1/n}$.
Here,the terms are $3^1, 3^2, 3^3, \dots, 3^n$.
Their product is $P = 3^1 \times 3^2 \times 3^3 \times \dots \times 3^n = 3^{(1+2+3+\dots+n)}$.
The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.
So,$P = 3^{n(n+1)/2}$.
The geometric mean is $P^{1/n} = (3^{n(n+1)/2})^{1/n} = 3^{(n+1)/2}$.

(A) The $n^{th}$ term of the geometric progression is given by $a r^{n-1} = 486$.
Given $r = 3$,we have $a(3)^{n-1} = 486$,which implies $a \cdot 3^n = 3 \times 486 = 1458$ (Equation $i$).
The sum of $n$ terms is $S_n = \frac{a(r^n - 1)}{r - 1} = 728$.
Substituting $r = 3$,we get $\frac{a(3^n - 1)}{3 - 1} = 728$,which simplifies to $a \cdot 3^n - a = 728 \times 2 = 1456$ (Equation $ii$).
Substituting the value of $a \cdot 3^n$ from Equation $i$ into Equation $ii$:
$1458 - a = 1456$.
Therefore,$a = 1458 - 1456 = 2$.

(D) Given that $x, 2x + 2$,and $3x + 3$ are in a geometric progression $(GP)$.
For a $GP$,the square of the middle term is equal to the product of the first and third terms: $(2x + 2)^2 = x(3x + 3)$.
Expanding this,we get $4x^2 + 8x + 4 = 3x^2 + 3x$.
Rearranging the terms gives the quadratic equation $x^2 + 5x + 4 = 0$.
Factoring the equation,we get $(x + 1)(x + 4) = 0$,which gives $x = -1$ or $x = -4$.
If $x = -1$,the terms are $-1, 0, 0$,which cannot form a $GP$.
If $x = -4$,the terms are $-4, -6, -9$. The common ratio $r = \frac{-6}{-4} = 1.5$.
The fourth term $t_4 = a \times r^3 = -4 \times (1.5)^3$.
$t_4 = -4 \times 3.375 = -13.5$.

(A) The given series is a geometric progression where the first term $a = 0.9 = \frac{9}{10}$ and the common ratio $r = \frac{0.09}{0.9} = 0.1 = \frac{1}{10}$.
The sum of the first $n$ terms of a geometric progression is given by $S_n = a \left( \frac{1 - r^n}{1 - r} \right)$.
For $n = 100$,we have:
$S_{100} = \frac{9}{10} \left( \frac{1 - (\frac{1}{10})^{100}}{1 - \frac{1}{10}} \right)$
$S_{100} = \frac{9}{10} \left( \frac{1 - (\frac{1}{10})^{100}}{\frac{9}{10}} \right)$
$S_{100} = 1 - \left( \frac{1}{10} \right)^{100}$.

(C) Given $y = x^{1/3} \cdot x^{1/9} \cdot x^{1/27} \cdot \dots \infty$.
Using the property of exponents $a^m \cdot a^n = a^{m+n}$,we get:
$y = x^{(1/3 + 1/9 + 1/27 + \dots \infty)}$.
The exponent is an infinite geometric series with first term $a = 1/3$ and common ratio $r = 1/3$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
$S = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = 1/2$.
Therefore,$y = x^{1/2}$.

(C) Since $x, y, z$ are in geometric progression,we have $y^2 = xz$.
Let $a^x = b^y = c^z = \lambda$.
Taking log on both sides,we get $x \log a = y \log b = z \log c = \log \lambda$.
Thus,$x = \frac{\log \lambda}{\log a}$,$y = \frac{\log \lambda}{\log b}$,and $z = \frac{\log \lambda}{\log c}$.
Substituting these values into $y^2 = xz$,we get:
$(\frac{\log \lambda}{\log b})^2 = (\frac{\log \lambda}{\log a}) \times (\frac{\log \lambda}{\log c})$.
This simplifies to $(\log b)^2 = \log a \cdot \log c$.
Dividing both sides by $\log a \cdot \log b$,we get $\frac{\log b}{\log a} = \frac{\log c}{\log b}$,which implies $\log_a b = \log_b c$.
Taking the reciprocal,we get $\frac{1}{\log_a b} = \frac{1}{\log_b c}$,which is $\log_b a = \log_c b$.

(A) Let the three numbers be $a, b,$ and $c$ in a geometric progression.
$\therefore b^2 = ac$
Taking the logarithm on both sides,we get:
$\log(b^2) = \log(ac)$
$2 \log b = \log a + \log c$
$\log b = \frac{\log a + \log c}{2}$
This condition shows that the logarithms $\log a, \log b,$ and $\log c$ satisfy the property of an arithmetic progression.

(C) The given inequality is $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \leq 0$.
This can be rewritten as $(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \leq 0$.
This simplifies to $(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \leq 0$.
Since the sum of squares of real numbers is non-negative,the only way for the sum to be $\leq 0$ is if each term is exactly $0$.
Thus,$ap - b = 0$,$bp - c = 0$,and $cp - d = 0$.
This implies $p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.
This shows that $a, b, c, d$ are in geometric progression.

(C) Given that $25, x - 6$,and $x - 12$ are in a geometric progression $(GP)$.
For three terms $a, b, c$ to be in $GP$,the condition is $b^2 = ac$.
Substituting the values: $(x - 6)^2 = 25(x - 12)$.
Expanding the equation: $x^2 - 12x + 36 = 25x - 300$.
Rearranging into a quadratic equation: $x^2 - 37x + 336 = 0$.
Factoring the quadratic: $(x - 16)(x - 21) = 0$.
Thus,$x = 16$ or $x = 21$.
Comparing with the given options,the correct value is $x = 16$.

(D) Given the infinite geometric series: $y = x - x^2 + x^3 - x^4 + \dots \infty$
This is a geometric series with first term $a = x$ and common ratio $r = -x$.
The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$.
Substituting the values: $y = \frac{x}{1 - (-x)} = \frac{x}{1 + x}$.
Now,solve for $x$:
$y(1 + x) = x$
$y + xy = x$
$y = x - xy$
$y = x(1 - y)$
$x = \frac{y}{1 - y}$

(B) Let the first term be $a = 1$ and the common ratio be $r$.
The $n^{th}$ term of a $GP$ is given by $a_n = a \cdot r^{n-1}$.
The third term is $a_3 = 1 \cdot r^{3-1} = r^2$.
The fifth term is $a_5 = 1 \cdot r^{5-1} = r^4$.
Given that $a_3 + a_5 = 90$,we have $r^2 + r^4 = 90$.
Rearranging the equation: $r^4 + r^2 - 90 = 0$.
Let $x = r^2$,then $x^2 + x - 90 = 0$.
Factoring the quadratic equation: $(x + 10)(x - 9) = 0$.
This gives $x = -10$ or $x = 9$.
Since $x = r^2$,$r^2 = -10$ (not possible for real $r$) or $r^2 = 9$.
Therefore,$r = \pm 3$.

(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r} = 3$.
This implies $a = 3(1-r)$.
The squares of the terms form a new geometric series: $a^2, a^2r^2, a^2r^4, \dots$
The sum of this series is $S' = \frac{a^2}{1-r^2} = 3$.
Substituting $a = 3(1-r)$ into the second equation:
$\frac{[3(1-r)]^2}{1-r^2} = 3$
$\frac{9(1-r)^2}{(1-r)(1+r)} = 3$
$\frac{3(1-r)}{1+r} = 1$
$3 - 3r = 1 + r$
$4r = 2 \implies r = \frac{1}{2}$.
Now,$a = 3(1 - \frac{1}{2}) = 3(\frac{1}{2}) = \frac{3}{2}$.
Thus,the first term is $\frac{3}{2}$ and the common ratio is $\frac{1}{2}$.

(B) Let the two geometric means be $a$ and $b$ such that $1, a, b, 64$ are in a Geometric Progression $(GP)$.
The common ratio $r$ is given by $a = 1 \times r$ and $b = 1 \times r^2$,with $64 = 1 \times r^3$.
From $r^3 = 64$,we get $r = \sqrt[3]{64} = 4$.
Thus,$a = 1 \times 4 = 4$ and $b = 1 \times 4^2 = 16$.
Therefore,the two geometric means are $4$ and $16$.

(A) Let the geometric progression be defined by the $k^{th}$ term $T_k = ar^{k-1}$.
Given $T_{m+n} = ar^{m+n-1} = 9$ and $T_{m-n} = ar^{m-n-1} = 4$.
We know that for a geometric progression,the $m^{th}$ term is the geometric mean of the $(m+n)^{th}$ and $(m-n)^{th}$ terms.
$T_m = \sqrt{T_{m+n} \times T_{m-n}}$
$T_m = \sqrt{9 \times 4}$
$T_m = \sqrt{36}$
$T_m = 6$.

(B) Let the first term be $a$ and the common ratio be $r$.
Given that the $(p + q)^{th}$ term is $m = a r^{p + q - 1}$ and the $(p - q)^{th}$ term is $n = a r^{p - q - 1}$.
Dividing the two equations: $\frac{m}{n} = \frac{a r^{p + q - 1}}{a r^{p - q - 1}} = r^{(p + q - 1) - (p - q - 1)} = r^{2q}$.
Thus,$r^{2q} = \frac{m}{n}$,which implies $r = (\frac{m}{n})^{1/(2q)}$.
The $p^{th}$ term is $T_p = a r^{p - 1}$.
Note that $m \times n = (a r^{p + q - 1}) \times (a r^{p - q - 1}) = a^2 r^{2p - 2} = (a r^{p - 1})^2$.
Therefore,$(T_p)^2 = mn$,which gives $T_p = \sqrt{mn}$.

(B) Let $b = ar, c = ar^2, d = ar^3$.
Then,$(a^3 + b^3)^{-1} = \frac{1}{a^3(1 + r^3)}$.
$(b^3 + c^3)^{-1} = \frac{1}{a^3r^3(1 + r^3)}$.
$(c^3 + d^3)^{-1} = \frac{1}{a^3r^6(1 + r^3)}$.
Let $T_1 = \frac{1}{a^3(1 + r^3)}$,$T_2 = \frac{1}{a^3r^3(1 + r^3)}$,and $T_3 = \frac{1}{a^3r^6(1 + r^3)}$.
We observe that $\frac{T_2}{T_1} = \frac{1}{r^3}$ and $\frac{T_3}{T_2} = \frac{1}{r^3}$.
Since the ratio is constant,the terms are in a geometric progression.

(D) The sum of an infinite geometric series is given by $S = \frac{a}{1 - r} = 4$,where $|r| < 1$.
The second term of a geometric series is $ar = 3/4$.
From the second equation,$a = \frac{3}{4r}$.
Substituting this into the sum formula: $\frac{3/4r}{1 - r} = 4$.
$\frac{3}{4r(1 - r)} = 4$
$3 = 16r(1 - r)$
$3 = 16r - 16r^2$
$16r^2 - 16r + 3 = 0$.
Factoring the quadratic equation: $16r^2 - 12r - 4r + 3 = 0$
$4r(4r - 3) - 1(4r - 3) = 0$
$(4r - 1)(4r - 3) = 0$.
So,$r = 1/4$ or $r = 3/4$.
If $r = 1/4$,then $a = \frac{3}{4(1/4)} = 3$.
If $r = 3/4$,then $a = \frac{3}{4(3/4)} = 1$.
Checking the options,the pair $(a = 3, r = 1/4)$ matches option $D$.

(C) Let the first term be $a$ and the common ratio be $r$,where $|r| < 1$.
The infinite geometric series is $a, ar, ar^2, ar^3, \dots$.
The sum of all terms subsequent to the first term is $S = ar + ar^2 + ar^3 + \dots = \frac{ar}{1-r}$.
According to the problem,the first term is twice the sum of the subsequent terms:
$a = 2 \times \left( \frac{ar}{1-r} \right)$.
Dividing both sides by $a$ (assuming $a \neq 0$):
$1 = \frac{2r}{1-r}$.
$1 - r = 2r$.
$1 = 3r$.
$r = 1/3$.

(B) The arithmetic mean of $b$ and $c$ is $a$,so $a = \frac{b+c}{2}$,which implies $b+c = 2a$.
Given $g_1$ and $g_2$ are two geometric means between $b$ and $c$,the sequence is $b, g_1, g_2, c$ in $G$.$P$.
Let the common ratio be $r$. Then $c = b r^3$,so $r^3 = \frac{c}{b}$.
$g_1 = br$ and $g_2 = br^2$.
Now,$g_1^3 + g_2^3 = (br)^3 + (br^2)^3 = b^3 r^3 + b^3 r^6$.
Substituting $r^3 = \frac{c}{b}$:
$g_1^3 + g_2^3 = b^3 \left( \frac{c}{b} \right) + b^3 \left( \frac{c}{b} \right)^2 = b^2 c + b^3 \left( \frac{c^2}{b^2} \right) = b^2 c + bc^2 = bc(b+c)$.
Since $b+c = 2a$,we have $g_1^3 + g_2^3 = bc(2a) = 2abc$.
Comparing this with $kabc$,we get $k = 2$.

(A) Given that $a, b, c$ are in geometric progression $(GP)$.
Therefore,$\frac{b}{a} = \frac{c}{b} = r$,where $r$ is the common ratio.
Squaring the terms,we get $\frac{b^2}{a^2} = \frac{c^2}{b^2} = r^2$.
This implies that $a^2, b^2, c^2$ are also in geometric progression with a common ratio of $r^2$.
Thus,option $A$ is correct.

(A) The first $n$ terms of the geometric progression are $a, ar, ar^2, \dots, ar^{n-1}$.
The arithmetic mean $A$ is given by:
$A = \frac{a + ar + ar^2 + \dots + ar^{n-1}}{n} = \frac{a(r^n - 1)}{n(r - 1)}$.
The harmonic mean $H$ is given by:
$H = \frac{n}{\frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}}} = \frac{n}{\frac{1}{a} \left( \frac{1 - (1/r)^n}{1 - 1/r} \right)} = \frac{n}{\frac{1}{a} \left( \frac{r^n - 1}{r^n} \cdot \frac{r}{r - 1} \right)} = \frac{n a r^{n-1} (r - 1)}{r^n - 1}$.
Multiplying $A$ and $H$:
$A \cdot H = \left( \frac{a(r^n - 1)}{n(r - 1)} \right) \cdot \left( \frac{n a r^{n-1} (r - 1)}{r^n - 1} \right) = a^2 r^{n-1}$.

(A) Let the first term be $a$ and the common ratio be $r$ for the geometric progression.
Given: $a_{10} = ar^9 = 9$ and $a_4 = ar^3 = 4$.
Dividing the two equations: $\frac{ar^9}{ar^3} = \frac{9}{4} \implies r^6 = \frac{9}{4}$.
The $7^{th}$ term is $a_7 = ar^6$.
Since $ar^3 = 4$,we have $a = \frac{4}{r^3}$.
From $r^6 = \frac{9}{4}$,we get $r^3 = \sqrt{\frac{9}{4}} = \frac{3}{2}$ (assuming $r > 0$).
Thus,$a = \frac{4}{3/2} = \frac{8}{3}$.
Therefore,$a_7 = ar^6 = \left(\frac{8}{3}\right) \times \left(\frac{9}{4}\right) = 2 \times 3 = 6$.

(B) Let the first term be $a$ and the common ratio be $r$.
$S_1 = a + ar + ar^2 + \dots + ar^9 = a\frac{r^{10}-1}{r-1}$.
$S_2 = ar^{10} + ar^{11} + \dots + ar^{19} = ar^{10}(1 + r + r^2 + \dots + r^9) = ar^{10}\frac{r^{10}-1}{r-1}$.
Dividing $S_2$ by $S_1$,we get:
$\frac{S_2}{S_1} = \frac{ar^{10}(\frac{r^{10}-1}{r-1})}{a(\frac{r^{10}-1}{r-1})} = r^{10}$.
Therefore,$r^{10} = \frac{S_2}{S_1}$,which implies $r = \pm \sqrt[10]{\frac{S_2}{S_1}}$.

(A) Let the three terms of the $GP$ be $\frac{a}{r}, a, ar$.
The product of these terms is $\frac{a}{r} \times a \times ar = a^3 = 216$.
Thus,$a = \sqrt[3]{216} = 6$.
The sum of the three terms is $\frac{6}{r} + 6 + 6r = 19$.
Subtracting $6$ from both sides,we get $\frac{6}{r} + 6r = 13$.
Multiplying by $r$,we get $6 + 6r^2 = 13r$,which simplifies to the quadratic equation $6r^2 - 13r + 6 = 0$.
Factoring the quadratic equation: $6r^2 - 9r - 4r + 6 = 0 \implies 3r(2r - 3) - 2(2r - 3) = 0$.
This gives $(3r - 2)(2r - 3) = 0$.
Therefore,$r = \frac{2}{3}$ or $r = \frac{3}{2}$.

(B) Let the first term of the geometric progression be $A$ and the common ratio be $R$.
Given that the $p^{th}$,$q^{th}$,and $r^{th}$ terms are $a, b, c$ respectively,we have:
$a = AR^{p-1}$,$b = AR^{q-1}$,and $c = AR^{r-1}$.
Now,consider the expression $a^{q-r} \cdot b^{r-p} \cdot c^{p-q}$:
$= (AR^{p-1})^{q-r} \cdot (AR^{q-1})^{r-p} \cdot (AR^{r-1})^{p-q}$
$= A^{(q-r) + (r-p) + (p-q)} \cdot R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$
$= A^0 \cdot R^{(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)}$
$= 1 \cdot R^0 = 1$.

(B) Given that $p, q, r$ are in a geometric progression,so $q^2 = pr$.
Also,$a, b, c$ are in a geometric progression,so $b^2 = ac$.
We need to check if $cp, bq, ar$ are in a geometric progression.
For $cp, bq, ar$ to be in a geometric progression,the condition $(bq)^2 = (cp)(ar)$ must hold.
Since $b^2 = ac$ and $q^2 = pr$,we have $b^2q^2 = (ac)(pr) = (cp)(ar)$.
Thus,$(bq)^2 = (cp)(ar)$,which confirms that $cp, bq, ar$ are in a geometric progression.

(B) Let the three terms be $\frac{b}{r}, b, br$ where $r$ is the common ratio.
Given that the harmonic mean of $a$ and $b$ is $12$:
$\frac{2ab}{a+b} = 12 \implies \frac{2(\frac{b}{r})b}{\frac{b}{r}+b} = 12 \implies \frac{2b}{1+r} = 12 \implies b = 6(1+r) \dots (1)$
Given that the harmonic mean of $b$ and $c$ is $36$:
$\frac{2bc}{b+c} = 36 \implies \frac{2b(br)}{b+br} = 36 \implies \frac{2br}{1+r} = 36 \implies br = 18(1+r) \dots (2)$
Dividing $(2)$ by $(1)$:
$\frac{br}{b} = \frac{18(1+r)}{6(1+r)} \implies r = 3$
Substituting $r=3$ into $(1)$:
$b = 6(1+3) = 24$
Since $a = \frac{b}{r} = \frac{24}{3} = 8$.

(A) The geometric mean of $n$ terms $a_1, a_2, \dots, a_n$ is given by $(a_1 \times a_2 \times \dots \times a_n)^{\frac{1}{n}}$.
For the sequence $7, 7^2, 7^3, \dots, 7^n$,the geometric mean is $(7^1 \times 7^2 \times 7^3 \times \dots \times 7^n)^{\frac{1}{n}}$.
Using the property of exponents,this becomes $(7^{1+2+3+\dots+n})^{\frac{1}{n}}$.
The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.
Thus,the expression becomes $(7^{\frac{n(n+1)}{2}})^{\frac{1}{n}}$.
Simplifying the exponents,we get $7^{\frac{n(n+1)}{2n}} = 7^{\frac{n+1}{2}}$.

(A) Given the infinite geometric series:
$x = \frac{a}{1 - 1/r} = \frac{ar}{r - 1}$
$y = \frac{b}{1 - (-1/r)} = \frac{b}{1 + 1/r} = \frac{br}{r + 1}$
$z = \frac{c}{1 - 1/r^2} = \frac{cr^2}{r^2 - 1}$
Now,calculate the product $xy$:
$xy = \left( \frac{ar}{r - 1} \right) \left( \frac{br}{r + 1} \right) = \frac{abr^2}{r^2 - 1}$
Now,divide by $z$:
$\frac{xy}{z} = \frac{abr^2}{r^2 - 1} \div \frac{cr^2}{r^2 - 1} = \frac{abr^2}{r^2 - 1} \times \frac{r^2 - 1}{cr^2} = \frac{ab}{c}$

(A) Let the three numbers in $GP$ be $\frac{a}{r}, a, ar$.
The product of these numbers is $\left( \frac{a}{r} \right) \times a \times (ar) = a^3 = 1728$.
Since $1728 = 12^3$,we have $a = 12$.
The sum of the numbers is $\frac{a}{r} + a + ar = 38$.
Substituting $a = 12$,we get $\frac{12}{r} + 12 + 12r = 38$.
$\frac{12}{r} + 12r = 26$.
Dividing by $2$,we get $\frac{6}{r} + 6r = 13$,which simplifies to $6r^2 - 13r + 6 = 0$.
Factoring the quadratic equation: $6r^2 - 9r - 4r + 6 = 0 \implies 3r(2r - 3) - 2(2r - 3) = 0$.
$(3r - 2)(2r - 3) = 0$,so $r = \frac{2}{3}$ or $r = \frac{3}{2}$.
If $r = \frac{3}{2}$,the numbers are $\frac{12}{3/2}, 12, 12(\frac{3}{2}) \implies 8, 12, 18$.
The largest number is $18$.

(A) Given the geometric progression $8, 12, 18, 27, \dots$
Here,the first term $a = 8$ and the common ratio $r = \frac{12}{8} = \frac{3}{2}$.
The $n^{th}$ term of a geometric progression is given by $T_n = a \cdot r^{n-1}$.
For the $9^{th}$ term $(n = 9)$:
$T_9 = 8 \cdot (\frac{3}{2})^{9-1}$
$T_9 = 8 \cdot (\frac{3}{2})^8$
$T_9 = 8 \cdot \frac{6561}{256}$
$T_9 = \frac{6561}{32}$

(C) Let the $n$ geometric means between $a$ and $b$ be $G_1, G_2, \dots, G_n$.
Then $a, G_1, G_2, \dots, G_n, b$ form a Geometric Progression $(GP)$.
Here,the total number of terms is $n+2$.
The last term is $b = a \cdot r^{n+1}$,where $r$ is the common ratio.
Thus,$r^{n+1} = \frac{b}{a}$,so $r = (\frac{b}{a})^{\frac{1}{n+1}}$.
The product of the $n$ geometric means is $P = G_1 \cdot G_2 \cdot \dots \cdot G_n$.
Since $G_k = a \cdot r^k$,we have $P = (a \cdot r^1) \cdot (a \cdot r^2) \dots (a \cdot r^n) = a^n \cdot r^{1+2+\dots+n} = a^n \cdot r^{\frac{n(n+1)}{2}}$.
Substituting $r = (\frac{b}{a})^{\frac{1}{n+1}}$,we get $P = a^n \cdot (\frac{b}{a})^{\frac{n(n+1)}{2(n+1)}} = a^n \cdot (\frac{b}{a})^{\frac{n}{2}} = a^n \cdot \frac{b^{n/2}}{a^{n/2}} = a^{n/2} \cdot b^{n/2} = (ab)^{n/2}$.

(B) Given that $x, y, z$ are in geometric progression,we have $y^2 = xz$.
Taking the natural logarithm on both sides,we get $\ln(y^2) = \ln(xz)$,which simplifies to $2 \ln y = \ln x + \ln z$.
This implies that $\ln x, \ln y, \ln z$ are in arithmetic progression.
Adding $1$ to each term,we get $1 + \ln x, 1 + \ln y, 1 + \ln z$,which are also in arithmetic progression.
Since the reciprocals of terms in an arithmetic progression form a harmonic progression,the terms $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$ are in harmonic progression.

(D) Let the geometric progression be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.
Given that each term is equal to the sum of the next two terms,we have:
$a_n = a_{n+1} + a_{n+2}$
$ar^{n-1} = ar^n + ar^{n+1}$
Dividing by $ar^{n-1}$ (since $a, r \neq 0$):
$1 = r + r^2$
$r^2 + r - 1 = 0$
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$
Since the terms are positive,the common ratio $r$ must be positive.
Therefore,$r = \frac{\sqrt{5} - 1}{2}$.

(C) Let the geometric progression be $a, ar, ar^2, \dots, ar^{49}$.
Then $a_n = ar^{n-1}$.
The numerator is $a_1 - a_3 + a_5 - \dots + a_{49} = a - ar^2 + ar^4 - \dots + ar^{48}$.
This is a geometric series with first term $A = a$,common ratio $R = -r^2$,and $n = 25$ terms.
The sum is $\frac{a(1 - (-r^2)^{25})}{1 - (-r^2)} = \frac{a(1 + r^{50})}{1 + r^2}$.
The denominator is $a_2 - a_4 + a_6 - \dots + a_{50} = ar - ar^3 + ar^5 - \dots + ar^{49}$.
This is a geometric series with first term $A' = ar$,common ratio $R = -r^2$,and $n = 25$ terms.
The sum is $\frac{ar(1 - (-r^2)^{25})}{1 - (-r^2)} = \frac{ar(1 + r^{50})}{1 + r^2}$.
Dividing the numerator by the denominator,we get $\frac{a(1 + r^{50}) / (1 + r^2)}{ar(1 + r^{50}) / (1 + r^2)} = \frac{a}{ar} = \frac{1}{r} = \frac{a_1}{a_2}$.