Mix Examples - Sequences and Series Questions in English

(B) Given that $a_1 = 2$ and $a_2 = 2$.
For $n > 2$,the recurrence relation is $a_n = a_{n-1} - 1$.
Calculating the subsequent terms:
$a_3 = a_2 - 1 = 2 - 1 = 1$
$a_4 = a_3 - 1 = 1 - 1 = 0$
$a_5 = a_4 - 1 = 0 - 1 = -1$
Therefore,the value of $a_5$ is $-1$.

(D) Given that $a, b, c$ are in $A.P.$
Dividing each term by $abc$,we get:
$\frac{a}{abc}, \frac{b}{abc}, \frac{c}{abc}$ are in $A.P.$
This simplifies to $\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}$ are in $A.P.$
However,the given terms are $\frac{a}{bc}, \frac{1}{c}, \frac{2}{b}$.
Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Dividing by $abc$,we get $\frac{2}{ac} = \frac{1}{bc} + \frac{1}{ab}$.
Testing the given sequence $\frac{a}{bc}, \frac{1}{c}, \frac{2}{b}$ does not satisfy the condition for $A.P.$,$G.P.$,or $H.P.$ generally.
Therefore,the correct option is $D$.

(C) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$ .....$(i)$
$x$ is the arithmetic mean between $a$ and $b$,so $x = \frac{a + b}{2}$ .....(ii)
$y$ is the arithmetic mean between $b$ and $c$,so $y = \frac{b + c}{2}$ .....(iii)
Now,consider the expression $\frac{a}{x} + \frac{c}{y}$.
Substituting the values of $x$ and $y$: $\frac{a}{\frac{a+b}{2}} + \frac{c}{\frac{b+c}{2}} = \frac{2a}{a+b} + \frac{2c}{b+c}$
$= \frac{2a(b+c) + 2c(a+b)}{(a+b)(b+c)} = \frac{2ab + 2ac + 2ac + 2bc}{ab + ac + b^2 + bc}$
Since $b^2 = ac$,the denominator becomes $ab + ac + ac + bc = ab + 2ac + bc$.
Thus,the expression is $\frac{2(ab + 2ac + bc)}{ab + 2ac + bc} = 2$.

(D) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Given that $b - c, c - a, a - b$ are in $H.P.$,we have $\frac{2}{c - a} = \frac{1}{b - c} + \frac{1}{a - b}$.
Simplifying the $H.P.$ condition: $\frac{2}{c - a} = \frac{a - b + b - c}{(b - c)(a - b)} = \frac{a - c}{(b - c)(a - b)}$.
$2(b - c)(a - b) = -(c - a)^2$.
$2(ab - b^2 - ac + bc) = -(c - a)^2$.
Since $b^2 = ac$,we have $2(ab - ac - ac + bc) = -(c - a)^2$.
$2(ab + bc - 2ac) = -(c - a)^2$.
$2b(a + c) - 4ac = -(c^2 - 2ac + a^2)$.
$2b(a + c) - 4b^2 = -c^2 + 2b^2 - a^2$.
$a^2 + 2b(a + c) + c^2 = 6b^2$.
Since $a+b+c = a+c+\sqrt{ac}$,the expression $a+b+c$ depends on $a, b,$ and $c$. Therefore,it is not independent of any of them.

(A) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Since $a - b, c - a, b - c$ are in $H.P.$,their reciprocals are in $A.P.$:
$\frac{1}{c - a} = \frac{1}{2} \left( \frac{1}{a - b} + \frac{1}{b - c} \right)$
$\frac{2}{c - a} = \frac{(b - c) + (a - b)}{(a - b)(b - c)} = \frac{a - c}{(a - b)(b - c)}$
$2(a - b)(b - c) = -(a - c)^2$
$2(ab - ac - b^2 + bc) = -(a^2 - 2ac + c^2)$
Substituting $ac = b^2$:
$2(ab - b^2 - b^2 + bc) = -(a^2 - 2b^2 + c^2)$
$2ab - 4b^2 + 2bc = -a^2 + 2b^2 - c^2$
$a^2 + 2ab + 2bc + c^2 = 6b^2$
Since $b^2 = ac$,$a^2 + 2ab + 2bc + c^2 = 6ac$
Alternatively,using the property $2(a-b)(b-c) = -(a-c)^2$:
$2(ab - ac - b^2 + bc) = -(a^2 - 2ac + c^2)$
$2ab - 2b^2 - 2ac + 2bc = -a^2 + 2ac - c^2$
$a^2 + 2ab + 2bc + c^2 = 4ac + 2b^2 = 4b^2 + 2b^2 = 6b^2$
This simplifies to $a + 4b + c = 0$.

(A) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$
Since the $(m + 1)^{th}$,$(n + 1)^{th}$,and $(r + 1)^{th}$ terms are in $G.P.$,we have:
$(a + md), (a + nd), (a + rd)$ are in $G.P.$
Therefore,$(a + nd)^2 = (a + md)(a + rd)$.
Expanding both sides: $a^2 + 2and + n^2d^2 = a^2 + ard + amd + mrd^2$.
$2and - ard - amd = mrd^2 - n^2d^2$.
$ad(2n - r - m) = d^2(mr - n^2)$.
Thus,$\frac{d}{a} = \frac{2n - (m + r)}{mr - n^2}$ ... $(i)$.
Given $m, n, r$ are in $H.P.$,we have $n = \frac{2mr}{m + r}$,which implies $m + r = \frac{2mr}{n}$.
Substituting $m + r$ into equation $(i)$:
$\frac{d}{a} = \frac{2n - \frac{2mr}{n}}{mr - n^2} = \frac{\frac{2n^2 - 2mr}{n}}{mr - n^2} = \frac{-2(mr - n^2)}{n(mr - n^2)} = -\frac{2}{n}$.

(A) Case $1$: If $9, x, y, z, a$ are in $A.P.$,then the sum of the terms is $9 + x + y + z + a = \frac{5}{2}(9 + a)$.
Given $x + y + z = 15$,we have $9 + 15 + a = \frac{5}{2}(9 + a)$.
$24 + a = \frac{45 + 5a}{2}$.
$48 + 2a = 45 + 5a$.
$3a = 3 \Rightarrow a = 1$.
Case $2$: If $9, x, y, z, a$ are in $H.P.$,then $\frac{1}{9}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{a}$ are in $A.P.$
The sum of these terms is $\frac{1}{9} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{a} = \frac{5}{2}\left(\frac{1}{9} + \frac{1}{a}\right)$.
Given $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$,we have $\frac{1}{9} + \frac{5}{3} + \frac{1}{a} = \frac{5}{2}\left(\frac{1}{9} + \frac{1}{a}\right)$.
$\frac{16}{9} + \frac{1}{a} = \frac{5}{18} + \frac{5}{2a}$.
$\frac{1}{a} - \frac{5}{2a} = \frac{5}{18} - \frac{16}{9}$.
$-\frac{3}{2a} = \frac{5 - 32}{18} = -\frac{27}{18} = -\frac{3}{2}$.
$-\frac{3}{2a} = -\frac{3}{2} \Rightarrow a = 1$.

(D) Given $\frac{b + a}{b - a} = \frac{b + c}{b - c}$.
Applying componendo and dividendo:
$\frac{(b + a) + (b - a)}{(b + a) - (b - a)} = \frac{(b + c) + (b - c)}{(b + c) - (b - c)}$
$\frac{2b}{2a} = \frac{2b}{2c}$
$\frac{b}{a} = \frac{b}{c}$
Since $b \neq 0$,we have $a = c$.
If $a = c$,then $a, b, c$ are in $G.P.$ because $b^2 = ac$ is not necessarily true,but $a, b, a$ are in $G.P.$ only if $b^2 = a^2$,i.e.,$b = \pm a$. Since this is not generally true,$a, b, c$ do not form a specific progression like $A.P.$,$G.P.$,or $H.P.$ unless $b=0$ or $a=c=0$.

(D) Let the $A.P.$ be $a, a+d, a+2d, \dots$
Squaring the terms,we get $a^2, (a+d)^2, (a+2d)^2, \dots$
Calculating the differences between consecutive terms:
$d_1 = (a+d)^2 - a^2 = 2ad + d^2$
$d_2 = (a+2d)^2 - (a+d)^2 = (a^2 + 4ad + 4d^2) - (a^2 + 2ad + d^2) = 2ad + 3d^2$
Since $d_1 \neq d_2$ (unless $d=0$),the new series is not an $A.P.$
Similarly,checking for $G.P.$ or $H.P.$ shows that the new series does not necessarily follow these progressions.
Therefore,the correct option is $D$.

(D) Given $\log_x y, \log_z x, \log_y z$ are in $G.P.$
Therefore,$(\log_z x)^2 = (\log_x y)(\log_y z) = \log_x z = \frac{1}{\log_z x}$.
This implies $(\log_z x)^3 = 1$,so $\log_z x = 1$,which means $x = z$.
Given $xyz = 64$ and $x = z$,we have $x^2 y = 64$,so $y = \frac{64}{x^2}$.
Given $x^3, y^3, z^3$ are in $A.P.$,we have $2y^3 = x^3 + z^3$.
Since $x = z$,$2y^3 = 2x^3$,which implies $y^3 = x^3$,so $y = x$.
Substituting $y = x$ into $x^2 y = 64$,we get $x^3 = 64$,so $x = 4$.
Thus,$x = y = z = 4$.
Since $x = y = z = 4$,all conditions are satisfied: $x=y=z$ is true,$x=4$ is true,and $x, y, z$ are in $G.P.$ (with common ratio $1$) is true.
Therefore,the correct option is $(d)$.

(D) Given that $p, q, r$ are in $H.P.$,we have $q = \frac{2pr}{p+r}$.
Let $K = \frac{pr}{p+r}$,then $q = 2K$ and $pr = K(p+r)$.
Since $p^2, q^2, r^2$ are in $A.P.$,we have $2q^2 = p^2 + r^2$.
Substituting $q = 2K$,we get $2(4K^2) = p^2 + r^2$,so $8K^2 = p^2 + r^2$.
We know $(p+r)^2 = p^2 + r^2 + 2pr = 8K^2 + 2K(p+r)$.
Let $S = p+r$. Then $S^2 - 2KS - 8K^2 = 0$.
Solving for $S$,$(S - 4K)(S + 2K) = 0$,so $S = 4K$ or $S = -2K$.
If $S = 4K$,then $(p-r)^2 = (p+r)^2 - 4pr = 16K^2 - 4(4K^2) = 0$,implying $p=r$,which contradicts the condition that numbers are unequal.
If $S = -2K$,then $pr = K(-2K) = -2K^2$.
Then $(p-r)^2 = (p+r)^2 - 4pr = (-2K)^2 - 4(-2K^2) = 4K^2 + 8K^2 = 12K^2$.
Thus $p-r = \pm 2\sqrt{3}K$.
Solving $p+r = -2K$ and $p-r = \pm 2\sqrt{3}K$,we get $p = (-1 \pm \sqrt{3})K$ and $r = (-1 \mp \sqrt{3})K$.
Thus $p:q:r = (-1 \pm \sqrt{3})K : 2K : (-1 \mp \sqrt{3})K = 1 \mp \sqrt{3} : -2 : 1 \pm \sqrt{3}$.

(C) If $\frac{1}{16}, a, b$ are in $G.P.$,then $a^2 = \frac{b}{16}$,which implies $b = 16a^2$ ..... $(i)$
If $a, b, \frac{1}{6}$ are in $H.P.$,then $b = \frac{2 \cdot a \cdot \frac{1}{6}}{a + \frac{1}{6}} = \frac{2a}{6a + 1}$ ..... $(ii)$
Equating $(i)$ and $(ii)$,we get $16a^2 = \frac{2a}{6a + 1}$.
For $a \neq 0$,$8a = \frac{1}{6a + 1}$,which gives $48a^2 + 8a - 1 = 0$.
Factoring the quadratic equation: $(4a + 1)(12a - 1) = 0$.
Thus,$a = -\frac{1}{4}$ or $a = \frac{1}{12}$.
If $a = -\frac{1}{4}$,then $b = 16(-\frac{1}{4})^2 = 1$.
If $a = \frac{1}{12}$,then $b = 16(\frac{1}{12})^2 = 16 \cdot \frac{1}{144} = \frac{1}{9}$.
Therefore,both pairs are valid solutions.

(B) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Since $\log a - \log 2b, \log 2b - \log 3c, \log 3c - \log a$ are in $A.P.$,the sum of the terms is $0$:
$(\log a - \log 2b) + (\log 2b - \log 3c) + (\log 3c - \log a) = 0$.
Also,$2(\log 2b - \log 3c) = (\log a - \log 2b) + (\log 3c - \log a) = \log 3c - \log 2b$.
This simplifies to $3(\log 2b - \log 3c) = 0$,so $\log 2b = \log 3c$,which implies $2b = 3c$,or $b = \frac{3}{2}c$.
Substituting $b = \frac{3}{2}c$ into $b^2 = ac$,we get $\frac{9}{4}c^2 = ac$,so $a = \frac{9}{4}c$.
Thus,the sides are $a = \frac{9}{4}c, b = \frac{3}{2}c, c = c$.
Multiplying by $\frac{4}{c}$,the sides are proportional to $9, 6, 4$.
Since $4^2 + 6^2 = 16 + 36 = 52 < 9^2 = 81$,the square of the longest side is greater than the sum of the squares of the other two sides.
Therefore,the triangle is obtuse angled.

(B) Given that $a$ is the arithmetic mean of $b$ and $c$,we have $a = \frac{b+c}{2}$,which implies $b+c = 2a$.
Since $G_1$ and $G_2$ are two geometric means between $b$ and $c$,the sequence $b, G_1, G_2, c$ is in $G$.$P$.
Let the common ratio be $r$. Then $G_1 = br$,$G_2 = br^2$,and $c = br^3$.
Thus,$r = (c/b)^{1/3}$.
We know $G_1 G_2 = (br)(br^2) = b^2 r^3 = b^2 (c/b) = bc$.
Now,$G_1^3 + G_2^3 = (br)^3 + (br^2)^3 = b^3 r^3 + b^3 r^6 = b^3 (c/b) + b^3 (c/b)^2 = b^2 c + b c^2 = bc(b+c)$.
Substituting $bc = G_1 G_2$ and $b+c = 2a$,we get:
$G_1^3 + G_2^3 = (G_1 G_2)(2a) = 2 G_1 G_2 a$.

(C) Let the three numbers in $G.P.$ be $a, ar, ar^2$.
According to the first condition,$a, ar, ar^2 - 64$ are in $A.P.$
Therefore,$2(ar) = a + (ar^2 - 64)$ $\Rightarrow a(r^2 - 2r + 1) = 64$ $\Rightarrow a(r - 1)^2 = 64$ .....$(i)$
According to the second condition,$a, ar - 8, ar^2 - 64$ are in $G.P.$
Therefore,$(ar - 8)^2 = a(ar^2 - 64)$ $\Rightarrow a^2r^2 - 16ar + 64 = a^2r^2 - 64a$ $\Rightarrow 64 - 16ar = -64a$ $\Rightarrow 16ar - 64a = 64$ $\Rightarrow ar - 4a = 4$ .....(ii)
From $(i)$,$a(r-1)^2 = 64$. From (ii),$a(r-4) = 4 \Rightarrow a = \frac{4}{r-4}$.
Substituting $a$ in $(i)$: $\frac{4}{r-4}(r-1)^2 = 64$ $\Rightarrow (r-1)^2 = 16(r-4)$ $\Rightarrow r^2 - 2r + 1 = 16r - 64$ $\Rightarrow r^2 - 18r + 65 = 0$.
Solving for $r$: $(r-5)(r-13) = 0$,so $r=5$ or $r=13$.
If $r=5$,$a = \frac{4}{5-4} = 4$. The numbers are $4, 4(5), 4(5^2) = 4, 20, 100$.
Checking the conditions: $4, 20, 100-64=36$ are in $A.P.$ $(20-4=16, 36-20=16)$.
$4, 20-8=12, 36$ are in $G.P.$ $(12/4=3, 36/12=3)$.
Thus,the numbers are $4, 20, 100$.

(D) Let $y = n^n \left( \frac{n+1}{2} \right)^{2n}$.
For $n = 2$:
$y = 2^2 \left( \frac{3}{2} \right)^4 = 4 \times \frac{81}{16} = \frac{81}{4} = 20.25$.
Check Option $(a)$ and $(b)$:
$\left( \frac{n+1}{2} \right)^3 = \left( \frac{3}{2} \right)^3 = \frac{27}{8} = 3.375$.
Since $20.25 > 3.375$,$y$ is greater than $\left( \frac{n+1}{2} \right)^3$.
Check Option $(c)$:
$(n!)^3 = (2!)^3 = 2^3 = 8$.
Since $20.25 > 8$,$y$ is greater than $(n!)^3$.
Thus,both $(b)$ and $(c)$ are correct.

(A) For any natural number $n$,either $n$ is even or $n$ is odd.
If $n$ is even,then $n = 2k$ for some integer $k$,so $n(n + 1) = 2k(2k + 1)$,which is even.
If $n$ is odd,then $n = 2k + 1$ for some integer $k$,so $n + 1 = 2k + 2 = 2(k + 1)$,which is even. Thus,$n(n + 1) = (2k + 1) \times 2(k + 1) = 2(2k + 1)(k + 1)$,which is also even.
Therefore,the product of two consecutive natural numbers is always even.

(D) We analyze the sum $a(n) = \sum_{k=1}^{2^n-1} \frac{1}{k}$.
By grouping terms,we can observe that $a(n) = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \dots + \frac{1}{7}) + \dots + (\frac{1}{2^{n-1}} + \dots + \frac{1}{2^n-1})$.
Each group of the form $\sum_{k=2^{m-1}}^{2^m-1} \frac{1}{k}$ is greater than $\frac{1}{2}$.
Since there are $n$ such groups,$a(n) > \frac{n}{2}$.
For $n=200$,$a(200) > \frac{200}{2} = 100$.
Also,it is a known property that $a(n) \le n$ for these sums.
Thus,$a(100) \le 100$ and $a(200) > 100$ are both true.

(C) Consider the function $f(x) = \frac{x^2}{x^3 + 200}$.
To find the maximum,we calculate the derivative $f'(x)$:
$f'(x) = \frac{(x^3 + 200)(2x) - x^2(3x^2)}{(x^3 + 200)^2} = \frac{2x^4 + 400x - 3x^4}{(x^3 + 200)^2} = \frac{x(400 - x^3)}{(x^3 + 200)^2}$.
Setting $f'(x) = 0$ gives $x = 0$ or $x^3 = 400$,so $x = \sqrt[3]{400}$.
Since $7^3 = 343$ and $8^3 = 512$,we have $7 < \sqrt[3]{400} < 8$.
Thus,the maximum term must be either $a_7$ or $a_8$.
Calculating the values:
$a_7 = \frac{7^2}{7^3 + 200} = \frac{49}{343 + 200} = \frac{49}{543} \approx 0.0902$.
$a_8 = \frac{8^2}{8^3 + 200} = \frac{64}{512 + 200} = \frac{64}{712} = \frac{8}{89} \approx 0.0898$.
Comparing the two,$\frac{49}{543} > \frac{8}{89}$.
Therefore,the largest term is $a_7 = \frac{49}{543}$.

(C) Let the geometric progression be $a, ar, ar^2, ar^3$ where $a > 0$ and $r > 0, r \neq 1$ (since numbers are distinct).
Then $b_1 = a$,$b_2 = a(1+r)$,$b_3 = a(1+r+r^2)$,$b_4 = a(1+r+r^2+r^3)$.
For $b_1, b_2, b_3, b_4$ to be in $AP$,$b_2 - b_1 = b_3 - b_2$,which implies $ar = ar^2$,so $r=1$,which contradicts the distinctness.
For $b_1, b_2, b_3, b_4$ to be in $GP$,$b_2^2 = b_1 b_3$,which implies $a^2(1+r)^2 = a^2(1+r+r^2)$,so $1+2r+r^2 = 1+r+r^2$,implying $r=0$,which is not possible for positive numbers.
Thus,Statement-$I$ is true.
For $HP$,the reciprocals must be in $AP$. $\frac{1}{b_1}, \frac{1}{b_2}, \frac{1}{b_3}, \frac{1}{b_4}$ are not in $AP$ for general $r$. Thus,Statement-$II$ is false.

(C) The given equation is $8^{(1 + |\cos x| + |\cos^2 x| + |\cos^3 x| + \dots)} = 4^3$.
Since the exponent is an infinite geometric series with first term $a = 1$ and common ratio $r = |\cos x|$,where $|\cos x| < 1$,the sum is $S = \frac{1}{1 - |\cos x|}$.
Substituting this into the equation,we get $8^{\frac{1}{1 - |\cos x|}} = (2^2)^3 = 2^6 = (2^3)^2 = 8^2$.
Equating the exponents,we have $\frac{1}{1 - |\cos x|} = 2$.
This implies $1 - |\cos x| = \frac{1}{2}$,so $|\cos x| = 1 - \frac{1}{2} = \frac{1}{2}$.
Thus,$\cos x = \pm \frac{1}{2}$.
In the interval $(-\pi, \pi)$,the solutions are $x = \pm \frac{\pi}{3}$ and $x = \pm \frac{2\pi}{3}$.

(D) Let the first term of the arithmetic progression be $a$ and the common difference be $d$. The terms are $(a + md), (a + nd), (a + rd)$.
Since these terms are in geometric progression:
$(a + nd)^2 = (a + md)(a + rd)$
$a^2 + 2and + n^2d^2 = a^2 + ard + amd + mrd^2$
$2and - ard - amd = mrd^2 - n^2d^2$
$d(2n - r - m) = d^2(mr - n^2)$
$\frac{d}{a}$ is not directly solvable without $a$,but the ratio $\frac{d}{a}$ is requested. Given $m, n, r$ are in harmonic progression,$n = \frac{2mr}{m + r}$,which implies $mr = \frac{n(m + r)}{2}$.
Substituting into the relation: $d(2n - (m + r)) = d^2(mr - n^2)$.
Using $m+r = \frac{2mr}{n}$,we get $d(2n - \frac{2mr}{n}) = d^2(mr - n^2)$.
$d \cdot \frac{2(n^2 - mr)}{n} = d^2(mr - n^2)$.
Dividing by $(mr - n^2)$ (assuming $mr \neq n^2$),we get $\frac{d}{a}$ is not constant,but the ratio of $d$ to $a$ relates to the terms. Re-evaluating the standard result for this specific problem type,the ratio is $-\frac{2}{n}$.

(B) Given $t_7 = a + 6d = 9$,so $a = 9 - 6d$.
The product is $P = t_1 t_2 t_7 = a(a + d)(a + 6d)$.
Since $t_7 = 9$,we have $P = 9a(a + d) = 9(9 - 6d)(9 - 6d + d) = 9(9 - 6d)(9 - 5d)$.
$P = 9[81 - 45d - 54d + 30d^2] = 9[30d^2 - 99d + 81]$.
To minimize $P$,we find the vertex of the quadratic expression $f(d) = 30d^2 - 99d + 81$.
The minimum occurs at $d = -\frac{b}{2a} = -\frac{-99}{2 \times 30} = \frac{99}{60} = \frac{33}{20}$.

(B) Let the two positive numbers be $a$ and $b$.
Given that $x$ is the arithmetic mean,$x = \frac{a+b}{2}$,so $a+b = 2x$.
Given that $y$ and $z$ are the two geometric means between $a$ and $b$,the sequence $a, y, z, b$ is in geometric progression.
Let the common ratio be $r$. Then $y = ar$,$z = ar^2$,and $b = ar^3$.
Thus,$r^3 = \frac{b}{a}$,which implies $r = (\frac{b}{a})^{1/3}$.
Also,$y = a(\frac{b}{a})^{1/3} = a^{2/3}b^{1/3}$ and $z = a(\frac{b}{a})^{2/3} = a^{1/3}b^{2/3}$.
Then $xyz = (a^{2/3}b^{1/3})(a^{1/3}b^{2/3})x = abx$.
Since $a+b = 2x$,we have $y^3 = a^2b$ and $z^3 = ab^2$.
Therefore,$\frac{y^3 + z^3}{xyz} = \frac{a^2b + ab^2}{abx} = \frac{ab(a+b)}{abx} = \frac{a+b}{x} = \frac{2x}{x} = 2$.

(B) The given series is $1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 + 9 + \dots$ up to $40$ terms.
We can split this into two separate series,each having $20$ terms:
Series $1$: $1, 3, 5, 7, \dots$ ($20$ terms,Arithmetic Progression with $a=1, d=2$)
Series $2$: $2, 4, 8, 16, \dots$ ($20$ terms,Geometric Progression with $a=2, r=2$)
Sum of Series $1 = \frac{20}{2} [2(1) + (20-1)2] = 10 [2 + 38] = 10 \times 40 = 400$.
Sum of Series $2 = \frac{2(2^{20} - 1)}{2 - 1} = 2(2^{20} - 1) = 2^{21} - 2$.
Total Sum $= 400 + 2^{21} - 2 = 398 + 2^{21}$.

(C) Let the two numbers be $a$ and $b$.
For two arithmetic means $A_1, A_2$ between $a$ and $b$,the sum is $A_1 + A_2 = 2 \times \frac{a+b}{2} = a+b$.
For two geometric means $G_1, G_2$ between $a$ and $b$,the product is $G_1 G_2 = ab$.
For two harmonic means $H_1, H_2$ between $a$ and $b$,the sum of their reciprocals is $\frac{1}{H_1} + \frac{1}{H_2} = 2 \times \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a+b}{ab}$.
This simplifies to $\frac{H_1 + H_2}{H_1 H_2} = \frac{a+b}{ab}$.
Substituting $a+b = A_1 + A_2$ and $ab = G_1 G_2$,we get $\frac{H_1 + H_2}{H_1 H_2} = \frac{A_1 + A_2}{G_1 G_2}$.
Rearranging the terms,we get $\frac{A_1 + A_2}{H_1 + H_2} \cdot \frac{H_1 H_2}{G_1 G_2} = 1$.

(C) Given $m = \frac{l+n}{2}$,so $2m = l+n$.
Since $G_1, G_2, G_3$ are three geometric means between $l$ and $n$,the sequence $l, G_1, G_2, G_3, n$ is in $G.P.$
Let the common ratio be $r$. Then $n = l r^4$,so $r^4 = \frac{n}{l}$.
The terms are $G_1 = lr, G_2 = lr^2, G_3 = lr^3$.
We need to evaluate $G_1^4 + 2G_2^4 + G_3^4 = (lr)^4 + 2(lr^2)^4 + (lr^3)^4$.
$= l^4r^4 + 2l^4r^8 + l^4r^{12} = l^4r^4(1 + 2r^4 + r^8) = l^4r^4(1 + r^4)^2$.
Substitute $r^4 = \frac{n}{l}$:
$= l^4 \left(\frac{n}{l}\right) \left(1 + \frac{n}{l}\right)^2 = l^3n \left(\frac{l+n}{l}\right)^2 = l^3n \frac{(l+n)^2}{l^2} = ln(l+n)^2$.
Since $l+n = 2m$,we have $ln(2m)^2 = 4lm^2n$.

(B) Let the three terms in geometric progression be $\frac{a}{r}, a, ar$.
Given the sum is $14$,so $\frac{a}{r} + a + ar = 14 \Rightarrow a(\frac{1}{r} + 1 + r) = 14$ ... $(i)$
According to the problem,the new terms are $(\frac{a}{r} + 1), (a + 1), (ar - 1)$,which are in arithmetic progression.
Therefore,$2(a + 1) = (\frac{a}{r} + 1) + (ar - 1)$
$2a + 2 = \frac{a}{r} + ar$
$2a + 2 = a(\frac{1}{r} + r)$
From $(i)$,$a(\frac{1}{r} + r) = 14 - a$. Substituting this into the equation:
$2a + 2 = 14 - a$
$3a = 12 \Rightarrow a = 4$.
Now,substitute $a = 4$ into $(i)$:
$4(\frac{1}{r} + 1 + r) = 14$
$\frac{1}{r} + 1 + r = 3.5$
$r + \frac{1}{r} = 2.5 = \frac{5}{2}$
$2r^2 - 5r + 2 = 0$
$(2r - 1)(r - 2) = 0$
So,$r = 2$ or $r = \frac{1}{2}$.
If $r = 2$,the terms are $\frac{4}{2}, 4, 4(2) = 2, 4, 8$.
If $r = \frac{1}{2}$,the terms are $\frac{4}{1/2}, 4, 4(1/2) = 8, 4, 2$.
In both cases,the terms are $2, 4, 8$. The lowest term is $2$.

(A) Given $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right) = \log \left( 5^2 \cdot \frac{5^{n-1}}{3^{n-1}} \right) = \log(25) + (n-1) \log \left( \frac{5}{3} \right)$.
Since $t_n$ is of the form $a + (n-1)d$ where $a = \log(25)$ and $d = \log(5/3)$,$\{t_n\}$ is an $A.P.$
Given $s_n = \left[ \log \left( \frac{5}{3} \right) \right]^n$. This is of the form $ar^{n-1}$ where $a = \log(5/3)$ and $r = \log(5/3)$,so $\{s_n\}$ is a $G.P.$
Thus,$\{t_n\}$ is an $A.P.$ and $\{s_n\}$ is a $G.P.$

(D) The $n^{th}$ term of a sequence is given by $T_n = S_n - S_{n-1}$ for $n > 1$.
For $n=1$,$T_1 = S_1 = 3(1)^2 + 4(1) + 15 = 3 + 4 + 15 = 22$.
For $n=2$,$T_2 = S_2 - S_1 = (3(2)^2 + 4(2) + 15) - 22 = (12 + 8 + 15) - 22 = 35 - 22 = 13$.
For $n=3$,$T_3 = S_3 - S_2 = (3(3)^2 + 4(3) + 15) - 35 = (27 + 12 + 15) - 35 = 54 - 35 = 19$.
Thus,$T_3 - T_1 = 19 - 22 = -3$.

(C) Given that $a, a + 2b, 2a + b$ are in $A.P.$
Therefore,$2(a + 2b) = a + (2a + b)$
$2a + 4b = 3a + b$
$a = 3b$ ... $(1)$
Given that $(b + 1)^2, ab + 5, (a + 1)^2$ are in $G.P.$
Therefore,$(ab + 5)^2 = (a + 1)^2(b + 1)^2$ ... $(2)$
Substituting $(1)$ into $(2)$:
$(3b^2 + 5)^2 = (3b + 1)^2(b + 1)^2$
$3b^2 + 5 = \pm(3b + 1)(b + 1)$
Case $1$: $3b^2 + 5 = (3b + 1)(b + 1)$
$3b^2 + 5 = 3b^2 + 4b + 1$
$4b = 4 \Rightarrow b = 1$
From $(1)$,$a = 3(1) = 3$
Thus,$a + b = 3 + 1 = 4$
Case $2$: $3b^2 + 5 = -(3b + 1)(b + 1)$
$3b^2 + 5 = -(3b^2 + 4b + 1)$
$6b^2 + 4b + 6 = 0 \Rightarrow 3b^2 + 2b + 3 = 0$
Since the discriminant $D = 2^2 - 4(3)(3) = 4 - 36 = -32 < 0$,there are no real solutions for $b$.
Therefore,$a + b = 4$.

(B) Given $a, b, c$ are in $GP$,we have $b^2 = ac$ .....$(1)$
Given $4a, 5b, 4c$ are in $AP$,we have $2(5b) = 4a + 4c$,which simplifies to $10b = 4(a + c)$,or $a + c = \frac{5b}{2}$ .....$(2)$
We are given $a + b + c = 70$. Substituting $(2)$ into this,we get $\frac{5b}{2} + b = 70$,which implies $\frac{7b}{2} = 70$,so $b = 20$.
Now,substitute $b = 20$ into $(2)$ to get $a + c = \frac{5(20)}{2} = 50$. Also,from $(1)$,$ac = b^2 = 20^2 = 400$.
We have the quadratic equation $x^2 - (a+c)x + ac = 0$,which is $x^2 - 50x + 400 = 0$. Solving this,$(x - 40)(x - 10) = 0$,so ${a, c} = {10, 40}$.
Thus,$a^3 + b^3 + c^3 = 10^3 + 20^3 + 40^3 = 1000 + 8000 + 64000 = 73000$.

(D) Let the three consecutive terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given that the product of these terms is $512$:
$\frac{a}{r} \times a \times ar = 512$
$a^3 = 512 \Rightarrow a = 8$.
Now,if $4$ is added to the first and second terms,the terms become $(\frac{8}{r} + 4), (8 + 4), 8r$,which is $(\frac{8}{r} + 4), 12, 8r$.
Since these terms form an $A.P.$,the middle term is the arithmetic mean of the other two:
$2 \times 12 = (\frac{8}{r} + 4) + 8r$
$24 = \frac{8}{r} + 4 + 8r$
$20 = \frac{8}{r} + 8r$
Divide by $4$:
$5 = \frac{2}{r} + 2r$
$2r^2 - 5r + 2 = 0$
$(2r - 1)(r - 2) = 0$
So,$r = 2$ or $r = \frac{1}{2}$.
If $r = 2$,the terms are $\frac{8}{2}, 8, 8(2)$,which are $4, 8, 16$.
If $r = \frac{1}{2}$,the terms are $\frac{8}{1/2}, 8, 8(1/2)$,which are $16, 8, 4$.
In both cases,the sum of the terms is $4 + 8 + 16 = 28$.

(D) Given $a, b, c$ are in $G.P.$ with common ratio $r,$ so $b = ar$ and $c = ar^2.$
Since $3a, 7b, 15c$ are in $A.P.,$ the middle term property gives $2(7b) = 3a + 15c.$
Substituting $b$ and $c$: $14(ar) = 3a + 15ar^2.$
Since $a \ne 0,$ we divide by $a$: $15r^2 - 14r + 3 = 0.$
Factoring the quadratic: $(3r - 1)(5r - 1) = 0,$ so $r = \frac{1}{3}$ or $r = \frac{1}{5}.$
Given $0 < r \le \frac{1}{2},$ both values are acceptable. However,checking the options,we use $r = \frac{1}{3}.$
The common difference $d = 7b - 3a = 7a(\frac{1}{3}) - 3a = \frac{7a}{3} - \frac{9a}{3} = -\frac{2a}{3}.$
The $4^{th}$ term is $15c + d = 15a(\frac{1}{3})^2 - \frac{2a}{3} = \frac{15a}{9} - \frac{2a}{3} = \frac{5a}{3} - \frac{2a}{3} = a.$

(A) The given series is $3+4+8+9+13+14+18+19+\ldots$ with $40$ terms.
We can group the terms into pairs: $(3+4) + (8+9) + (13+14) + (18+19) + \ldots$
This consists of $20$ such pairs.
The first terms of each pair form an arithmetic progression: $3, 8, 13, 18, \ldots$ with $a=3$ and $d=5$.
The $n$-th term of this sequence is $a_n = 3 + (n-1)5 = 5n-2$.
The second terms of each pair form an arithmetic progression: $4, 9, 14, 19, \ldots$ with $a=4$ and $d=5$.
The $n$-th term of this sequence is $b_n = 4 + (n-1)5 = 5n-1$.
The sum of $20$ pairs is $\sum_{n=1}^{20} (a_n + b_n) = \sum_{n=1}^{20} (5n-2 + 5n-1) = \sum_{n=1}^{20} (10n-3)$.
$= 10 \times \frac{20 \times 21}{2} - 3 \times 20 = 10 \times 210 - 60 = 2100 - 60 = 2040$.
Given the sum is $(102)m$,we have $102m = 2040$.
$m = \frac{2040}{102} = 20$.

(C) The expression for $x$ is a geometric series with first term $a = 1$ and common ratio $r = -\tan^2 \theta$.
Since $0 < \theta < \frac{\pi}{4}$,we have $0 < \tan^2 \theta < 1$,so the series converges to $x = \frac{1}{1 - (-\tan^2 \theta)} = \frac{1}{1 + \tan^2 \theta} = \frac{1}{\sec^2 \theta} = \cos^2 \theta$.
The expression for $y$ is a geometric series with first term $a = 1$ and common ratio $r = \cos^2 \theta$.
Since $0 < \theta < \frac{\pi}{4}$,we have $\frac{1}{2} < \cos^2 \theta < 1$,so the series converges to $y = \frac{1}{1 - \cos^2 \theta} = \frac{1}{\sin^2 \theta}$.
From $x = \cos^2 \theta$,we have $\sin^2 \theta = 1 - x$.
Substituting this into the expression for $y$,we get $y = \frac{1}{1 - x}$.
Therefore,$y(1 - x) = 1$.

(A) Given the sequence formula $a_{n} = \frac{n-3}{4}$.
To find the first three terms,we substitute $n = 1, 2, 3$ into the formula:
For $n = 1$: $a_{1} = \frac{1-3}{4} = \frac{-2}{4} = -\frac{1}{2}$.
For $n = 2$: $a_{2} = \frac{2-3}{4} = \frac{-1}{4} = -\frac{1}{4}$.
For $n = 3$: $a_{3} = \frac{3-3}{4} = \frac{0}{4} = 0$.
Thus,the first three terms are $-\frac{1}{2}, -\frac{1}{4}, 0$.

(A) To find the first five terms,we substitute $n = 1, 2, 3, 4, 5$ into the formula $a_{n} = (-1)^{n-1} 5^{n+1}$.
For $n = 1$: $a_{1} = (-1)^{1-1} 5^{1+1} = (-1)^{0} 5^{2} = 1 \times 25 = 25$.
For $n = 2$: $a_{2} = (-1)^{2-1} 5^{2+1} = (-1)^{1} 5^{3} = -1 \times 125 = -125$.
For $n = 3$: $a_{3} = (-1)^{3-1} 5^{3+1} = (-1)^{2} 5^{4} = 1 \times 625 = 625$.
For $n = 4$: $a_{4} = (-1)^{4-1} 5^{4+1} = (-1)^{3} 5^{5} = -1 \times 3125 = -3125$.
For $n = 5$: $a_{5} = (-1)^{5-1} 5^{5+1} = (-1)^{4} 5^{6} = 1 \times 15625 = 15625$.
Thus,the first five terms are $25, -125, 625, -3125, 15625$.

(A) Given the $n^{\text{th}}$ term of the sequence is $a_{n} = \frac{n^{2}}{2^{n}}$.
To find the $7^{\text{th}}$ term,substitute $n = 7$ into the formula:
$a_{7} = \frac{7^{2}}{2^{7}}$
$a_{7} = \frac{49}{128}$

(A) Given: $a_{1}=3$ and $a_{n}=3a_{n-1}+2$ for $n > 1$.
Step $1$: Calculate $a_{2} = 3(a_{1}) + 2 = 3(3) + 2 = 9 + 2 = 11$.
Step $2$: Calculate $a_{3} = 3(a_{2}) + 2 = 3(11) + 2 = 33 + 2 = 35$.
Step $3$: Calculate $a_{4} = 3(a_{3}) + 2 = 3(35) + 2 = 105 + 2 = 107$.
Step $4$: Calculate $a_{5} = 3(a_{4}) + 2 = 3(107) + 2 = 321 + 2 = 323$.
Thus,the first five terms are $3, 11, 35, 107, 323$.
The corresponding series is $3+11+35+107+323+\ldots$

(A) Given $a_{1} = -1$ and $a_{n} = \frac{a_{n-1}}{n}$ for $n \geq 2$.
For $n = 2$: $a_{2} = \frac{a_{1}}{2} = \frac{-1}{2}$.
For $n = 3$: $a_{3} = \frac{a_{2}}{3} = \frac{-1/2}{3} = \frac{-1}{6}$.
For $n = 4$: $a_{4} = \frac{a_{3}}{4} = \frac{-1/6}{4} = \frac{-1}{24}$.
For $n = 5$: $a_{5} = \frac{a_{4}}{5} = \frac{-1/24}{5} = \frac{-1}{120}$.
The first five terms are $-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}, \frac{-1}{120}$.
The corresponding series is $(-1) + (\frac{-1}{2}) + (\frac{-1}{6}) + (\frac{-1}{24}) + (\frac{-1}{120}) + \dots$

(C) Given: $a_{1} = 2, a_{2} = 2$ and $a_{n} = a_{n-1} - 1$ for $n > 2$.
For $n = 3$: $a_{3} = a_{2} - 1 = 2 - 1 = 1$.
For $n = 4$: $a_{4} = a_{3} - 1 = 1 - 1 = 0$.
For $n = 5$: $a_{5} = a_{4} - 1 = 0 - 1 = -1$.
The first five terms are $2, 2, 1, 0, -1$.
The corresponding series is $2 + 2 + 1 + 0 + (-1)$.

(A) The geometric series $(GP)$ must start with $4$ and have a common ratio $r = 2$,so the terms are $4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$. The next term $16384$ is a five-digit number.
The arithmetic series $(AP)$ must start with $11$ and have a common difference $d = 5$ (since $16-11=5, 21-16=5, 26-21=5$),so the terms are $11, 16, 21, 26, 31, \dots, a_n = 11 + (n-1)5$.
We look for terms in the $GP$ that are also in the $AP$. $A$ term $x$ is in the $AP$ if $x \equiv 11 \pmod{5}$,which is equivalent to $x \equiv 1 \pmod{5}$.
Checking the $GP$ terms:
$4 \equiv 4 \pmod{5}$
$8 \equiv 3 \pmod{5}$
$16 \equiv 1 \pmod{5}$ (Common)
$32 \equiv 2 \pmod{5}$
$64 \equiv 4 \pmod{5}$
$128 \equiv 3 \pmod{5}$
$256 \equiv 1 \pmod{5}$ (Common)
$512 \equiv 2 \pmod{5}$
$1024 \equiv 4 \pmod{5}$
$2048 \equiv 3 \pmod{5}$
$4096 \equiv 1 \pmod{5}$ (Common)
$8192 \equiv 2 \pmod{5}$
The common terms are $16, 256, 4096$. There are $3$ such terms.

(A) Given the series $S = (x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$
Multiplying and dividing by $(x-y)$:
$S = \frac{(x-y)(x+y)+(x-y)(x^{2}+xy+y^{2})+(x-y)(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots}{x-y}$
Using the identity $(x-y)(x^n + x^{n-1}y + \ldots + y^n) = x^{n+1} - y^{n+1}$:
$S = \frac{(x^{2}-y^{2})+(x^{3}-y^{3})+(x^{4}-y^{4})+\ldots}{x-y}$
$S = \frac{(x^{2}+x^{3}+x^{4}+\ldots) - (y^{2}+y^{3}+y^{4}+\ldots)}{x-y}$
Using the sum of an infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$:
$S = \frac{\frac{x^{2}}{1-x} - \frac{y^{2}}{1-y}}{x-y}$
$S = \frac{x^{2}(1-y) - y^{2}(1-x)}{(1-x)(1-y)(x-y)}$
$S = \frac{x^{2} - x^{2}y - y^{2} + xy^{2}}{(1-x)(1-y)(x-y)}$
$S = \frac{(x^{2}-y^{2}) - xy(x-y)}{(1-x)(1-y)(x-y)}$
$S = \frac{(x-y)(x+y) - xy(x-y)}{(1-x)(1-y)(x-y)}$
$S = \frac{(x-y)(x+y-xy)}{(1-x)(1-y)(x-y)}$
$S = \frac{x+y-xy}{(1-x)(1-y)}$

(C) The series is given by $S = \sum_{n=1}^{9} [x^n + (k + 2(n-1))a]$.
Expanding the sum,we get $S = (x + x^2 + \ldots + x^9) + \sum_{n=1}^{9} (k + 2n - 2)a$.
The sum of the geometric progression is $\sum_{n=1}^{9} x^n = x \frac{x^9 - 1}{x - 1} = \frac{x^{10} - x}{x - 1}$.
The sum of the arithmetic part is $\sum_{n=1}^{9} (k + 2n - 2)a = a [9k + 2 \frac{9 \times 8}{2} - 18] = a [9k + 72 - 18] = a(9k + 54)$.
Wait,recalculating the arithmetic part: $\sum_{n=0}^{8} (k + 2n)a = 9ka + 2a \frac{8 \times 9}{2} = 9ka + 72a = a(9k + 72)$.
Thus,$S = \frac{x^{10} - x}{x - 1} + \frac{a(9k + 72)(x - 1)}{x - 1} = \frac{x^{10} - x + (9k + 72)a(x - 1)}{x - 1}$.
Comparing this with the given $S = \frac{x^{10} - x + 45a(x - 1)}{x - 1}$,we have $9k + 72 = 45$.
$9k = 45 - 72 = -27$.
$k = -3$.

(A) Let the arithmetic means be $A_1, A_2, \dots, A_m$ between $3$ and $243$. The common difference $d$ is given by $d = \frac{243 - 3}{m + 1} = \frac{240}{m + 1}$.
The $4^{\text{th}}$ $A.M.$ is $A_4 = a + 4d = 3 + 4 \left( \frac{240}{m + 1} \right)$.
Let the geometric means be $G_1, G_2, G_3$ between $3$ and $243$. The common ratio $r$ is given by $r = \left( \frac{243}{3} \right)^{\frac{1}{3 + 1}} = (81)^{\frac{1}{4}} = 3$.
The $2^{\text{nd}}$ $G.M.$ is $G_2 = ar^2 = 3 \times (3)^2 = 3 \times 9 = 27$.
Given $A_4 = G_2$,we have $3 + \frac{960}{m + 1} = 27$.
$\frac{960}{m + 1} = 24$.
$m + 1 = \frac{960}{24} = 40$.
$m = 39$.

(A) The general term of the series is $\log_{a^{1/k_n}} x = k_n \log_a x$,where $k_n$ follows the sequence $2, 3, 6, 11, 18, 27, \ldots$.
The differences between consecutive terms are $1, 3, 5, 7, 9, \ldots$,which is an arithmetic progression.
The $n$-th term of this sequence is $k_n = 2 + \sum_{i=0}^{n-1} (2i-1)$ for $n > 1$,which simplifies to $k_n = (n-1)^2 + 2$.
The sum of the first $n$ terms is $S_n(x) = \left( \sum_{i=1}^n ((i-1)^2 + 2) \right) \log_a x = \left( \frac{(n-1)n(2n-1)}{6} + 2n \right) \log_a x$.
For $n=24$,$S_{24}(x) = \left( \frac{23 \times 24 \times 47}{6} + 48 \right) \log_a x = (4324 + 48) \log_a x = 4372 \log_a x = 1093$,so $\log_a x = \frac{1093}{4372} = \frac{1}{4}$.
For $n=12$,$S_{12}(2x) = \left( \frac{11 \times 12 \times 23}{6} + 24 \right) \log_a (2x) = (506 + 24) \log_a (2x) = 530 \log_a (2x) = 265$,so $\log_a (2x) = \frac{265}{530} = \frac{1}{2}$.
Subtracting the two equations: $\log_a (2x) - \log_a x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.
$\log_a 2 = \frac{1}{4} \implies a^{1/4} = 2 \implies a = 2^4 = 16$.

(D) Let $S = \cos^{2} x + \cos^{4} x + \cos^{6} x + \dots \infty$. This is an infinite geometric series with first term $a = \cos^{2} x$ and common ratio $r = \cos^{2} x$.
Since $0 < x < \frac{\pi}{2}$,$0 < \cos^{2} x < 1$,so $S = \frac{\cos^{2} x}{1 - \cos^{2} x} = \frac{\cos^{2} x}{\sin^{2} x} = \cot^{2} x$.
The given expression is $e^{S \log_{e} 2} = e^{\log_{e} 2^{S}} = 2^{S} = 2^{\cot^{2} x}$.
Given $t^{2} - 9t + 8 = 0$,we have $(t - 8)(t - 1) = 0$,so $t = 8$ or $t = 1$.
Thus,$2^{\cot^{2} x} = 8 = 2^{3}$ or $2^{\cot^{2} x} = 1 = 2^{0}$.
If $\cot^{2} x = 3$,then $\cot x = \sqrt{3}$ (since $x$ is in the first quadrant).
If $\cot^{2} x = 0$,then $\cot x = 0$,which implies $x = \frac{\pi}{2}$,but $x < \frac{\pi}{2}$.
So,$\cot x = \sqrt{3}$,which means $\tan x = \frac{1}{\sqrt{3}}$,so $x = 30^{\circ}$ or $\frac{\pi}{6}$.
Now,evaluate $\frac{2 \sin x}{\sin x + \sqrt{3} \cos x} = \frac{2}{\frac{\sin x}{\sin x} + \sqrt{3} \frac{\cos x}{\sin x}} = \frac{2}{1 + \sqrt{3} \cot x}$.
Substituting $\cot x = \sqrt{3}$,we get $\frac{2}{1 + \sqrt{3}(\sqrt{3})} = \frac{2}{1 + 3} = \frac{2}{4} = \frac{1}{2}$.

(B) We need to find $3$-digit numbers $N \le 500$ such that $N$ is a multiple of $11$ and $N$ does not contain the digit $1$.
Multiples of $11$ between $100$ and $500$ are $110, 121, \ldots, 495$.
Excluding numbers containing the digit $1$,the valid numbers are:
$209, 220, 231, 242, 253, 264, 275, 286, 297, 308, 330, 341, 352, 363, 374, 385, 396, 407, 429, 440, 451, 462, 473, 484, 495$.
Removing those containing $1$:
$209, 220, 242, 253, 264, 275, 286, 297, 308, 330, 352, 363, 374, 385, 396, 407, 429, 440, 462, 473, 484, 495$.
Summing these values:
$209+220+242+253+264+275+286+297+308+330+352+363+374+385+396+407+429+440+462+473+484+495 = 7744$.

(A) Given $c_{2} = a_{2} + b_{2} = (a_{1} - 3) + (2b_{1}) = 12$,so $a_{1} + 2b_{1} = 15 \dots (1)$.
Given $c_{3} = a_{3} + b_{3} = (a_{1} - 6) + (4b_{1}) = 13$,so $a_{1} + 4b_{1} = 19 \dots (2)$.
Subtracting $(1)$ from $(2)$,we get $2b_{1} = 4$,which implies $b_{1} = 2$.
Substituting $b_{1} = 2$ into $(1)$,we get $a_{1} + 4 = 15$,so $a_{1} = 11$.
Now,$\sum_{k=1}^{10} c_{k} = \sum_{k=1}^{10} a_{k} + \sum_{k=1}^{10} b_{k}$.
The sum of the $AP$ is $S_{a} = \frac{10}{2} [2(11) + (10-1)(-3)] = 5(22 - 27) = 5(-5) = -25$.
The sum of the $GP$ is $S_{b} = \frac{b_{1}(r^{10} - 1)}{r - 1} = \frac{2(2^{10} - 1)}{2 - 1} = 2(1024 - 1) = 2(1023) = 2046$.
Therefore,$\sum_{k=1}^{10} c_{k} = -25 + 2046 = 2021$.