Geometric progression Questions in English

(B) If $a, b, c$ are in $G.P.$,then $b^2 = ac$.
Consider the expression $a(b^2 + c^2) = ab^2 + ac^2$.
Since $b^2 = ac$,we can substitute $b^2$ with $ac$:
$a(ac) + ac^2 = a^2c + ac^2 = c(a^2 + ac)$.
Alternatively,starting from $b^2 = ac$:
$b^2(a - c) = ac(a - c)$
$b^2a - b^2c = a^2c - ac^2$
$b^2a + ac^2 = a^2c + b^2c$
$a(b^2 + c^2) = c(a^2 + b^2)$.
Verification: Let $a = 1, b = 2, c = 4$.
$LHS = a(b^2 + c^2) = 1(4 + 16) = 20$.
$RHS = c(a^2 + b^2) = 4(1 + 4) = 4(5) = 20$.
Since $LHS = RHS$,option $B$ is correct.

(D) The given sequence is $\sqrt{2}, \sqrt{10}, 5\sqrt{2}, \dots$
This is a Geometric Progression $(GP)$ where the first term $a = \sqrt{2}$.
The common ratio $r = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5}$.
The $n^{th}$ term of a $GP$ is given by $t_n = a \cdot r^{n-1}$.
For the $7^{th}$ term $(n=7)$:
$t_7 = \sqrt{2} \cdot (\sqrt{5})^{7-1}$
$t_7 = \sqrt{2} \cdot (\sqrt{5})^6$
$t_7 = \sqrt{2} \cdot (5)^3$
$t_7 = 125\sqrt{2}$.

(C) Let the first term of the $G.P.$ be $A$ and the common ratio be $r$.
We know that the $n^{th}$ term of a $G.P.$ is given by $T_n = Ar^{n-1}$.
Given that $T_4 = a$,$T_7 = b$,and $T_{10} = c$,we have:
$a = Ar^3$
$b = Ar^6$
$c = Ar^9$
Now,consider the product $ac$:
$ac = (Ar^3)(Ar^9) = A^2r^{12}$
Also,consider $b^2$:
$b^2 = (Ar^6)^2 = A^2r^{12}$
Since $b^2 = A^2r^{12}$ and $ac = A^2r^{12}$,it follows that $b^2 = ac$.
Alternatively,if the indices $p, q, r$ are in $A.P.$,then the $p^{th}, q^{th}, r^{th}$ terms of a $G.P.$ are in $G.P.$. Here,$4, 7, 10$ are in $A.P.$ with a common difference of $3$,therefore $a, b, c$ are in $G.P.$,which implies $b^2 = ac$.

(B) Given that the first term $a = 5$ and the common ratio $r = -5$.
Let the $n^{th}$ term be $3125$.
The formula for the $n^{th}$ term of a $G.P.$ is $a_n = a r^{n-1}$.
Substituting the given values: $5(-5)^{n-1} = 3125$.
Dividing both sides by $5$: $(-5)^{n-1} = 625$.
Since $625 = (-5)^4$,we have $(-5)^{n-1} = (-5)^4$.
Comparing the exponents,$n - 1 = 4$,which gives $n = 5$.
Thus,the $5^{th}$ term is $3125$.

(B) Let the number to be added be $x$.
Then,the numbers $x + 2, x + 14, x + 62$ are in $G.P.$
For three numbers $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.
Therefore,$(x + 14)^2 = (x + 2)(x + 62)$.
Expanding both sides: $x^2 + 28x + 196 = x^2 + 64x + 124$.
Subtracting $x^2$ from both sides: $28x + 196 = 64x + 124$.
Rearranging the terms: $196 - 124 = 64x - 28x$.
$72 = 36x$.
$x = 2$.
Thus,adding $2$ to the given numbers results in $4, 16, 64$,which are in $G.P.$ as $16^2 = 4 \times 64$ $(256 = 256)$.

(B) Given that the $(p + q)^{th}$ term $T_{p+q} = ar^{p+q-1} = m$ and the $(p - q)^{th}$ term $T_{p-q} = ar^{p-q-1} = n$.
Dividing the two equations:
$\frac{T_{p+q}}{T_{p-q}} = \frac{ar^{p+q-1}}{ar^{p-q-1}} = r^{(p+q-1) - (p-q-1)} = r^{2q} = \frac{m}{n}$.
Thus,$r^{2q} = \frac{m}{n} \Rightarrow r = (\frac{m}{n})^{1/(2q)}$.
The $p^{th}$ term is $T_p = ar^{p-1}$.
Since $T_{p+q} = ar^{p-1} \cdot r^q = m$,we have $T_p \cdot r^q = m \Rightarrow T_p = \frac{m}{r^q}$.
Since $T_{p-q} = ar^{p-1} \cdot r^{-q} = n$,we have $T_p \cdot r^{-q} = n \Rightarrow T_p = n \cdot r^q$.
Multiplying these two expressions for $T_p$:
$T_p^2 = (\frac{m}{r^q}) \cdot (n \cdot r^q) = mn$.
Therefore,$T_p = \sqrt{mn}$.
Alternatively: In a $G.P.$,the $p^{th}$ term is the geometric mean of the terms equidistant from it. Since $(p+q)$ and $(p-q)$ are equidistant from $p$,the $p^{th}$ term is $\sqrt{T_{p+q} \cdot T_{p-q}} = \sqrt{mn}$.

(A) Let the terms of the $G.P.$ be $a, ar, ar^2, ar^3, \dots$ where $a > 0$ and $r > 0$.
Given that each term is equal to the sum of the two terms that follow it,we have:
$T_n = T_{n+1} + T_{n+2}$
$ar^{n-1} = ar^n + ar^{n+1}$
Dividing both sides by $ar^{n-1}$ (since $a \neq 0$ and $r \neq 0$):
$1 = r + r^2$
$r^2 + r - 1 = 0$
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,$r = \frac{\sqrt{5} - 1}{2}$.

(D) Given that $x, 2x + 2, 3x + 3$ are in $G.P.$
Therefore,$(2x + 2)^2 = x(3x + 3)$
$4(x + 1)^2 = 3x(x + 1)$
$4(x + 1)^2 - 3x(x + 1) = 0$
$(x + 1)(4x + 4 - 3x) = 0$
$(x + 1)(x + 4) = 0$
So,$x = -1$ or $x = -4$.
If $x = -1$,the terms are $-1, 0, 0$,which is not a valid $G.P.$ (as common ratio is undefined).
If $x = -4$,the terms are $-4, -6, -9$.
Here,the first term $a = -4$ and the common ratio $r = \frac{-6}{-4} = 1.5$.
The fourth term $T_4 = ar^3 = (-4)(1.5)^3 = (-4)(3.375) = -13.5$.

(A) Given the sum of the first $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.
We are given $\frac{S_3}{S_6} = \frac{125}{152}$.
Substituting the formula: $\frac{a(r^3 - 1)/(r - 1)}{a(r^6 - 1)/(r - 1)} = \frac{125}{152}$.
This simplifies to $\frac{r^3 - 1}{r^6 - 1} = \frac{125}{152}$.
Since $r^6 - 1 = (r^3 - 1)(r^3 + 1)$,we have $\frac{1}{r^3 + 1} = \frac{125}{152}$.
$152 = 125(r^3 + 1) \Rightarrow 152 = 125r^3 + 125$.
$125r^3 = 152 - 125 = 27$.
$r^3 = \frac{27}{125} = (\frac{3}{5})^3$.
Therefore,$r = \frac{3}{5}$.

(B) Given that $x, y, z$ are in $G.P.$,we have $y^2 = xz$.
Let $a^x = b^y = c^z = m$.
Taking logarithm on both sides,we get $x \log a = y \log b = z \log c = \log m$.
This implies $x = \frac{\log m}{\log a} = \log_a m$,$y = \log_b m$,and $z = \log_c m$.
Since $x, y, z$ are in $G.P.$,we have $\frac{y}{x} = \frac{z}{y}$.
Substituting the values,we get $\frac{\log_b m}{\log_a m} = \frac{\log_c m}{\log_b m}$.
Using the change of base formula $\frac{\log_b m}{\log_a m} = \log_b a$,we get $\log_b a = \log_c b$.

(B) Let the first term of the $G.P.$ be $A$ and the common ratio be $R$.
Then,the $p^{th}$,$q^{th}$,and $r^{th}$ terms are given by:
$a = A R^{p-1}$ $(i)$
$b = A R^{q-1}$ $(ii)$
$c = A R^{r-1}$ $(iii)$
Substituting these values into the expression $a^{q - r} b^{r - p} c^{p - q}$:
$= (A R^{p-1})^{q-r} (A R^{q-1})^{r-p} (A R^{r-1})^{p-q}$
$= A^{(q-r) + (r-p) + (p-q)} \times R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$
$= A^0 \times R^{(pq - pr - q + r) + (qr - pq - r + p) + (rp - rq - p + q)}$
$= A^0 \times R^0 = 1 \times 1 = 1$.

(C) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$.
Given that the third term $T_3 = ar^2 = 4$.
We need to find the product of the first $5$ terms:
$P = a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)$
$P = a^5 \times r^{(1+2+3+4)} = a^5 \times r^{10}$
$P = (ar^2)^5$
Substituting the value $ar^2 = 4$:
$P = 4^5$.

(B) Let the first term be $a$ and the common ratio be $r$. The $n^{th}$ term of a $G.P.$ is given by $T_n = ar^{n-1}$.
Given $T_5 = ar^4 = \frac{1}{3}$ $(i)$
Given $T_9 = ar^8 = \frac{16}{243}$ $(ii)$
Dividing $(ii)$ by $(i)$:
$\frac{ar^8}{ar^4} = \frac{16/243}{1/3}$
$r^4 = \frac{16}{243} \times 3 = \frac{16}{81}$
$r^4 = (\frac{2}{3})^4$,so $r = \frac{2}{3}$ (taking the positive root).
Substituting $r$ in $(i)$:
$a(\frac{2}{3})^4 = \frac{1}{3}$
$a(\frac{16}{81}) = \frac{1}{3}$
$a = \frac{1}{3} \times \frac{81}{16} = \frac{27}{16}$
Now,the $4^{th}$ term $T_4 = ar^3 = \frac{27}{16} \times (\frac{2}{3})^3 = \frac{27}{16} \times \frac{8}{27} = \frac{8}{16} = \frac{1}{2}$.

(A) Let the first term of the $G.P.$ be $A$ and the common ratio be $R$.
Then,$a = A R^{p-1}$,$b = A R^{q-1}$,and $c = A R^{r-1}$.
Substituting these into the expression:
$\left( \frac{c}{b} \right)^p \left( \frac{b}{a} \right)^r \left( \frac{a}{c} \right)^q = \left( \frac{A R^{r-1}}{A R^{q-1}} \right)^p \left( \frac{A R^{q-1}}{A R^{p-1}} \right)^r \left( \frac{A R^{p-1}}{A R^{r-1}} \right)^q$
$= (R^{r-q})^p (R^{q-p})^r (R^{p-r})^q$
$= R^{pr - pq + qr - pr + pq - qr}$
$= R^0 = 1$.

(D) The $n^{th}$ term of a $G.P.$ is given by $l = a r^{n-1}$.
Dividing both sides by $a$,we get $\frac{l}{a} = r^{n-1}$.
Taking the logarithm on both sides,we have $\log(\frac{l}{a}) = \log(r^{n-1})$.
Using the properties of logarithms,$\log l - \log a = (n-1) \log r$.
Therefore,$n-1 = \frac{\log l - \log a}{\log r}$.
Adding $1$ to both sides,we get $n = 1 + \frac{\log l - \log a}{\log r}$.

(C) Given that $\log_x a, a^{x/2}, \log_b x$ are in $G.P.$
Therefore,$(a^{x/2})^2 = (\log_x a) \cdot (\log_b x)$.
Using the change of base formula $\log_x a \cdot \log_b x = \frac{\log a}{\log x} \cdot \frac{\log x}{\log b} = \frac{\log a}{\log b} = \log_b a$.
So,$a^x = \log_b a$.
Taking $\log_a$ on both sides,we get $x = \log_a(\log_b a)$.
Using the change of base formula for $\log_b a = \frac{\log_e a}{\log_e b}$,we have $x = \log_a\left(\frac{\log_e a}{\log_e b}\right)$.
By the property of logarithms,$\log_a\left(\frac{M}{N}\right) = \log_a M - \log_a N$.
Thus,$x = \log_a(\log_e a) - \log_a(\log_e b)$.

(A) Let the roots of the equation $ax^3 + bx^2 + cx + d = 0$ be $\frac{A}{R}, A, AR$.
The product of the roots is given by $\frac{-d}{a}$.
So,$(\frac{A}{R}) \cdot A \cdot (AR) = A^3 = -\frac{d}{a}$.
Since $A$ is a root of the equation,it must satisfy $aA^3 + bA^2 + cA + d = 0$.
Substituting $A^3 = -\frac{d}{a}$,we get $a(-\frac{d}{a}) + bA^2 + cA + d = 0$,which simplifies to $bA^2 + cA = 0$.
Since $A \neq 0$ (assuming $d \neq 0$),we have $bA + c = 0$,so $A = -\frac{c}{b}$.
Substituting $A = -\frac{c}{b}$ into $A^3 = -\frac{d}{a}$,we get $(-\frac{c}{b})^3 = -\frac{d}{a}$.
This implies $-\frac{c^3}{b^3} = -\frac{d}{a}$,which simplifies to $c^3a = b^3d$.

(A) Let the first term be $a$ and the common ratio be $r$.
Given that the $10^{th}$ term $a_{10} = ar^9 = 9$ and the $4^{th}$ term $a_4 = ar^3 = 4$.
Dividing the two equations: $\frac{ar^9}{ar^3} = \frac{9}{4} \Rightarrow r^6 = \frac{9}{4}$.
We need to find the $7^{th}$ term,$a_7 = ar^6$.
Since $ar^9 = 9$,we have $a = \frac{9}{r^9}$.
Alternatively,using the property of geometric progression: $a_7 = \sqrt{a_{10} \times a_4} = \sqrt{9 \times 4} = \sqrt{36} = 6$.
Thus,the $7^{th}$ term is $6$.

(B) Given that the $6^{th}$ term $T_6 = 32$ and the $8^{th}$ term $T_8 = 128$.
Using the formula for the $n^{th}$ term of a $G.P.$,$T_n = ar^{n-1}$,we have:
$T_6 = ar^5 = 32$ .....$(i)$
$T_8 = ar^7 = 128$ .....$(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{ar^7}{ar^5} = \frac{128}{32}$
$r^2 = 4$
$r = \pm 2$
Since the options provided include $2$,the common ratio is $2$.

(A) The $n^{th}$ term of a geometric progression is given by $T_n = ar^{n-1}$.
Here,the first term $a = 5$ and the common ratio $r = \frac{-5/2}{5} = -\frac{1}{2}$.
Given $T_n = \frac{5}{1024}$,we have:
$\frac{5}{1024} = 5 \times (-\frac{1}{2})^{n-1}$
Dividing both sides by $5$:
$\frac{1}{1024} = (-\frac{1}{2})^{n-1}$
Since $1024 = 2^{10}$,we can write:
$(-\frac{1}{2})^{10} = (-\frac{1}{2})^{n-1}$
Comparing the exponents:
$10 = n - 1$
$n = 11$.

(C) Let the first term be $a$ and the common ratio be $r$.
Given that the third term is the square of the first term: $ar^2 = a^2$.
Since $a \neq 0$,we have $a = r^2$.
Given the second term is $8$: $ar = 8$.
Substituting $a = r^2$ into the equation $ar = 8$,we get $(r^2)r = 8$,which implies $r^3 = 8$,so $r = 2$.
Then $a = r^2 = 2^2 = 4$.
The $6^{th}$ term is given by $T_6 = ar^5$.
$T_6 = 4 \times 2^5 = 4 \times 32 = 128$.

(B) Let the $9$ terms of a $G.P.$ be $\frac{a}{r^4}, \frac{a}{r^3}, \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2, ar^3, ar^4$.
The fifth term is $a = 2$.
The product of these $9$ terms is $P = \left(\frac{a}{r^4}\right) \times \left(\frac{a}{r^3}\right) \times \left(\frac{a}{r^2}\right) \times \left(\frac{a}{r}\right) \times a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)$.
Simplifying the product,we get $P = a^9$.
Since $a = 2$,the product is $P = 2^9 = 512$.

(D) Let the infinite $G.P.$ be $a, ar, ar^2, \dots, \infty$.
The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 9$,where $|r| < 1$.
$\Rightarrow a = 9(1-r) \dots (i)$
The sum of the first two terms is $a + ar = 5$.
$\Rightarrow a(1+r) = 5 \dots (ii)$
Substituting $(i)$ into $(ii)$:
$9(1-r)(1+r) = 5$
$9(1-r^2) = 5$
$1-r^2 = \frac{5}{9}$
$r^2 = 1 - \frac{5}{9} = \frac{4}{9}$
$r = \pm \frac{2}{3}$.
Since the sum of the infinite series exists,$|r| < 1$,both values are valid. Given the options,the correct choice is $2/3$.

(A) The given series is $3 + \frac{9}{2} + \frac{27}{4} + \dots$
This is a Geometric Progression $(G.P.)$ where the first term $a = 3$ and the common ratio $r = \frac{9/2}{3} = \frac{3}{2}$.
The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
For $n = 5$,$S_5 = \frac{3((\frac{3}{2})^5 - 1)}{\frac{3}{2} - 1}$.
$S_5 = \frac{3(\frac{243}{32} - 1)}{\frac{1}{2}} = 6 \times \frac{243 - 32}{32} = 6 \times \frac{211}{32} = 3 \times \frac{211}{16} = \frac{633}{16}$.
Converting to a mixed fraction,$\frac{633}{16} = 39\frac{9}{16}$.

(A) The given series is a Geometric Progression $(G.P.)$ 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 $G.P.$ is given by $S_n = a \frac{1 - r^n}{1 - r}$.
For $n = 100$,we have:
$S_{100} = \frac{9}{10} \left( \frac{1 - (\frac{1}{10})^{100}}{1 - \frac{1}{10}} \right)$
$S_{100} = \frac{9}{10} \left( \frac{1 - (\frac{1}{10})^{100}}{\frac{9}{10}} \right)$
$S_{100} = 1 - \left( \frac{1}{10} \right)^{100}$.

(B) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given that the product of the terms is $216$:
$\frac{a}{r} \times a \times ar = 216$
$a^3 = 216 \Rightarrow a = 6$.
Given that the sum of the terms is $19$:
$\frac{6}{r} + 6 + 6r = 19$
$\frac{6}{r} + 6r = 13$
Multiply by $r$:
$6 + 6r^2 = 13r$
$6r^2 - 13r + 6 = 0$
$6r^2 - 9r - 4r + 6 = 0$
$3r(2r - 3) - 2(2r - 3) = 0$
$(3r - 2)(2r - 3) = 0$
Thus,$r = \frac{2}{3}$ or $r = \frac{3}{2}$.

(D) Let the first term be $a$ and the common ratio be $r$ for the $G.P.$ with positive terms.
According to the given condition,every term is the sum of its two previous terms:
$T_n = T_{n-1} + T_{n-2}$
Substituting the general term formula $T_n = ar^{n-1}$:
$ar^{n-1} = ar^{n-2} + ar^{n-3}$
Dividing both sides by $ar^{n-3}$ (since $a > 0$ and $r > 0$):
$r^2 = r + 1$
$r^2 - r - 1 = 0$
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. Thus,we take the positive root:
$r = \frac{1 + \sqrt{5}}{2}$

(B) Let the first term be $a$ and the common ratio be $r$.
Given that every term is twice its previous term,the common ratio $r = 2$.
The sum of the first two terms is $a + ar = 1$.
Substituting $r = 2$ into the equation:
$a + a(2) = 1$
$3a = 1$
$a = \frac{1}{3}$.

(A) Given that the sum of $n$ terms $S_n = \frac{a(r^n - 1)}{r - 1} = 255$ (since $r > 1$) .....$(i)$
The $n^{th}$ term $a_n = ar^{n-1} = 128$ .....$(ii)$
Common ratio $r = 2$ .....$(iii)$
From $(ii)$ and $(iii)$,we have $a(2^{n-1}) = 128$ .....$(iv)$
From $(i)$ and $(iii)$,we have $\frac{a(2^n - 1)}{2 - 1} = 255 \Rightarrow a(2^n - 1) = 255$ .....$(v)$
Dividing $(v)$ by $(iv)$:
$\frac{a(2^n - 1)}{a(2^{n-1})} = \frac{255}{128}$
$\frac{2^n - 1}{2^{n-1}} = \frac{255}{128}$
$\frac{2^n}{2^{n-1}} - \frac{1}{2^{n-1}} = \frac{255}{128}$
$2 - \frac{1}{2^{n-1}} = \frac{255}{128}$
$\frac{1}{2^{n-1}} = 2 - \frac{255}{128} = \frac{256 - 255}{128} = \frac{1}{128}$
$2^{n-1} = 128 = 2^7$
$n - 1 = 7 \Rightarrow n = 8$
Substituting $n = 8$ into $(iv)$:
$a(2^{8-1}) = 128$
$a(2^7) = 128$
$a(128) = 128$
$a = 1$.

(B) Let the first term be $a$ and the common ratio be $r$. The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Given that $S_6 = 9 \times S_3$,we have:
$\frac{a(r^6 - 1)}{r - 1} = 9 \times \frac{a(r^3 - 1)}{r - 1}$
Assuming $r \neq 1$,we can simplify this to:
$r^6 - 1 = 9(r^3 - 1)$
$(r^3)^2 - 1 = 9(r^3 - 1)$
Using the identity $x^2 - y^2 = (x - y)(x + y)$,we get:
$(r^3 - 1)(r^3 + 1) = 9(r^3 - 1)$
Since $r \neq 1$,$r^3 - 1 \neq 0$,so we can divide both sides by $(r^3 - 1)$:
$r^3 + 1 = 9$
$r^3 = 8$
$r = 2$
Thus,the common ratio is $2$.

(C) The number $S$ consists of $91$ ones,which can be written as a geometric series:
$S = 1 + 10 + 10^2 + \dots + 10^{90} = \frac{10^{91} - 1}{10 - 1} = \frac{10^{91} - 1}{9}$.
Since $91 = 7 \times 13$,we can use the identity $x^n - 1 = (x^k - 1)(x^{n-k} + x^{n-2k} + \dots + 1)$ where $k$ is a divisor of $n$.
Let $x = 10^{13}$,then $10^{91} - 1 = (10^{13})^7 - 1 = (10^{13} - 1)((10^{13})^6 + (10^{13})^5 + \dots + 1)$.
Thus,$S = \frac{10^{13} - 1}{9} \times ((10^{13})^6 + (10^{13})^5 + \dots + 1)$.
Since $S$ is the product of two integers greater than $1$,it is a composite number and therefore not prime.

(B) The given sequence is a Geometric Progression $(G.P.)$ with the first term $a = 2$ and common ratio $r = \frac{1}{3}$.
The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.
For $n = 20$,$S_{20} = \frac{2 \left(1 - (\frac{1}{3})^{20}\right)}{1 - \frac{1}{3}}$.
$S_{20} = \frac{2 \left(1 - \frac{1}{3^{20}}\right)}{\frac{2}{3}}$.
$S_{20} = 2 \times \frac{3}{2} \left(1 - \frac{1}{3^{20}}\right) = 3 \left(1 - \frac{1}{3^{20}}\right)$.

(C) Given the equation: $1 + a + a^2 + \dots + a^x = (1 + a)(1 + a^2)(1 + a^4)$
The left side is a geometric series with $x+1$ terms,so its sum is $\frac{1 - a^{x+1}}{1 - a}$.
Thus,$\frac{1 - a^{x+1}}{1 - a} = (1 + a)(1 + a^2)(1 + a^4)$
Multiplying both sides by $(1 - a)$,we get:
$1 - a^{x+1} = (1 - a)(1 + a)(1 + a^2)(1 + a^4)$
Using the identity $(1 - a)(1 + a) = (1 - a^2)$,we have:
$1 - a^{x+1} = (1 - a^2)(1 + a^2)(1 + a^4)$
Continuing the pattern $(1 - a^2)(1 + a^2) = (1 - a^4)$:
$1 - a^{x+1} = (1 - a^4)(1 + a^4)$
Finally,$(1 - a^4)(1 + a^4) = (1 - a^8)$:
$1 - a^{x+1} = 1 - a^8$
Equating the exponents,$x + 1 = 8$,which gives $x = 7$.

(B) Given: $a_1 = 3$,$a_n = 96$,and $S_n = 189$.
We know that $a_n = a_1 r^{n-1}$,so $3 r^{n-1} = 96$,which implies $r^{n-1} = 32$.
Thus,$r^n = 32r$.
The sum of $n$ terms of a geometric progression is $S_n = \frac{a_1(r^n - 1)}{r - 1} = 189$.
Substituting the values: $\frac{3(32r - 1)}{r - 1} = 189$.
Dividing by $3$: $\frac{32r - 1}{r - 1} = 63$.
$32r - 1 = 63r - 63$.
$62 = 31r$,so $r = 2$.
Since $r^{n-1} = 32$,we have $2^{n-1} = 2^5$.
Therefore,$n - 1 = 5$,which gives $n = 6$.

(A) Let the first term be $a$,common ratio $r = 3$,and the number of terms be $n$.
Given the $n^{th}$ term $l = a r^{n-1} = 486$.
So,$a(3)^{n-1} = 486$,which implies $a \cdot \frac{3^n}{3} = 486$,or $a \cdot 3^n = 1458$ $(i)$.
The sum of $n$ terms is $S_n = \frac{a(r^n - 1)}{r - 1} = 728$.
Substituting $r = 3$: $\frac{a(3^n - 1)}{3 - 1} = 728$.
$a(3^n - 1) = 728 \times 2 = 1456$.
$a \cdot 3^n - a = 1456$ $(ii)$.
Substituting $(i)$ into $(ii)$: $1458 - a = 1456$.
$a = 1458 - 1456 = 2$.

(A) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$.
Given that their product is $1728$,so $(\frac{a}{r}) \times a \times (ar) = 1728$.
$a^3 = 1728$,which implies $a = \sqrt[3]{1728} = 12$.
Given that their sum is $38$,so $\frac{a}{r} + a + ar = 38$.
Substituting $a = 12$,we get $\frac{12}{r} + 12 + 12r = 38$.
$\frac{12}{r} + 12r = 26$.
Dividing by $2$,we get $\frac{6}{r} + 6r = 13$.
$6 + 6r^2 = 13r$,which gives $6r^2 - 13r + 6 = 0$.
Solving the quadratic equation $6r^2 - 9r - 4r + 6 = 0$,we get $3r(2r - 3) - 2(2r - 3) = 0$.
$(3r - 2)(2r - 3) = 0$,so $r = \frac{2}{3}$ or $r = \frac{3}{2}$.
If $r = \frac{3}{2}$,the numbers are $\frac{12}{3/2}, 12, 12 \times \frac{3}{2} = 8, 12, 18$.
If $r = \frac{2}{3}$,the numbers are $\frac{12}{2/3}, 12, 12 \times \frac{2}{3} = 18, 12, 8$.
In both cases,the numbers are $8, 12, 18$. The greatest number is $18$.

(D) Given: First term $a = 7$,last term $l = a{r^{n - 1}} = 448$,and sum of terms $S_n = 889$.
The formula for the sum of a $G.P.$ is $S_n = \frac{lr - a}{r - 1}$.
Substituting the given values: $889 = \frac{448r - 7}{r - 1}$.
$889(r - 1) = 448r - 7$
$889r - 889 = 448r - 7$
$889r - 448r = 889 - 7$
$441r = 882$
$r = \frac{882}{441} = 2$.
Thus,the common ratio is $2$.

(A) The sum of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Given $S_n = 364$,$r = 3$,and the last term $l = a r^{n-1} = 243$.
We can rewrite the sum formula as $S_n = \frac{a r^{n-1} \cdot r - a}{r - 1}$.
Substituting the known values: $364 = \frac{243 \cdot 3 - a}{3 - 1}$.
$364 = \frac{729 - a}{2}$.
$728 = 729 - a$,which gives $a = 1$.
Now,use the last term formula: $a r^{n-1} = 243$.
$1 \cdot 3^{n-1} = 243$.
$3^{n-1} = 3^5$.
Therefore,$n - 1 = 5$,which means $n = 6$.

(C) If $n$ geometric means $g_1, g_2, \dots, g_n$ are inserted between two positive real numbers $a$ and $b$,then the sequence $a, g_1, g_2, \dots, g_n, b$ forms a $G.P.$
Let $r$ be the common ratio. The total number of terms in this $G.P.$ is $n+2$.
Thus,the last term $b = a \cdot r^{(n+2)-1} = a \cdot r^{n+1}$.
This implies $r^{n+1} = \frac{b}{a}$,so $r = \left( \frac{b}{a} \right)^{\frac{1}{n+1}}$.
The $n^{th}$ geometric mean is $g_n = a \cdot r^n$.
Substituting the value of $r$,we get $g_n = a \left( \left( \frac{b}{a} \right)^{\frac{1}{n+1}} \right)^n = a \left( \frac{b}{a} \right)^{\frac{n}{n+1}}$.

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

(C) Let the three geometric means be $g_1, g_2, g_3$ between $2$ and $32$.
Then the sequence $2, g_1, g_2, g_3, 32$ forms a Geometric Progression $(GP)$.
Here,the first term $a = 2$ and the fifth term $ar^4 = 32$.
$2 \times r^4 = 32 \Rightarrow r^4 = 16$.
Since $r^4 = 2^4$,we have $r = 2$ (taking the positive common ratio).
The third geometric mean is $g_3 = ar^3$.
$g_3 = 2 \times (2)^3 = 2 \times 8 = 16$.

(B) Let $G_1, G_2, G_3, G_4, G_5$ be the five $G.M.s$ inserted between $a = 486$ and $b = 2/3$.
The total number of terms in the sequence is $n = 5 + 2 = 7$.
The $7^{th}$ term is given by $T_7 = ar^{7-1} = ar^6$.
Substituting the values: $2/3 = 486 \times r^6$.
$r^6 = \frac{2}{3 \times 486} = \frac{2}{1458} = \frac{1}{729} = (1/3)^6$.
Thus,$r = 1/3$.
The fourth $G.M.$ is the $5^{th}$ term of the sequence,$T_5 = ar^{5-1} = ar^4$.
$T_5 = 486 \times (1/3)^4 = 486 \times \frac{1}{81} = 6$.

(B) The given numbers are $3^1, 3^2, 3^3, ..., 3^n$.
The Geometric Mean $(G.M.)$ of $n$ numbers $x_1, x_2, ..., x_n$ is given by $(x_1 \cdot x_2 \cdot ... \cdot x_n)^{1/n}$.
$G.M. = (3^1 \cdot 3^2 \cdot 3^3 \cdot ... \cdot 3^n)^{1/n}$
Using the property of exponents $a^m \cdot a^n = a^{m+n}$,we get:
$G.M. = (3^{1+2+3+...+n})^{1/n}$
The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.
$G.M. = (3^{\frac{n(n+1)}{2}})^{1/n}$
$G.M. = 3^{\frac{n(n+1)}{2n}} = 3^{\frac{n+1}{2}}$.

(D) Let the three geometric means between $4$ and $\frac{1}{4}$ be $g_1, g_2, g_3$.
Then $4, g_1, g_2, g_3, \frac{1}{4}$ form a $G.P.$
Here,the first term $a = 4$ and the fifth term $ar^4 = \frac{1}{4}$.
$4r^4 = \frac{1}{4} \Rightarrow r^4 = \frac{1}{16} = \left(\frac{1}{2}\right)^4$.
Thus,$r = \frac{1}{2}$ (taking the positive common ratio).
The product of the three geometric means is $g_1 \times g_2 \times g_3 = (ar) \times (ar^2) \times (ar^3) = a^3r^6$.
Substituting the values: $a^3r^6 = (4)^3 \times \left(\frac{1}{2}\right)^6 = 64 \times \frac{1}{64} = 1$.
Alternatively,the product of $n$ geometric means between $a$ and $b$ is $(ab)^{n/2}$. Here $a=4, b=1/4, n=3$,so $(4 \times \frac{1}{4})^{3/2} = 1^{3/2} = 1$.

(B) Let the two geometric means be $a$ and $b$ such that $1, a, b, 64$ form a Geometric Progression $(GP)$.
In a $GP$,the common ratio $r$ is given by $a_n = a_1 \cdot r^{n-1}$.
Here,$a_1 = 1$ and $a_4 = 64$.
$64 = 1 \cdot r^{4-1} \Rightarrow r^3 = 64$.
Since $64 = 4^3$,we have $r = 4$.
The terms are $a = 1 \cdot 4 = 4$ and $b = 4 \cdot 4 = 16$.
Thus,the two geometric means are $4$ and $16$.

(A) Given that $a, b, c$ are in $G.P.$
Therefore,$\frac{b}{a} = \frac{c}{b} = r$,where $r$ is the common ratio.
Squaring both sides,we get $\frac{b^2}{a^2} = \frac{c^2}{b^2} = r^2$.
This implies that $a^2, b^2, c^2$ are in $G.P.$ with a common ratio of $r^2$.

(C) Given that $x, G_1, G_2, y$ are in a $G.P.$
Let the common ratio be $r$.
Then $G_1 = xr$,$G_2 = xr^2$,and $y = xr^3$.
We need to find the value of $G_1 G_2$.
$G_1 G_2 = (xr)(xr^2) = x^2r^3$.
Since $y = xr^3$,we can write $x^2r^3 = x(xr^3) = xy$.
Therefore,$G_1 G_2 = xy$.

(A) Let the three numbers in geometric progression be $\frac{a}{r}, a, ar$.
Given that their product is $1728$,we have:
$(\frac{a}{r}) \times a \times (ar) = 1728$
$a^3 = 1728$
$a = \sqrt[3]{1728} = 12$
Thus,the middle number is $12$.

(B) Let the three consecutive terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given that the product of the terms is $216$:
$\frac{a}{r} \times a \times ar = 216$
$a^3 = 216$
$a = 6$
Given that the sum of the products taken two at a time is $156$:
$(\frac{a}{r} \times a) + (a \times ar) + (\frac{a}{r} \times ar) = 156$
$\frac{a^2}{r} + a^2r + a^2 = 156$
Substitute $a = 6$:
$\frac{36}{r} + 36r + 36 = 156$
$\frac{36}{r} + 36r = 120$
Divide by $12$:
$\frac{3}{r} + 3r = 10$
$3r^2 - 10r + 3 = 0$
$(3r - 1)(r - 3) = 0$
So,$r = 3$ or $r = \frac{1}{3}$.
If $r = 3$,the terms are $\frac{6}{3}, 6, 6 \times 3$,which are $2, 6, 18$.
If $r = \frac{1}{3}$,the terms are $\frac{6}{1/3}, 6, 6 \times \frac{1}{3}$,which are $18, 6, 2$.
Thus,the numbers are $2, 6, 18$.