Progression and Sequence Questions in English

(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
$11^{th}$ term of $A.P. = a + 10d$.
$21^{st}$ term of $A.P. = a + 20d$.
According to the problem,$2 \times (11^{th} \text{ term}) = 7 \times (21^{st} \text{ term})$.
$2(a + 10d) = 7(a + 20d)$.
$2a + 20d = 7a + 140d$.
$5a + 120d = 0$.
Dividing by $5$,we get $a + 24d = 0$.
The $25^{th}$ term of the $A.P.$ is $a_{25} = a + (25-1)d = a + 24d$.
Since $a + 24d = 0$,the $25^{th}$ term is $0$.

(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$.
Also,$\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A.P.$,so $2 \tan^{-1} y = \tan^{-1} x + \tan^{-1} z$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right)$,we get $\tan^{-1} \left( \frac{2y}{1-y^2} \right) = \tan^{-1} \left( \frac{x+z}{1-xz} \right)$.
This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
Substituting $x+z = 2y$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.
This equation holds if $2y = 0$ (which implies $x = -z$,but then $y=0$ and $x, y, z$ would be $x, 0, -x$,which satisfies the condition) or if $1-y^2 = 1-xz$,which means $y^2 = xz$.
If $x, y, z$ are in $A.P.$ and $G.P.$,then $x = y = z$.

(B) If $a, b, c$ are in $G.P.$,then the common ratio is constant,so $b/a = c/b$,which implies $b^2 = ac$.
Now,consider the expression $a(b^2 + c^2)$:
$a(b^2 + c^2) = ab^2 + ac^2$
Since $b^2 = ac$,substitute $b^2$ with $ac$:
$a(ac) + ac^2 = a^2c + ac^2 = ac(a + c)$.
Now,consider the expression $c(a^2 + b^2)$:
$c(a^2 + b^2) = ca^2 + cb^2$
Since $b^2 = ac$,substitute $b^2$ with $ac$:
$ca^2 + c(ac) = ca^2 + ac^2 = ac(a + c)$.
Since both expressions equal $ac(a + c)$,we have $a(b^2 + c^2) = c(a^2 + b^2)$.
Alternatively,let $a=1, b=2, c=4$. Then $a(b^2 + c^2) = 1(4 + 16) = 20$ and $c(a^2 + b^2) = 4(1 + 4) = 20$. Thus,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 $r$ is calculated as $r = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5}$.
The $n^{th}$ term of a $GP$ is given by the formula $t_n = a \cdot r^{n-1}$.
For the $7^{th}$ term $(n = 7)$:
$t_7 = \sqrt{2} \cdot (\sqrt{5})^{7-1}$
$t_7 = \sqrt{2} \cdot (\sqrt{5})^6$
$t_7 = \sqrt{2} \cdot (5)^3$
$t_7 = \sqrt{2} \cdot 125 = 125\sqrt{2}$.

(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 $T_n = A r^{n-1}$.
Given terms are:
$T_4 = a = A r^3$
$T_7 = b = A r^6$
$T_{10} = c = A r^9$
Now,consider the product $ac$:
$ac = (A r^3) \times (A r^9) = A^2 r^{12}$
Also,consider $b^2$:
$b^2 = (A r^6)^2 = A^2 r^{12}$
Since $b^2 = A^2 r^{12}$ and $ac = A^2 r^{12}$,it follows that $b^2 = ac$.
Therefore,$a, b, c$ are in $G.P.$

(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 $n = 5$.
Therefore,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 $b^2 = ac$.
Applying this condition: $(x + 14)^2 = (x + 2)(x + 62)$.
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 form a $G.P.$ with a common ratio of $4$.

(B) Given that the $(p + q)^{th}$ term is $m = ar^{p+q-1}$ and the $(p - q)^{th}$ term is $n = ar^{p-q-1}$.
Dividing the two equations: $\frac{m}{n} = \frac{ar^{p+q-1}}{ar^{p-q-1}} = r^{(p+q-1) - (p-q-1)} = r^{2q}$.
Thus,$r^{2q} = \frac{m}{n}$,which implies $r = (\frac{m}{n})^{1/(2q)}$.
The $p^{th}$ term is $T_p = ar^{p-1}$.
We can express $m$ as $T_p \cdot r^q$ and $n$ as $T_p \cdot r^{-q}$.
Multiplying $m$ and $n$: $m \cdot n = (T_p \cdot r^q) \cdot (T_p \cdot r^{-q}) = T_p^2 \cdot r^0 = T_p^2$.
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 $p$ (at a distance of $q$),the $p^{th}$ term is $\sqrt{mn}$.

(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:
$T_n = T_{n+1} + T_{n+2}$
Substituting the general term formula $T_n = ar^{n-1}$:
$ar^{n-1} = ar^n + ar^{n+1}$
Since $a > 0$ and $r > 0$,we can divide both sides by $ar^{n-1}$:
$1 = r + r^2$
Rearranging the terms,we get 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 the terms of the $G.P.$ are positive,the common ratio $r$ must be positive.
Therefore,we discard the negative root:
$r = \frac{\sqrt{5} - 1}{2}$

(D) Given that $x, 2x + 2, 3x + 3$ are in $G.P.$
For three terms $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.
Therefore,$(2x + 2)^2 = x(3x + 3)$.
Expanding the terms: $4(x + 1)^2 = 3x(x + 1)$.
Case $1$: If $x + 1 = 0$,then $x = -1$. The sequence becomes $-1, 0, 0$,which is not a valid $G.P.$ as the common ratio is undefined.
Case $2$: If $x + 1 \neq 0$,we can divide by $(x + 1)$: $4(x + 1) = 3x$.
$4x + 4 = 3x \Rightarrow x = -4$.
The terms are $x = -4$,$2(-4) + 2 = -6$,$3(-4) + 3 = -9$.
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) The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
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 $\frac{r^3 - 1}{(r^3 - 1)(r^3 + 1)} = \frac{125}{152}$.
$\frac{1}{r^3 + 1} = \frac{125}{152}$.
$r^3 + 1 = \frac{152}{125}$.
$r^3 = \frac{152}{125} - 1 = \frac{152 - 125}{125} = \frac{27}{125}$.
Taking the cube root on both sides,$r = \sqrt[3]{\frac{27}{125}} = \frac{3}{5}$.

(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 $x \log a = y \log b = z \log c = \log m$.
Thus,$x = \frac{\log m}{\log a} = \log_a m$,$y = \log_b m$,and $z = \log_c m$.
Since $x, y, z$ are in $G.P.$,the common ratio is constant,so $\frac{y}{x} = \frac{z}{y}$.
Substituting the values of $x, y, z$ in terms of logarithms,we get $\frac{\log_b m}{\log_a m} = \frac{\log_c m}{\log_b m}$.
Using the change of base formula $\log_k n = \frac{\ln n}{\ln k}$,we have $\frac{\ln m / \ln b}{\ln m / \ln a} = \frac{\ln m / \ln c}{\ln m / \ln b}$.
This simplifies to $\frac{\ln a}{\ln b} = \frac{\ln b}{\ln c}$.
Therefore,$\log_b a = \log_c b$.

(B) Let the first term of the $G.P.$ be $A$ and the common ratio be $R$.
The $p^{th}$ term is $a = A R^{p-1}$ ---$(i)$
The $q^{th}$ term is $b = A R^{q-1}$ ---(ii)
The $r^{th}$ term is $c = A R^{r-1}$ ---(iii)
Now,consider the expression $E = a^{q - r} b^{r - p} c^{p - q}$.
Substituting the values of $a, b,$ and $c$:
$E = (A R^{p-1})^{q-r} (A R^{q-1})^{r-p} (A R^{r-1})^{p-q}$
$E = A^{(q-r) + (r-p) + (p-q)} \cdot R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$
Calculating the exponent of $A$:
$(q-r) + (r-p) + (p-q) = 0$
Calculating the exponent of $R$:
$(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q) = 0$
Thus,$E = A^0 \cdot R^0 = 1 \cdot 1 = 1$.

(C) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$.
The terms of the $G.P.$ are $a, ar, ar^2, ar^3, ar^4, \dots$
Given that the third term is $4$,so $ar^2 = 4$.
The product of the first $5$ terms is $P = a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)$.
$P = a^5 \times r^{(1+2+3+4)} = a^5 \times r^{10}$.
$P = (ar^2)^5$.
Substituting the value $ar^2 = 4$,we get $P = 4^5$.

(B) Let the first term be $a$ and the common ratio be $r$ for the $G.P.$
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 equation $(ii)$ by equation $(i)$:
$\frac{ar^8}{ar^4} = \frac{16/243}{1/3}$
$r^4 = \frac{16}{243} \times 3 = \frac{16}{81}$
$r^4 = (\frac{2}{3})^4$,which implies $r = \frac{2}{3}$ (taking the positive root).
Substituting $r = \frac{2}{3}$ into equation $(i)$:
$a(\frac{2}{3})^4 = \frac{1}{3}$
$a(\frac{16}{81}) = \frac{1}{3}$
$a = \frac{1}{3} \times \frac{81}{16} = \frac{27}{16}$
Now,the $4^{th}$ term $T_4 = ar^3$:
$T_4 = \frac{27}{16} \times (\frac{2}{3})^3$
$T_4 = \frac{27}{16} \times \frac{8}{27} = \frac{8}{16} = \frac{1}{2}$.

(B) The given series is $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$
The ${n^{th}}$ term of the series can be written as ${T_n} = (2n) \times (2n + 2) = 4n(n + 1)$.
To find the ${20^{th}}$ term,we substitute $n = 20$ into the formula:
${T_{20}} = 4 \times 20 \times (20 + 1)$
${T_{20}} = 80 \times 21$
${T_{20}} = 1680$.
Thus,the correct option is $B$.

(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 the formula $l = a r^{n-1}$,where $a$ is the first term,$r$ is the common ratio,and $l$ is the last term.
Dividing both sides by $a$,we get $\frac{l}{a} = r^{n-1}$.
Taking the logarithm on both sides,we have $\log\left(\frac{l}{a}\right) = \log(r^{n-1})$.
Using the property $\log(x/y) = \log x - \log y$ and $\log(x^k) = k \log x$,we get $\log l - \log a = (n-1) \log r$.
Rearranging the terms to solve for $n$,we get $n-1 = \frac{\log l - \log a}{\log r}$.
Therefore,$n = 1 + \frac{\log l - \log a}{\log r}$.

(C) Given that $\log _x a, a^{x/2}, \log _b x$ are in $G.P.$
By the property of $G.P.$,the square of the middle term is equal to the product of the extremes:
$(a^{x/2})^2 = (\log _x a) \cdot (\log _b x)$
$a^x = \frac{\log a}{\log x} \cdot \frac{\log x}{\log b}$
$a^x = \frac{\log a}{\log b} = \log _b a$
Taking $\log _a$ on both sides:
$x = \log _a(\log _b a)$
Using the change of base formula $\log _b a = \frac{\log _e a}{\log _e b}$:
$x = \log _a\left( \frac{\log _e a}{\log _e b} \right)$
Using the logarithmic property $\log (m/n) = \log m - \log n$:
$x = \log _a(\log _e a) - \log _a(\log _e b)$.

(A) Let the roots of the equation $ax^3 + bx^2 + cx + d = 0$ be $\frac{A}{R}, A, AR$ where $A$ and $R$ are constants.
Since the roots are in $G.P.$,their product is $\frac{A}{R} \cdot A \cdot AR = A^3 = -\frac{d}{a}$.
Thus,$A = -\left(\frac{d}{a}\right)^{1/3}$.
Since $A$ is a root of the equation,it must satisfy $aA^3 + bA^2 + cA + d = 0$.
Substituting $A^3 = -\frac{d}{a}$ and $A = -\left(\frac{d}{a}\right)^{1/3}$:
$a\left(-\frac{d}{a}\right) + b\left(-\frac{d}{a}\right)^{2/3} + c\left(-\frac{d}{a}\right)^{1/3} + d = 0$.
$-d + b\left(\frac{d}{a}\right)^{2/3} - c\left(\frac{d}{a}\right)^{1/3} + d = 0$.
$b\left(\frac{d}{a}\right)^{2/3} = c\left(\frac{d}{a}\right)^{1/3}$.
Cubing both sides:
$b^3 \cdot \frac{d^2}{a^2} = c^3 \cdot \frac{d}{a}$.
Multiplying both sides by $a^2$:
$b^3d^2 = c^3ad$.
Dividing by $d$ (assuming $d \neq 0$):
$b^3d = c^3a$.

(A) Let the first term be $a$ and the common ratio be $r$.
According to the problem,the $n^{th}$ term of a geometric progression is given by $T_n = ar^{n-1}$.
Given: $T_{10} = ar^9 = 9$ and $T_4 = ar^3 = 4$.
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,$T_7 = ar^6$.
From $ar^3 = 4$,we have $a = \frac{4}{r^3}$.
Thus,$T_7 = \left(\frac{4}{r^3}\right)r^6 = 4r^3$.
Since $r^6 = \frac{9}{4}$,then $r^3 = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
Therefore,$T_7 = 4 \times \frac{3}{2} = 6$.
Alternatively,for a geometric progression,the $7^{th}$ term is the geometric mean of the $4^{th}$ and $10^{th}$ terms because $7$ is the arithmetic mean of $4$ and $10$: $T_7 = \sqrt{T_4 \times T_{10}} = \sqrt{4 \times 9} = \sqrt{36} = 6$.

(B) The $n^{th}$ term of a $G.P.$ is given by $T_n = ar^{n-1}$,where $a$ is the first term and $r$ is the common ratio.
Given $T_6 = 32$,we have $ar^5 = 32$ .....$(i)$
Given $T_8 = 128$,we have $ar^7 = 128$ .....$(ii)$
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 given geometric progression is $5, - \frac{5}{2}, \frac{5}{4}, - \frac{5}{8}, \dots$
Here,the first term $a = 5$ and the common ratio $r = \frac{-5/2}{5} = -\frac{1}{2}$.
The $n^{th}$ term of a geometric progression is given by $T_n = ar^{n-1}$.
Given $T_n = \frac{5}{1024}$,we have:
$\frac{5}{1024} = 5 \left( -\frac{1}{2} \right)^{n-1}$
Dividing both sides by $5$,we get:
$\frac{1}{1024} = \left( -\frac{1}{2} \right)^{n-1}$
Since $1024 = 2^{10}$,we can write $\frac{1}{1024} = \left( -\frac{1}{2} \right)^{10}$ because $(-1)^{10} = 1$.
Thus,$\left( -\frac{1}{2} \right)^{10} = \left( -\frac{1}{2} \right)^{n-1}$.
Comparing the exponents,we get $10 = n - 1$,which implies $n = 11$.

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

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

(D) Let the infinite $G.P.$ be $a, ar, ar^2, \dots, \infty$.
The sum of an infinite $G.P.$ is given by $S = \frac{a}{1 - r} = 9$.
From this,we get $a = 9(1 - r)$ $\dots(i)$.
The sum of the first two terms is $a + ar = 5$,which can be written as $a(1 + r) = 5$ $\dots(ii)$.
Substituting the value of $a$ from equation $(i)$ into equation $(ii)$:
$9(1 - r)(1 + r) = 5$
$9(1 - r^2) = 5$
$1 - r^2 = \frac{5}{9}$
$r^2 = 1 - \frac{5}{9} = \frac{4}{9}$
$r = \pm \frac{2}{3}$.
Since the options provided include $2/3$,the correct option is $D$.

(A) The given series is $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots = 3 + \frac{9}{2} + \frac{27}{4} + \dots$
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 $n = 5$ terms:
$S_5 = \frac{3 \left[ (\frac{3}{2})^5 - 1 \right]}{\frac{3}{2} - 1} = \frac{3 \left[ \frac{243}{32} - 1 \right]}{\frac{1}{2}}$
$S_5 = 6 \left[ \frac{243 - 32}{32} \right] = 6 \left[ \frac{211}{32} \right] = \frac{3 \times 211}{16} = \frac{633}{16}$
Converting to a mixed fraction: $\frac{633}{16} = 39\frac{9}{16}$.

(A) The given series is $0.9 + 0.09 + 0.009 + \dots$ up to $100$ terms.
This 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 the formula $S_n = a \frac{1 - r^n}{1 - r}$.
Substituting the values $a = \frac{9}{10}$,$r = \frac{1}{10}$,and $n = 100$:
$S_{100} = \frac{9}{10} \left( \frac{1 - (\frac{1}{10})^{100}}{1 - \frac{1}{10}} \right)$
$S_{100} = \frac{9}{10} \left( \frac{1 - (\frac{1}{10})^{100}}{\frac{9}{10}} \right)$
$S_{100} = 1 - \left( \frac{1}{10} \right)^{100}$.

(A) Let $x = 0.2343434...$
Multiply by $10$ to isolate the repeating part: $10x = 2.343434...$
Multiply by $1000$ to shift the decimal point: $1000x = 234.343434...$
Subtract the first equation from the second: $1000x - 10x = 234.343434... - 2.343434...$
$990x = 232$
$x = \frac{232}{990}$

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

(B) Let $S_n = 6 + 66 + 666 + \dots$ up to $n$ terms.
$S_n = 6(1 + 11 + 111 + \dots \text{ up to } n \text{ terms})$
$S_n = \frac{6}{9}(9 + 99 + 999 + \dots \text{ up to } n \text{ terms})$
$S_n = \frac{2}{3}((10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1))$
$S_n = \frac{2}{3}((10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + 1 \text{ up to } n \text{ terms}))$
Using the sum of a geometric progression formula $S = a(r^n - 1)/(r - 1)$:
$S_n = \frac{2}{3} \left( \frac{10(10^n - 1)}{10 - 1} - n \right)$
$S_n = \frac{2}{3} \left( \frac{10(10^n - 1)}{9} - n \right)$
$S_n = \frac{2}{3} \left( \frac{10^{n+1} - 10 - 9n}{9} \right)$
$S_n = \frac{2(10^{n+1} - 9n - 10)}{27}$.

(D) Let the first term be $a$ and the common ratio be $r$ for the $G.P.$
Given that every term is the sum of its two previous terms,we have the relation: $T_n = T_{n-1} + T_{n-2}$.
Substituting the general term formula $T_n = ar^{n-1}$,we get: $ar^{n-1} = ar^{n-2} + ar^{n-3}$.
Dividing both sides by $ar^{n-3}$ (since $a > 0$ and $r > 0$),we obtain: $r^2 = r + 1$.
Rearranging the terms gives the quadratic equation: $r^2 - r - 1 = 0$.
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $r = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$.
Since the terms of the $G.P.$ are positive,the common ratio $r$ must be positive. Therefore,we take the positive root: $r = \frac{1 + \sqrt{5}}{2}$.

(B) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$.
According to the problem,every term is twice its previous term,so $r = 2$.
The sum of the first two terms is given as $a + ar = 1$.
Substituting $r = 2$ into the equation:
$a + a(2) = 1$
$3a = 1$
$a = 1/3$.
Therefore,the first term is $1/3$.

(A) Given that the sum of $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1} = 255$ (since $r > 1$) .....$(i)$
The $n^{th}$ term is $a_n = ar^{n-1} = 128$ .....$(ii)$
The common ratio is $r = 2$ .....$(iii)$
Substitute $r = 2$ into equation $(ii)$:
$a(2^{n-1}) = 128$ .....$(iv)$
Substitute $r = 2$ into equation $(i)$:
$\frac{a(2^n - 1)}{2 - 1} = 255 \Rightarrow a(2^n - 1) = 255$ .....$(v)$
Divide equation $(v)$ by equation $(iv)$:
$\frac{a(2^n - 1)}{a(2^{n-1})} = \frac{255}{128}$
$\frac{2^n - 1}{2^{n-1}} = \frac{255}{128}$
$\frac{2^n}{2^{n-1}} - \frac{1}{2^{n-1}} = \frac{255}{128}$
$2 - \frac{1}{2^{n-1}} = \frac{255}{128}$
$\frac{1}{2^{n-1}} = 2 - \frac{255}{128} = \frac{256 - 255}{128} = \frac{1}{128}$
Since $128 = 2^7$,we have $2^{n-1} = 2^7$,which implies $n - 1 = 7$,so $n = 8$.
Substitute $n = 8$ into equation $(iv)$:
$a(2^{8-1}) = 128$
$a(2^7) = 128$
$a(128) = 128$
$a = 1$.

(C) The $k$-th term of the series is given by $T_k = 1 + x + x^2 + \dots + x^{k-1} = \frac{1 - x^k}{1 - x}$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1 - x^k}{1 - x}$.
$S_n = \frac{1}{1 - x} \left[ \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} x^k \right]$.
$S_n = \frac{1}{1 - x} \left[ n - \frac{x(1 - x^n)}{1 - x} \right]$.
$S_n = \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2}$.

(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 $S_6 = 9 \times S_3$.
Substituting the formula: $\frac{a(r^6 - 1)}{r - 1} = 9 \times \frac{a(r^3 - 1)}{r - 1}$.
Assuming $r \neq 1$,we can cancel $\frac{a}{r - 1}$ from both sides:
$r^6 - 1 = 9(r^3 - 1)$.
Using the identity $x^2 - 1 = (x - 1)(x + 1)$ where $x = r^3$:
$(r^3 - 1)(r^3 + 1) = 9(r^3 - 1)$.
Since $r \neq 1$,$r^3 - 1 \neq 0$,so we can divide by $(r^3 - 1)$:
$r^3 + 1 = 9$.
$r^3 = 8$.
$r = 2$.

(C) Let $S$ be the number consisting of $91$ ones. This can be written as a geometric series: $S = 1 + 10 + 10^2 + \dots + 10^{90}$.
Using the sum formula for a geometric progression,$S = \frac{10^{91} - 1}{10 - 1} = \frac{10^{91} - 1}{9}$.
Since $91 = 7 \times 13$,we can write $10^{91} - 1 = (10^{13})^7 - 1$.
Using the algebraic identity $x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + \dots + 1)$,we have $(10^{13})^7 - 1 = (10^{13} - 1)( (10^{13})^6 + (10^{13})^5 + \dots + 1)$.
Thus,$S = \frac{10^{13} - 1}{9} \times (10^{78} + 10^{65} + \dots + 1)$.
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 $(G.P.)$ because the ratio of consecutive terms is constant.
Here,the first term $a = a_1 = 2$ and the common ratio $r = \frac{1}{3}$.
The sum of the first $n$ terms of a $G.P.$ is given by the formula $S_n = \frac{a(1 - r^n)}{1 - r}$.
For $n = 20$,we have:
$S_{20} = \frac{2(1 - (1/3)^{20})}{1 - 1/3}$
$S_{20} = \frac{2(1 - 1/3^{20})}{2/3}$
$S_{20} = 2 \times \frac{3}{2} \times (1 - \frac{1}{3^{20}})$
$S_{20} = 3 \left( 1 - \frac{1}{3^{20}} \right)$.

(C) We have the equation $1 + a + a^2 + a^3 + \dots + a^x = (1 + a)(1 + a^2)(1 + a^4)$.
The left side is a geometric series with $x+1$ terms,which sums to $\frac{1 - a^{x+1}}{1 - a}$.
So,$\frac{1 - a^{x+1}}{1 - a} = (1 + a)(1 + a^2)(1 + a^4)$.
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:
$1 - a^{x+1} = (1 - a^2)(1 + a^2)(1 + a^4)$.
Using $(1 - a^2)(1 + a^2) = (1 - a^4)$,we have:
$1 - a^{x+1} = (1 - a^4)(1 + a^4)$.
Using $(1 - a^4)(1 + a^4) = (1 - a^8)$,we have:
$1 - a^{x+1} = 1 - a^8$.
Comparing the exponents,$x + 1 = 8$,which gives $x = 7$.

(B) Given: $a_1 = 3$,$a_n = 96$,and $S_n = 189$.
In a geometric progression,the $n^{th}$ term is given by $a_n = a_1 r^{n-1}$.
Substituting the values: $3 r^{n-1} = 96 \Rightarrow r^{n-1} = 32$ ..... $(i)$.
The sum of $n$ terms is $S_n = \frac{a_1(r^n - 1)}{r - 1} = 189$.
We can write $S_n = \frac{a_1 r^n - a_1}{r - 1} = \frac{r(a_1 r^{n-1}) - a_1}{r - 1} = 189$.
Substituting $a_1 r^{n-1} = 96$ and $a_1 = 3$: $\frac{r(96) - 3}{r - 1} = 189$.
$96r - 3 = 189(r - 1) \Rightarrow 96r - 3 = 189r - 189$.
$186 = 93r \Rightarrow r = 2$.
From $(i)$,$2^{n-1} = 32 = 2^5$.
Therefore,$n - 1 = 5 \Rightarrow n = 6$.

(A) Let the first term be $a$,common ratio $r = 3$,and the number of terms be $n$.
The $n^{th}$ term of a geometric progression is given by $T_n = a r^{n-1}$.
Given $T_n = 486$,we have $a(3)^{n-1} = 486$. Multiplying both sides by $3$,we get $a(3)^n = 1458$ .....$(i)$.
The sum of $n$ terms is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Given $S_n = 728$ and $r = 3$,we have $728 = \frac{a(3^n - 1)}{3 - 1}$.
$728 = \frac{a(3^n) - a}{2}$.
$1456 = a(3^n) - a$ .....$(ii)$.
Substituting the value of $a(3^n)$ from equation $(i)$ into equation $(ii)$:
$1456 = 1458 - a$.
$a = 1458 - 1456 = 2$.
Thus,the first term is $2$.

(D) Given that the product of $n$ positive numbers $x_1, x_2, \dots, x_n$ is $1$,i.e.,$x_1 \cdot x_2 \cdot \dots \cdot x_n = 1$.
According to the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,for any set of positive real numbers,the Arithmetic Mean is always greater than or equal to the Geometric Mean.
$\frac{x_1 + x_2 + \dots + x_n}{n} \ge (x_1 \cdot x_2 \cdot \dots \cdot x_n)^{1/n}$.
Substituting the given product value:
$\frac{x_1 + x_2 + \dots + x_n}{n} \ge (1)^{1/n} = 1$.
Therefore,$x_1 + x_2 + \dots + x_n \ge n$.
This implies that the sum of these $n$ positive numbers can never be less than $n$.

(A) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$.
Given that their product is $1728$,so $(\frac{a}{r}) \cdot a \cdot (ar) = 1728$,which implies $a^3 = 1728$,so $a = 12$.
Given that their sum is $38$,so $\frac{a}{r} + a + ar = 38$.
Substituting $a = 12$,we get $\frac{12}{r} + 12 + 12r = 38$.
Dividing by $2$,we get $\frac{6}{r} + 6 + 6r = 19$,which simplifies to $6r^2 - 13r + 6 = 0$.
Solving the quadratic equation $6r^2 - 9r - 4r + 6 = 0$,we get $3r(2r - 3) - 2(2r - 3) = 0$,so $(3r - 2)(2r - 3) = 0$.
Thus,$r = \frac{2}{3}$ or $r = \frac{3}{2}$.
If $r = \frac{3}{2}$,the numbers are $\frac{12}{3/2}, 12, 12(\frac{3}{2}) = 8, 12, 18$.
If $r = \frac{2}{3}$,the numbers are $\frac{12}{2/3}, 12, 12(\frac{2}{3}) = 18, 12, 8$.
In both cases,the numbers are $8, 12, 18$. The greatest number is $18$.

(B) Let the sum be $S_n = 3 + 33 + 333 + \dots + n$ terms.
We can write $S_n = 3(1 + 11 + 111 + \dots + n \text{ terms})$.
Multiply and divide by $9$:
$S_n = \frac{3}{9}(9 + 99 + 999 + \dots + n \text{ terms})$
$S_n = \frac{1}{3}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$
$S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n \text{ terms})]$
The first part is a geometric progression with $a = 10$,$r = 10$,and $n$ terms. The sum is $\frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$.
$S_n = \frac{1}{3} \left[ \frac{10(10^n - 1)}{9} - n \right]$
$S_n = \frac{1}{3} \left[ \frac{10^{n+1} - 10 - 9n}{9} \right]$
$S_n = \frac{1}{27}(10^{n+1} - 9n - 10)$.

(D) Given: First term $a = 7$,last term $l = a r^{n-1} = 448$,and sum of $n$ terms $S_n = 889$.
The formula for the sum of $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.
We can rewrite this as $S_n = \frac{a r^n - a}{r - 1} = \frac{(a r^{n-1})r - a}{r - 1}$.
Substituting the given values: $889 = \frac{448r - 7}{r - 1}$.
Multiplying both sides by $(r - 1)$: $889(r - 1) = 448r - 7$.
$889r - 889 = 448r - 7$.
$889r - 448r = 889 - 7$.
$441r = 882$.
$r = \frac{882}{441} = 2$.
Therefore,the common ratio is $2$.

(A) Let the first term be $a$, the number of terms be $n$, and the common ratio be $r = 3$.
The sum of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1} = 364$.
The last term $l$ is given by $l = ar^{n-1} = 243$.
We can rewrite the sum formula as $S_n = \frac{ar^{n-1} \cdot r - a}{r - 1} = 364$.
Substituting $ar^{n-1} = 243$ and $r = 3$ into the equation:
$\frac{243 \cdot 3 - a}{3 - 1} = 364$
$\frac{729 - a}{2} = 364$
$729 - a = 728$
$a = 1$.
Now, using the last term formula $ar^{n-1} = 243$:
$1 \cdot 3^{n-1} = 243$
$3^{n-1} = 3^5$
$n - 1 = 5$
$n = 6$.
Thus, the number of terms is $6$.

(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 Geometric Progression $(G.P.)$.
Let $r$ be the common ratio of this $G.P.$ The total number of terms in this sequence is $n+2$.
Thus,the last term $b$ can be expressed as $b = a \cdot r^{(n+2)-1} = a \cdot r^{n+1}$.
Solving for $r$,we get $r = \left( \frac{b}{a} \right)^{\frac{1}{n+1}}$.
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) 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: $a^{n + 1} - a^{n + 1/2}b^{1/2} + b^{n + 1} - a^{1/2}b^{n + 1/2} = 0$.
$a^{n + 1/2}(a^{1/2} - b^{1/2}) - b^{n + 1/2}(a^{1/2} - b^{1/2}) = 0$.
$(a^{n + 1/2} - b^{n + 1/2})(a^{1/2} - b^{1/2}) = 0$.
Since $a \neq b$,$a^{1/2} - b^{1/2} \neq 0$,so we must have $a^{n + 1/2} - b^{n + 1/2} = 0$.
$a^{n + 1/2} = b^{n + 1/2}$.
$(a/b)^{n + 1/2} = 1 = (a/b)^0$.
Equating the exponents,$n + 1/2 = 0$,which gives $n = -1/2$.

(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 $ar^4 = 32$.
Substituting $a = 2$ in the equation $ar^4 = 32$,we get $2r^4 = 32$,which implies $r^4 = 16$.
Since $16 = 2^4$,we have $r = 2$.
The third geometric mean is the fourth term of the sequence,which is $g_3 = ar^3$.
Substituting the values,$g_3 = 2 \times (2)^3 = 2 \times 8 = 16$.