Progression and Sequence Questions in English

(C) Let the first term be $a$ and the common difference be $d$.
The terms of an $A.P.$ are given by $a_n = a + (n-1)d$.
Given the equation: $a_4 - a_7 + a_{10} = m$.
Substituting the general form:
$(a + 3d) - (a + 6d) + (a + 9d) = m$
$a + 3d - a - 6d + a + 9d = m$
$a + 6d = m$
Note that $a_7 = a + (7-1)d = a + 6d$,so $a_7 = m$.
The sum of the first $13$ terms $(S_{13})$ is given by the formula:
$S_{13} = \frac{13}{2} [2a + (13-1)d]$
$S_{13} = \frac{13}{2} [2a + 12d]$
$S_{13} = 13(a + 6d)$
Since $a + 6d = m$,we have:
$S_{13} = 13m$.

(B) Let the sequence be $a, b, c, d$.
Since the first three terms $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Since the last three terms $b, c, d$ are in $A.P.$ with a common difference of $6$,we have $c - b = 6$ and $d - c = 6$.
Given that the first and last terms are equal,$a = d$.
From $d - c = 6$,we have $d = c + 6$. Since $a = d$,then $a = c + 6$,which implies $c = a - 6$.
From $c - b = 6$,we have $b = c - 6 = (a - 6) - 6 = a - 12$.
Substituting $b = a - 12$ and $c = a - 6$ into the $G.P.$ condition $b^2 = ac$:
$(a - 12)^2 = a(a - 6)$
$a^2 - 24a + 144 = a^2 - 6a$
$144 = 18a$
$a = 8$.
Since $a = d$,the last term $d = 8$.

(A) The given series is $1^2 + 3^2 + 5^2 + ....... + 25^2$.
The $n^{th}$ term is $T_n = (2n - 1)^2$. For the last term $25$,we have $2n - 1 = 25$,which gives $n = 13$.
The sum $S_{13} = \sum_{n=1}^{13} (2n - 1)^2 = \sum_{n=1}^{13} (4n^2 - 4n + 1)$.
Using the summation formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$:
$S_{13} = 4 \sum_{n=1}^{13} n^2 - 4 \sum_{n=1}^{13} n + \sum_{n=1}^{13} 1$
$S_{13} = 4 \left[ \frac{13(13+1)(2 \times 13 + 1)}{6} \right] - 4 \left[ \frac{13(13+1)}{2} \right] + 13$
$S_{13} = 4 \left[ \frac{13 \times 14 \times 27}{6} \right] - 2(13 \times 14) + 13$
$S_{13} = 4 \times (13 \times 7 \times 9) - 2(182) + 13$
$S_{13} = 4 \times 819 - 364 + 13$
$S_{13} = 3276 - 364 + 13 = 2925$.

(C) Let the terms of the Geometric Progression $(G.P.)$ be $a, ar, ar^2, ar^3, ar^4, ar^5$,where $a$ is the first term and $r$ is the common ratio.
According to the given conditions:
$1$) The difference between the fourth term and the first term is $52$: $ar^3 - a = 52 \Rightarrow a(r^3 - 1) = 52$ ... $(1)$
$2$) The sum of the first three terms is $26$: $a + ar + ar^2 = 26 \Rightarrow a(1 + r + r^2) = 26$ ... $(2)$
We know that $r^3 - 1 = (r - 1)(r^2 + r + 1)$.
Dividing equation $(1)$ by equation $(2)$:
$\frac{a(r^3 - 1)}{a(1 + r + r^2)} = \frac{52}{26}$
$\frac{(r - 1)(r^2 + r + 1)}{(r^2 + r + 1)} = 2$
$r - 1 = 2 \Rightarrow r = 3$.
Substituting $r = 3$ into equation $(2)$:
$a(1 + 3 + 3^2) = 26 \Rightarrow a(1 + 3 + 9) = 26 \Rightarrow 13a = 26 \Rightarrow a = 2$.
The sum of the first six terms is $S_6 = a(1 + r + r^2 + r^3 + r^4 + r^5) = a(1 + r + r^2)(1 + r^3)$.
$S_6 = 2(1 + 3 + 9)(1 + 3^3) = 2(13)(1 + 27) = 26 \times 28 = 728$.

(A) The given series is $S = 1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \dots + 2(2m)^2$.
We can group the odd and even terms:
$S = (1^2 + 3^2 + 5^2 + \dots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \dots + (2m)^2)$.
$S = \sum_{k=1}^{m} (2k-1)^2 + 2 \sum_{k=1}^{m} (2k)^2$.
$S = \sum_{k=1}^{m} (4k^2 - 4k + 1) + 8 \sum_{k=1}^{m} k^2$.
$S = 4 \sum k^2 - 4 \sum k + \sum 1 + 8 \sum k^2 = 12 \sum_{k=1}^{m} k^2 - 4 \sum_{k=1}^{m} k + \sum_{k=1}^{m} 1$.
Using standard summation formulas:
$S = 12 \frac{m(m+1)(2m+1)}{6} - 4 \frac{m(m+1)}{2} + m$.
$S = 2m(m+1)(2m+1) - 2m(m+1) + m$.
$S = m [2(m+1)(2m+1) - 2(m+1) + 1]$.
$S = m [2(2m^2 + 3m + 1) - 2m - 2 + 1]$.
$S = m [4m^2 + 6m + 2 - 2m - 1] = m(4m^2 + 4m + 1) = m(2m+1)^2$.

(C) The given series is $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots$
The $n^{th}$ term is $T_n = \frac{1}{\sqrt{n} + \sqrt{n+1}}$.
By rationalizing the denominator,we get:
$T_n = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1} + \sqrt{n})(\sqrt{n+1} - \sqrt{n})} = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}$.
Sum of $15$ terms $S_{15} = \sum_{n=1}^{15} (\sqrt{n+1} - \sqrt{n})$.
Expanding the sum:
$S_{15} = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{16} - \sqrt{15})$.
This is a telescoping series where intermediate terms cancel out:
$S_{15} = -\sqrt{1} + \sqrt{16} = -1 + 4 = 3$.
Hence,the required sum is $3$.

(C) The given series is $S_n = 1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots$ up to $n$ terms.
We can rewrite each term as $1 + \frac{1}{3^k}$ for $k = 0, 1, 2, \dots, n-1$.
Thus,$S_n = \sum_{k=0}^{n-1} (1 + \frac{1}{3^k}) = \sum_{k=0}^{n-1} 1 + \sum_{k=0}^{n-1} (\frac{1}{3})^k$.
The first part is a sum of $n$ ones,which is $n$.
The second part is a geometric progression with first term $a = 1$ and common ratio $r = \frac{1}{3}$.
The sum of the geometric progression is $S = \frac{a(1 - r^n)}{1 - r} = \frac{1(1 - (1/3)^n)}{1 - 1/3} = \frac{1 - 1/3^n}{2/3} = \frac{3}{2}(1 - \frac{1}{3^n}) = \frac{3}{2} - \frac{1}{2 \cdot 3^{n-1}}$.
Wait,simplifying the expression: $S_n = n + \frac{1 - (1/3)^n}{2/3} = n + \frac{3}{2}(1 - \frac{1}{3^n}) = n + \frac{3}{2} - \frac{3}{2 \cdot 3^n} = n + \frac{3}{2} - \frac{1}{2 \cdot 3^{n-1}}$.
Re-evaluating the provided options: The expression $n + \frac{1}{2} - \frac{1}{2 \cdot 3^n}$ is equivalent to $n + \frac{3^n - 1}{2 \cdot 3^n}$.
Let's check $n=1$: $S_1 = 1$. Formula: $1 + 1/2 - 1/6 = 1 + 2/6 = 1.33$ (Incorrect).
Let's re-examine the series: $1, 4/3, 10/9, 28/27$. These are $(3^0+0)/3^0, (3^1+1)/3^1, (3^2+1)/3^2, (3^3+1)/3^3$.
Actually,the terms are $1 + 0, 1 + 1/3, 1 + 1/9, 1 + 1/27$.
Sum $= n + (1/3^0 + 1/3^1 + \dots + 1/3^{n-1}) = n + \frac{1(1 - (1/3)^n)}{1 - 1/3} = n + \frac{3}{2}(1 - \frac{1}{3^n}) = n + \frac{3}{2} - \frac{1}{2 \cdot 3^{n-1}}$.
Given the options,option $C$ is $n + 1/2 - 1/(2 \cdot 3^n)$. This matches if the series starts from $1/3$.

(D) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $p^{th}$ term is $a_p = a + (p - 1)d$.
The $q^{th}$ term is $a_q = a + (q - 1)d$.
The $r^{th}$ term is $a_r = a + (r - 1)d$.
The $s^{th}$ term is $a_s = a + (s - 1)d$.
Given that the $A.M.$ of $a_p$ and $a_q$ is equal to the $A.M.$ of $a_r$ and $a_s$:
$\frac{a_p + a_q}{2} = \frac{a_r + a_s}{2}$
$\Rightarrow a_p + a_q = a_r + a_s$
Substituting the values:
$[a + (p - 1)d] + [a + (q - 1)d] = [a + (r - 1)d] + [a + (s - 1)d]$
$2a + (p + q - 2)d = 2a + (r + s - 2)d$
Subtracting $2a$ from both sides:
$(p + q - 2)d = (r + s - 2)d$
Assuming $d \neq 0$,we can divide by $d$:
$p + q - 2 = r + s - 2$
Therefore,$p + q = r + s$.

(C) The series is $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$
When $n$ is odd,the sum up to $n$ terms is $S_n = (1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot (n-1)^2) + n^2$.
Since $(n-1)$ is even,we use the given formula for the sum of the first $(n-1)$ terms:
$S_{n-1} = \frac{(n-1)((n-1)+1)^2}{2} = \frac{(n-1)n^2}{2}$.
Therefore,the sum for $n$ terms is $S_n = \frac{(n-1)n^2}{2} + n^2$.
$S_n = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.

(D) Let the three distinct numbers in $G.P.$ be $a, ar, ar^2$ where $r \neq 1$ and $r \neq 0$.
Given $a + b + c = xb$,substituting the terms: $a + ar + ar^2 = x(ar)$.
Since $a \neq 0$,we can divide by $a$: $1 + r + r^2 = xr$.
Dividing by $r$: $r + 1 + \frac{1}{r} = x$,which simplifies to $x = r + \frac{1}{r} + 1$.
We know that for any real $r > 0$,$r + \frac{1}{r} \geq 2$,and for $r < 0$,$r + \frac{1}{r} \leq -2$.
Since the numbers are distinct,$r \neq 1$ and $r \neq -1$.
If $r > 0$ and $r \neq 1$,then $r + \frac{1}{r} > 2$,so $x > 3$.
If $r < 0$ and $r \neq -1$,then $r + \frac{1}{r} < -2$,so $x < -1$.
Thus,$x$ cannot lie in the interval $(-1, 3)$.
Among the given options,$2$ lies in the interval $(-1, 3)$,so $x$ cannot be $2$.

(A) Given that $a_1, a_2, ......., a_{30}$ is an $A.P.$ with first term $a$ and common difference $D$.
$S = a_1 + a_2 + a_3 + ....... + a_{30} = \frac{30}{2} [2a + 29D] = 15(2a + 29D) = 30a + 435D$.
$T = a_1 + a_3 + a_5 + ....... + a_{29}$ is an $A.P.$ with $15$ terms,first term $a_1 = a$,and common difference $2D$.
$T = \frac{15}{2} [2a + (15-1)(2D)] = \frac{15}{2} [2a + 28D] = 15(a + 14D) = 15a + 210D$.
Now,$S - 2T = (30a + 435D) - 2(15a + 210D) = 30a + 435D - 30a - 420D = 15D$.
Given $S - 2T = 75$,so $15D = 75$,which implies $D = 5$.
Given $a_5 = 27$,we have $a + 4D = 27$.
Substituting $D = 5$,$a + 4(5) = 27 \Rightarrow a + 20 = 27 \Rightarrow a = 7$.
We need to find $a_{10} = a + 9D$.
$a_{10} = 7 + 9(5) = 7 + 45 = 52$.

(A) The $n^{th}$ term of the series is given by $T_n = \frac{(3 + (n-1) \times 3)(1^2 + 2^2 + ... + n^2)}{(2n + 1)}$.
Using the formula for the sum of squares,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$T_n = \frac{3n \times \frac{n(n+1)(2n+1)}{6}}{2n+1} = \frac{n^2(n+1)}{2} = \frac{n^3 + n^2}{2}$.
To find the sum of $15$ terms,we calculate $S_{15} = \sum_{n=1}^{15} T_n = \frac{1}{2} \sum_{n=1}^{15} (n^3 + n^2)$.
Using the summation formulas $\sum_{n=1}^N n^3 = [\frac{N(N+1)}{2}]^2$ and $\sum_{n=1}^N n^2 = \frac{N(N+1)(2N+1)}{6}$:
$S_{15} = \frac{1}{2} [(\frac{15 \times 16}{2})^2 + \frac{15 \times 16 \times 31}{6}]$
$S_{15} = \frac{1}{2} [120^2 + 1240] = \frac{1}{2} [14400 + 1240] = \frac{15640}{2} = 7820$.

(B) Let the first term of the $A.P.$ be $A$ and the common difference be $d$. Since the $A.P.$ is non-constant,$d \neq 0$.
Given terms are:
$a = A + 6d$
$b = A + 10d$
$c = A + 12d$
Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
$(A + 10d)^2 = (A + 6d)(A + 12d)$
$A^2 + 20Ad + 100d^2 = A^2 + 18Ad + 72d^2$
$2Ad = -28d^2$
Since $d \neq 0$,we divide by $2d$ to get $A = -14d$,or $\frac{A}{d} = -14$.
Now,we calculate $\frac{a}{c}$:
$\frac{a}{c} = \frac{A + 6d}{A + 12d}$
Dividing numerator and denominator by $d$:
$\frac{a}{c} = \frac{\frac{A}{d} + 6}{\frac{A}{d} + 12} = \frac{-14 + 6}{-14 + 12} = \frac{-8}{-2} = 4$.

(D) We know the identity for binomial coefficients: ${}^{n}{C_r} + {}^{n}{C_{r-1}} = {}^{n+1}{C_r}$.
Applying this to the denominator: ${}^{20}{C_i} + {}^{20}{C_{i-1}} = {}^{21}{C_i}$.
Thus,the term inside the summation becomes $\frac{{}^{20}{C_{i-1}}}{{}^{21}{C_i}}$.
Using the formula ${}^{n}{C_r} = \frac{n}{r} \cdot {}^{n-1}{C_{r-1}}$,we have ${}^{21}{C_i} = \frac{21}{i} \cdot {}^{20}{C_{i-1}}$.
Substituting this back: $\frac{{}^{20}{C_{i-1}}}{\frac{21}{i} \cdot {}^{20}{C_{i-1}}} = \frac{i}{21}$.
The summation becomes $\sum\limits_{i=1}^{20} \left( \frac{i}{21} \right)^3 = \frac{1}{21^3} \sum\limits_{i=1}^{20} i^3$.
Using the sum of cubes formula $\sum i^3 = \left( \frac{n(n+1)}{2} \right)^2$,for $n=20$ we get $\left( \frac{20 \times 21}{2} \right)^2 = (10 \times 21)^2 = 100 \times 21^2$.
Therefore,$S = \frac{100 \times 21^2}{21^3} = \frac{100}{21}$.
Given $S = \frac{k}{21}$,we find $k = 100$.

(D) Two-digit numbers of the form $7n + 2$ are $16, 23, \dots, 93$. This is an arithmetic progression with first term $a = 16$, last term $l = 93$, and common difference $d = 7$. The number of terms $n_1$ is given by $93 = 16 + (n_1 - 1)7$, which gives $77 = (n_1 - 1)7$, so $n_1 = 12$. The sum $S_1 = \frac{12}{2}(16 + 93) = 6 \times 109 = 654$.
Two-digit numbers of the form $7n + 5$ are $12, 19, \dots, 96$. This is an arithmetic progression with first term $a = 12$, last term $l = 96$, and common difference $d = 7$. The number of terms $n_2$ is given by $96 = 12 + (n_2 - 1)7$, which gives $84 = (n_2 - 1)7$, so $n_2 = 13$. The sum $S_2 = \frac{13}{2}(12 + 96) = \frac{13}{2} \times 108 = 13 \times 54 = 702$.
The total sum is $S_1 + S_2 = 654 + 702 = 1356$.

(A) We know that $^{n}{C_r} \cdot ^{n-r}{C_k} = ^{n}{C_k} \cdot ^{n-k}{C_{r}}$.
Applying this identity to the given expression:
$^{50}{C_r} \cdot ^{50-r}{C_{25-r}} = ^{50}{C_{25}} \cdot ^{50-25}{C_{r}} = ^{50}{C_{25}} \cdot ^{25}{C_r}$.
Now,substitute this into the summation:
$\sum\limits_{r = 0}^{25} {\left( {^{50}{C_r} \cdot ^{50 - r}{C_{25 - r}}} \right)} = \sum\limits_{r = 0}^{25} {\left( {^{50}{C_{25}} \cdot ^{25}{C_r}} \right)}$.
Since $^{50}{C_{25}}$ is independent of $r$,we can take it out of the summation:
$= ^{50}{C_{25}} \sum\limits_{r = 0}^{25} {^{25}{C_r}}$.
We know that $\sum\limits_{r = 0}^{n} {^{n}{C_r} = 2^n}$. Therefore,$\sum\limits_{r = 0}^{25} {^{25}{C_r} = 2^{25}}$.
Thus,the expression becomes $^{50}{C_{25}} \cdot 2^{25}$.
Comparing this with $K \cdot ^{50}{C_{25}}$,we get $K = 2^{25}$.

(B) Let the infinite geometric series be $a, ar, ar^2, \dots$ where $a > 0$ and $|r| < 1$.
The sum of the series is given by $S = \frac{a}{1-r} = 3 \implies a = 3(1-r)$.
The cubes of the terms are $a^3, (ar)^3, (ar^2)^3, \dots$,which is a new geometric series with first term $a^3$ and common ratio $r^3$.
The sum of the cubes is given by $S' = \frac{a^3}{1-r^3} = \frac{27}{19}$.
Substituting $a = 3(1-r)$ into the second equation:
$\frac{[3(1-r)]^3}{1-r^3} = \frac{27}{19}$
$\frac{27(1-r)^3}{(1-r)(1+r+r^2)} = \frac{27}{19}$
$\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$
$19(1 - 2r + r^2) = 1 + r + r^2$
$19 - 38r + 19r^2 = 1 + r + r^2$
$18r^2 - 39r + 18 = 0$
Dividing by $3$: $6r^2 - 13r + 6 = 0$
$6r^2 - 9r - 4r + 6 = 0$
$3r(2r - 3) - 2(2r - 3) = 0$
$(3r - 2)(2r - 3) = 0$
Since $|r| < 1$,we must have $r = \frac{2}{3}$.

(A) In a $G.P.$,the $n^{th}$ term is given by $a_n = a_1 r^{n-1}$.
Given $\frac{a_3}{a_1} = 25$.
Substituting the formula,we get $\frac{a_1 r^2}{a_1} = r^2 = 25$.
We need to find the value of $\frac{a_9}{a_5}$.
Using the formula,$\frac{a_9}{a_5} = \frac{a_1 r^8}{a_1 r^4} = r^4$.
Since $r^2 = 25$,then $r^4 = (r^2)^2 = (25)^2 = (5^2)^2 = 5^4$.
Therefore,the value is $5^4$.

(D) Given $S_n = \frac{q^{n+1}-1}{q-1}$. The expression is $\sum_{r=1}^{101} {^{101}C_r} S_{r-1}$.
Substituting $S_{r-1} = \frac{q^r-1}{q-1}$,we get $\sum_{r=1}^{101} {^{101}C_r} \frac{q^r-1}{q-1} = \frac{1}{q-1} \left( \sum_{r=1}^{101} {^{101}C_r} q^r - \sum_{r=1}^{101} {^{101}C_r} \right)$.
Using the binomial theorem,$\sum_{r=1}^{101} {^{101}C_r} q^r = (1+q)^{101} - 1$ and $\sum_{r=1}^{101} {^{101}C_r} = 2^{101} - 1$.
Thus,the expression becomes $\frac{1}{q-1} ((1+q)^{101} - 1 - (2^{101} - 1)) = \frac{(1+q)^{101} - 2^{101}}{q-1}$.
Now,$T_{100} = \frac{(\frac{q+1}{2})^{101} - 1}{\frac{q+1}{2} - 1} = \frac{(\frac{q+1}{2})^{101} - 1}{\frac{q-1}{2}} = \frac{2((q+1)^{101} - 2^{101})}{2^{101}(q-1)} = \frac{(q+1)^{101} - 2^{101}}{2^{100}(q-1)}$.
Comparing $\frac{(1+q)^{101} - 2^{101}}{q-1} = \alpha \cdot \frac{(1+q)^{101} - 2^{101}}{2^{100}(q-1)}$,we find $\alpha = 2^{100}$.

(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 $t_n = a + (n - 1)d$.
Given that the $19^{th}$ term is zero,we have $t_{19} = a + 18d = 0$,which implies $a = -18d$.
We need to find the ratio of the $49^{th}$ term to the $29^{th}$ term:
$\frac{t_{49}}{t_{29}} = \frac{a + 48d}{a + 28d}$.
Substituting $a = -18d$ into the expression:
$\frac{t_{49}}{t_{29}} = \frac{-18d + 48d}{-18d + 28d} = \frac{30d}{10d} = 3$.
Thus,the ratio is $3 : 1$.

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

(C) The sum of the first $k$ natural numbers is given by $\frac{k(k+1)}{2}$.
Thus,$S_k = \frac{k(k+1)}{2k} = \frac{k+1}{2}$.
We are given the equation $\sum_{k=1}^{10} S_k^2 = \frac{5}{12}A$.
Substituting $S_k$,we get $\sum_{k=1}^{10} \left(\frac{k+1}{2}\right)^2 = \frac{5}{12}A$.
This expands to $\frac{1}{4} \sum_{k=1}^{10} (k+1)^2 = \frac{5}{12}A$.
Let $n = k+1$. When $k=1, n=2$; when $k=10, n=11$. So,$\frac{1}{4} \sum_{n=2}^{11} n^2 = \frac{5}{12}A$.
We know $\sum_{n=1}^{11} n^2 = \frac{11(11+1)(2 \times 11 + 1)}{6} = \frac{11 \times 12 \times 23}{6} = 11 \times 2 \times 23 = 506$.
Therefore,$\sum_{n=2}^{11} n^2 = 506 - 1^2 = 505$.
Substituting this back: $\frac{1}{4} \times 505 = \frac{5}{12}A$.
$A = 505 \times \frac{12}{4 \times 5} = 505 \times \frac{3}{5} = 101 \times 3 = 303$.

(B) Given that $^nC_4, ^nC_5,$ and $^nC_6$ are in $A.P.$
Therefore,$2(^nC_5) = ^nC_4 + ^nC_6$.
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$
Dividing both sides by $n!$ and multiplying by $6!(n-4)!$:
$2 \cdot \frac{6 \cdot (n-4)}{5} = 6 \cdot 5 + (n-4)(n-5)$
$12(n-4) = 30 + n^2 - 9n + 20$
$12n - 48 = n^2 - 9n + 50$
$n^2 - 21n + 98 = 0$
$(n-7)(n-14) = 0$
Thus,$n = 7$ or $n = 14$. Since $n \ge 6$ is required for $^nC_6$ to be defined,both are possible,but $14$ is given in the options.

(B) The given series is ${\left( {\frac{3}{4}} \right)^3} + {\left( {\frac{6}{4}} \right)^3} + {\left( {\frac{9}{4}} \right)^3} + {\left( {\frac{12}{4}} \right)^3} + \dots$ up to $15$ terms.
This can be written as $\sum_{r=1}^{15} {\left( \frac{3r}{4} \right)^3}$.
Taking the constant out,we get $\left( \frac{3}{4} \right)^3 \sum_{r=1}^{15} r^3 = \frac{27}{64} \sum_{r=1}^{15} r^3$.
Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} \right]^2$.
For $n=15$,$\sum_{r=1}^{15} r^3 = \left[ \frac{15 \times 16}{2} \right]^2 = (15 \times 8)^2 = 120^2 = 14400$.
Thus,the sum $S = \frac{27}{64} \times 14400$.
$S = 27 \times \frac{14400}{64} = 27 \times 225$.
Given that $S = 225k$,we have $225k = 27 \times 225$.
Therefore,$k = 27$.

(D) We need to find the sum of all natural numbers $n$ such that $100 < n < 200$ and $H.C.F. (91, n) > 1$.
Since $91 = 7 \times 13$,the condition $H.C.F. (91, n) > 1$ implies that $n$ must be divisible by $7$ or $13$.
Let $S_A$ be the sum of numbers between $100$ and $200$ divisible by $7$.
The numbers are $105, 112, \dots, 196$.
This is an arithmetic progression with $a = 105$,$l = 196$,and $d = 7$.
Number of terms $n_A = \frac{196 - 105}{7} + 1 = 14$.
$S_A = \frac{14}{2} (105 + 196) = 7 \times 301 = 2107$.
Let $S_B$ be the sum of numbers between $100$ and $200$ divisible by $13$.
The numbers are $104, 117, \dots, 195$.
This is an arithmetic progression with $a = 104$,$l = 195$,and $d = 13$.
Number of terms $n_B = \frac{195 - 104}{13} + 1 = 8$.
$S_B = \frac{8}{2} (104 + 195) = 4 \times 299 = 1196$.
Let $S_C$ be the sum of numbers between $100$ and $200$ divisible by both $7$ and $13$ (i.e.,divisible by $91$).
The only such number is $182$.
So,$S_C = 182$.
By the Principle of Inclusion-Exclusion,the required sum is $S = S_A + S_B - S_C$.
$S = 2107 + 1196 - 182 = 3121$.

(A) Let $S = \sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}} = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + \dots + \frac{{20}}{{{2^{20}}}}$
Multiply by $\frac{1}{2}$:
$\frac{1}{2}S = \frac{1}{{{2^2}}} + \frac{2}{{{2^3}}} + \dots + \frac{{19}}{{{2^{20}}}} + \frac{{20}}{{{2^{21}}}}$
Subtract the two equations:
$S - \frac{1}{2}S = \frac{1}{2} + \left( \frac{2-1}{{{2^2}}} \right) + \left( \frac{3-2}{{{2^3}}} \right) + \dots + \left( \frac{20-19}{{{2^{20}}}} \right) - \frac{{20}}{{{2^{21}}}}$
$\frac{1}{2}S = \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \dots + \frac{1}{{{2^{20}}}} \right) - \frac{{20}}{{{2^{21}}}}$
The term in the bracket is a Geometric Progression with $a = \frac{1}{2}$,$r = \frac{1}{2}$,and $n = 20$:
Sum $= \frac{\frac{1}{2}(1 - (\frac{1}{2})^{20})}{1 - \frac{1}{2}} = 1 - \frac{1}{{{2^{20}}}}$
So,$\frac{1}{2}S = 1 - \frac{1}{{{2^{20}}}} - \frac{20}{{{2^{21}}}} = 1 - \frac{2}{{{2^{21}}}} - \frac{20}{{{2^{21}}}} = 1 - \frac{22}{{{2^{21}}}} = 1 - \frac{11}{{{2^{20}}}}$
$S = 2 - \frac{11}{{{2^{19}}}}$

(A) Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
The equation $ax^2 + 2bx + c = 0$ can be written as $ax^2 + 2\sqrt{ac}x + c = 0$,which is $(\sqrt{a}x + \sqrt{c})^2 = 0$.
Thus,the root of this equation is $x = -\frac{\sqrt{c}}{\sqrt{a}} = -\frac{b}{a}$.
Since this is a common root for $dx^2 + 2ex + f = 0$,we substitute $x = -\frac{b}{a}$ into the second equation:
$d(-\frac{b}{a})^2 + 2e(-\frac{b}{a}) + f = 0$
$d(\frac{b^2}{a^2}) - \frac{2eb}{a} + f = 0$
Multiplying by $a^2$,we get $db^2 - 2eab + fa^2 = 0$.
Since $b^2 = ac$,we substitute it:
$dac - 2eab + fa^2 = 0$.
Dividing the entire equation by $ac$,we get:
$\frac{d}{a} - \frac{2e}{b} + \frac{f}{c} = 0$,which implies $\frac{d}{a} + \frac{f}{c} = 2(\frac{e}{b})$.
This condition indicates that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in $A.P.$

(A) Given the sum of the first $n$ terms: $S_n = 50n + \frac{n(n - 7)}{2}A$.
The $n^{th}$ term $a_n$ is given by $a_n = S_n - S_{n-1}$.
$a_n = 50n + \frac{n(n - 7)}{2}A - [50(n - 1) + \frac{(n - 1)(n - 8)}{2}A]$
$a_n = 50n + \frac{A}{2}(n^2 - 7n) - 50n + 50 - \frac{A}{2}(n^2 - 9n + 8)$
$a_n = 50 + \frac{A}{2}(n^2 - 7n - n^2 + 9n - 8)$
$a_n = 50 + \frac{A}{2}(2n - 8) = 50 + A(n - 4)$.
The common difference $d = a_n - a_{n-1} = [50 + A(n - 4)] - [50 + A(n - 5)] = A(n - 4 - n + 5) = A$.
To find $a_{50}$,substitute $n = 50$ into the expression for $a_n$:
$a_{50} = 50 + A(50 - 4) = 50 + 46A$.
Thus,the ordered pair $(d, a_{50})$ is $(A, 50 + 46A)$.

(A) Let the number of rows in the equilateral triangle be $n$. The total number of balls in the triangle is given by the sum of the first $n$ natural numbers: $S = \frac{n(n+1)}{2}$.
According to the problem,if $99$ more balls are added,the total number of balls becomes $S + 99 = \frac{n(n+1)}{2} + 99$.
These balls form a square with side length $(n-2)$. Thus,the total number of balls is $(n-2)^2$.
Equating the two expressions: $\frac{n(n+1)}{2} + 99 = (n-2)^2$.
Multiplying by $2$: $n^2 + n + 198 = 2(n^2 - 4n + 4)$.
$n^2 + n + 198 = 2n^2 - 8n + 8$.
Rearranging the terms: $n^2 - 9n - 190 = 0$.
Factoring the quadratic equation: $(n - 19)(n + 10) = 0$.
Since $n$ must be positive,$n = 19$.
The number of balls used to form the triangle is $\frac{19(19+1)}{2} = \frac{19 \times 20}{2} = 190$.

(A) Let the three numbers in $A.P.$ be $a-d, a, a+d$.
Given that the sum is $(a-d) + a + (a+d) = 33$.
$3a = 33 \Rightarrow a = 11$.
Given that the product is $(a-d)(a)(a+d) = 1155$.
$a(a^2 - d^2) = 1155$.
$11(121 - d^2) = 1155$.
$121 - d^2 = 105$.
$d^2 = 16 \Rightarrow d = \pm 4$.
Case $1$: If $d = 4$,the first term $A = a-d = 11-4 = 7$. The $11^{th}$ term $T_{11} = A + 10d = 7 + 10(4) = 47$.
Case $2$: If $d = -4$,the first term $A = a-d = 11 - (-4) = 15$. The $11^{th}$ term $T_{11} = A + 10d = 15 + 10(-4) = 15 - 40 = -25$.
Thus,one possible value for the $11^{th}$ term is $-25$.

(B) The $n^{th}$ term of the series is given by $T_n = n(2n - 1) = 2n^2 - n$.
The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (2k^2 - k) = 2 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k$.
Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$,we get:
$S_n = 2 \left[ \frac{n(n+1)(2n+1)}{6} \right] - \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} - \frac{n(n+1)}{2}$.
For $n = 11$:
$S_{11} = \frac{11(12)(23)}{3} - \frac{11(12)}{2}$.
$S_{11} = 11 \times 4 \times 23 - 11 \times 6$.
$S_{11} = 1012 - 66 = 946$.

(B) The $n^{th}$ term of the series is given by:
$T_n = \frac{(2n + 1) \sum_{k=1}^n k^3}{\sum_{k=1}^n k^2}$
Using the formulas $\sum k^3 = \frac{n^2(n+1)^2}{4}$ and $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$T_n = \frac{(2n + 1) \cdot \frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(2n+1)}{6}}$
$T_n = \frac{n^2(n+1)^2}{4} \cdot \frac{6}{n(n+1)} = \frac{3}{2} n(n+1)$
Now,the sum of the first $10$ terms is:
$S_{10} = \sum_{n=1}^{10} \frac{3}{2} (n^2 + n) = \frac{3}{2} \left[ \sum_{n=1}^{10} n^2 + \sum_{n=1}^{10} n \right]$
$S_{10} = \frac{3}{2} \left[ \frac{10(11)(21)}{6} + \frac{10(11)}{2} \right]$
$S_{10} = \frac{3}{2} [385 + 55] = \frac{3}{2} [440] = 3 \times 220 = 660$.

(A) The given sequence is an $A.P.$ with $6$ terms: $a_1, a_4, a_7, a_{10}, a_{13}, a_{16}$.
The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(\text{first term} + \text{last term})$.
Here, $n = 6$, so $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = \frac{6}{2}(a_1 + a_{16}) = 114$.
$3(a_1 + a_{16}) = 114 \Rightarrow a_1 + a_{16} = 38$.
Now, we need to find the sum $a_1 + a_6 + a_{11} + a_{16}$.
This is an $A.P.$ with $4$ terms, where the sum is $\frac{4}{2}(a_1 + a_{16})$.
Substituting the value: $2 \times 38 = 76$.

(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 $c,$ we get $14(ar) = 3a + 15(ar^2).$
Since $a \ne 0,$ we divide by $a$ to get $15r^2 - 14r + 3 = 0.$
Factoring the quadratic equation: $(3r - 1)(5r - 1) = 0,$ which gives $r = \frac{1}{3}$ or $r = \frac{1}{5}.$
Given $0 < r \le \frac{1}{2},$ both values are valid. However,checking the options,we proceed with $r = \frac{1}{3}.$
The common difference $d = 7b - 3a = 7ar - 3a = a(7r - 3).$
For $r = \frac{1}{3},$ $d = a(7/3 - 3) = -\frac{2}{3}a.$
The $4^{th}$ term is $15c + d = 15ar^2 - \frac{2}{3}a = 15a(1/9) - \frac{2}{3}a = \frac{5}{3}a - \frac{2}{3}a = a.$

(A) The $n^{th}$ term of the series is $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n k}$.
Using the formulas $\sum_{k=1}^n k^3 = \left[ \frac{n(n+1)}{2} \right]^2$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,we get:
$T_n = \frac{[n(n+1)/2]^2}{n(n+1)/2} = \frac{n(n+1)}{2}$.
The total sum is $S = \sum_{n=1}^{15} T_n - \frac{1}{2} \sum_{n=1}^{15} n$.
$S = \sum_{n=1}^{15} \frac{n(n+1)}{2} - \frac{1}{2} \cdot \frac{15 \cdot 16}{2}$.
$S = \frac{1}{2} \left[ \sum_{n=1}^{15} n^2 + \sum_{n=1}^{15} n \right] - 60$.
$S = \frac{1}{2} \left[ \frac{15(16)(31)}{6} + \frac{15(16)}{2} \right] - 60$.
$S = \frac{1}{2} [1240 + 120] - 60 = \frac{1360}{2} - 60 = 680 - 60 = 620$.

(C) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.
For $S_4 = 16$,we have $\frac{4}{2} [2a + 3d] = 16$,which simplifies to $2a + 3d = 8$ (Equation $1$).
For $S_6 = -48$,we have $\frac{6}{2} [2a + 5d] = -48$,which simplifies to $3(2a + 5d) = -48$,or $2a + 5d = -16$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(2a + 5d) - (2a + 3d) = -16 - 8$,which gives $2d = -24$,so $d = -12$.
Substituting $d = -12$ into Equation $1$: $2a + 3(-12) = 8$,so $2a - 36 = 8$,which means $2a = 44$,so $a = 22$.
Now,$S_{10} = \frac{10}{2} [2a + 9d] = 5 [2(22) + 9(-12)]$.
$S_{10} = 5 [44 - 108] = 5 [-64] = -320$.

(C) Let the series be $S = \sum_{k=0}^{99} \left[ -\frac{1}{3} - \frac{k}{100} \right]$.
For $0 \le k \le 66$,we have $-\frac{1}{3} - \frac{k}{100} \ge -\frac{1}{3} - \frac{66}{100} = -\frac{1}{3} - 0.66 = -0.9966$. Since $-1 < -0.9966 < 0$,the greatest integer value is $-1$. There are $67$ such terms (from $k=0$ to $k=66$).
Sum for these terms $= 67 \times (-1) = -67$.
For $67 \le k \le 99$,we have $-\frac{1}{3} - \frac{k}{100} \le -\frac{1}{3} - \frac{67}{100} = -0.3333 - 0.67 = -1.0033$. Also,$-\frac{1}{3} - \frac{99}{100} = -0.3333 - 0.99 = -1.3233$. Since $-2 < -1.3233 \le -1.0033 < -1$,the greatest integer value is $-2$. There are $99 - 67 + 1 = 33$ such terms.
Sum for these terms $= 33 \times (-2) = -66$.
Total sum $S = -67 + (-66) = -133$.

(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The terms are given by $a_n = a + (n-1)d$.
Given the equation: $a_1 + a_7 + a_{16} = 40$.
Substituting the general form: $a + (a + 6d) + (a + 15d) = 40$.
This simplifies to: $3a + 21d = 40$.
Dividing by $3$: $a + 7d = \frac{40}{3}$.
The sum of the first $15$ terms $(S_{15})$ is given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$.
For $n = 15$: $S_{15} = \frac{15}{2}[2a + 14d]$.
Factoring out $2$: $S_{15} = 15(a + 7d)$.
Substituting the value of $(a + 7d)$: $S_{15} = 15 \times \frac{40}{3} = 5 \times 40 = 200$.

(A) The given series is $3+4+8+9+13+14+18+19+\ldots$ up to $40$ terms.
This can be split into two arithmetic progressions,each having $20$ terms:
Series $1$: $3, 8, 13, 18, \ldots$ ($20$ terms)
Series $2$: $4, 9, 14, 19, \ldots$ ($20$ terms)
Sum of Series $1$ $(S_1)$ = $\frac{20}{2} [2(3) + (20-1)5] = 10 [6 + 95] = 10 \times 101 = 1010$.
Sum of Series $2$ $(S_2)$ = $\frac{20}{2} [2(4) + (20-1)5] = 10 [8 + 95] = 10 \times 103 = 1030$.
Total Sum = $S_1 + S_2 = 1010 + 1030 = 2040$.
Given that the total sum is $(102)m$,we have $2040 = 102m$.
Therefore,$m = \frac{2040}{102} = 20$.

(A) Given that $a_{1}, a_{2}, a_{3}, \ldots$ is a $G$.$P$. with common ratio $r$.
We have $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$.
Since $a_{3} = a_{1}r^{2}$ and $a_{4} = a_{2}r^{2}$,we can write $a_{3} + a_{4} = r^{2}(a_{1} + a_{2}) = 16$.
Substituting $a_{1} + a_{2} = 4$,we get $r^{2}(4) = 16$,which implies $r^{2} = 4$,so $r = 2$ or $r = -2$.
Given $a_{1} + a_{2} = 4$,we have $a_{1}(1 + r) = 4$.
If $r = 2$,$a_{1}(3) = 4 \Rightarrow a_{1} = 4/3$ (not possible as $a_{1} < 0$).
If $r = -2$,$a_{1}(1 - 2) = 4 \Rightarrow -a_{1} = 4 \Rightarrow a_{1} = -4$.
Now,the sum of the first $9$ terms is $S_{9} = a_{1} \frac{r^{9} - 1}{r - 1} = (-4) \frac{(-2)^{9} - 1}{-2 - 1} = (-4) \frac{-512 - 1}{-3} = (-4) \frac{-513}{-3} = -4 \times 171 = -684$.
Given $S_{9} = 4 \lambda$,we have $4 \lambda = -684$,so $\lambda = -171$.

(C) Let the five numbers in $A.P.$ be $(a-2d, a-d, a, a+d, a+2d)$.
Sum $= (a-2d) + (a-d) + a + (a+d) + (a+2d) = 5a = 25$,so $a = 5$.
The numbers are $(5-2d, 5-d, 5, 5+d, 5+2d)$.
The product is $(5-2d)(5+2d)(5-d)(5+d)(5) = 2520$.
$(25-4d^2)(25-d^2) = 504$.
$625 - 25d^2 - 100d^2 + 4d^4 = 504$.
$4d^4 - 125d^2 + 121 = 0$.
Let $x = d^2$,then $4x^2 - 125x + 121 = 0$.
$(4x - 121)(x - 1) = 0$,so $d^2 = 1$ or $d^2 = \frac{121}{4}$.
If $d^2 = 1$,$d = \pm 1$,the terms are $(3, 4, 5, 6, 7)$ or $(7, 6, 5, 4, 3)$,none of which is $-\frac{1}{2}$.
If $d^2 = \frac{121}{4}$,$d = \pm \frac{11}{2}$.
For $d = \frac{11}{2}$,the terms are $(5-11, 5-5.5, 5, 5+5.5, 5+11) = (-6, -0.5, 5, 10.5, 16)$.
Since $-\frac{1}{2}$ is present,this is the correct sequence.
The greatest number is $16$.

(B) The $n^{\text{th}}$ term of an $A$.$P$. is given by $T_n = a + (n-1)d$.
Given $T_{10} = a + 9d = \frac{1}{20} \quad \dots(i)$
Given $T_{20} = a + 19d = \frac{1}{10} \quad \dots(ii)$
Subtracting equation $(i)$ from $(ii)$:
$(a + 19d) - (a + 9d) = \frac{1}{10} - \frac{1}{20}$
$10d = \frac{2-1}{20} = \frac{1}{20}$
$d = \frac{1}{200}$
Substituting $d$ in equation $(i)$:
$a + 9(\frac{1}{200}) = \frac{1}{20}$
$a = \frac{10}{200} - \frac{9}{200} = \frac{1}{200}$
Now,the sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.
For $n = 200$:
$S_{200} = \frac{200}{2} [2(\frac{1}{200}) + (200-1)(\frac{1}{200})]$
$S_{200} = 100 [\frac{2}{200} + \frac{199}{200}]$
$S_{200} = 100 [\frac{201}{200}] = \frac{201}{2} = 100 \frac{1}{2}$.

(C) The given sum is $S = \sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$.
We know that the sum of the first $n$ squares is $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
Therefore,$n(n+1)(2n+1) = 6 \sum_{k=1}^{n} k^2$.
Substituting this into the expression,we get $S = \frac{1}{4} \sum_{n=1}^{7} (6 \sum_{k=1}^{n} k^2) = \frac{6}{4} \sum_{n=1}^{7} \sum_{k=1}^{n} k^2$.
Alternatively,we can expand the numerator: $n(n+1)(2n+1) = 2n^3 + 3n^2 + n$.
So,$S = \frac{1}{4} \left( 2 \sum_{n=1}^{7} n^3 + 3 \sum_{n=1}^{7} n^2 + \sum_{n=1}^{7} n \right)$.
Using the standard summation formulas:
$\sum_{n=1}^{7} n^3 = \left( \frac{7 \times 8}{2} \right)^2 = 28^2 = 784$.
$\sum_{n=1}^{7} n^2 = \frac{7 \times 8 \times 15}{6} = 140$.
$\sum_{n=1}^{7} n = \frac{7 \times 8}{2} = 28$.
Substituting these values: $S = \frac{1}{4} (2(784) + 3(140) + 28) = \frac{1}{4} (1568 + 420 + 28) = \frac{2016}{4} = 504$.

(B) Given that $(2^{1+x}+2^{1-x})$,$f(x)$,and $(3^x+3^{-x})$ are in $A.P$.
By the definition of $A.P.$,$2f(x) = (2^{1+x}+2^{1-x}) + (3^x+3^{-x})$.
$f(x) = \frac{2(2^x+2^{-x}) + (3^x+3^{-x})}{2} = (2^x+2^{-x}) + \frac{1}{2}(3^x+3^{-x})$.
Using the $A.M. \geq G.M.$ inequality,we know that for any $a > 0$,$a^x + a^{-x} \geq 2\sqrt{a^x \cdot a^{-x}} = 2$.
Thus,$2^x+2^{-x} \geq 2$ and $3^x+3^{-x} \geq 2$.
Substituting these minimum values into the expression for $f(x)$:
$f(x) \geq 2 + \frac{1}{2}(2) = 2 + 1 = 3$.
Therefore,the minimum value of $f(x)$ is $3$.

(C) The sum of the first $k$ natural numbers is given by the formula $\frac{k(k+1)}{2}$.
We need to calculate $\sum_{k=1}^{20} \frac{k(k+1)}{2}$.
This can be written as $\frac{1}{2} \sum_{k=1}^{20} (k^2 + k)$.
Using the summation formulas $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ for $n=20$:
$\sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = 2870$.
$\sum_{k=1}^{20} k = \frac{20(21)}{2} = 210$.
Therefore,the sum is $\frac{1}{2} (2870 + 210) = \frac{1}{2} (3080) = 1540$.

(C) The given series 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 given series 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 the expression for $x$,we have $\cos^2 \theta = x$,which implies $\sin^2 \theta = 1 - x$.
Substituting this into the expression for $y$,we get $y = \frac{1}{1 - x}$.
Rearranging this gives $y(1 - x) = 1$.

(D) Given that $a_n$ is a $G$.$P$. with common ratio $r$. The terms are $a, ar, ar^2, \dots$.
$\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + \dots + a_{201} = ar^2 + ar^4 + \dots + ar^{200} = ar^2 \frac{(r^{200}-1)}{(r^2-1)} = 200$.
$\sum_{n=1}^{100} a_{2n} = a_2 + a_4 + \dots + a_{200} = ar + ar^3 + \dots + ar^{199} = ar \frac{(r^{200}-1)}{(r^2-1)} = 100$.
Dividing the first equation by the second equation: $\frac{ar^2}{ar} = \frac{200}{100} \Rightarrow r = 2$.
Now,$\sum_{n=1}^{200} a_n = a_1 + a_2 + \dots + a_{200} = a \frac{(r^{200}-1)}{(r-1)}$.
From the second equation,$ar \frac{(r^{200}-1)}{(r^2-1)} = 100$. Since $r=2$,$2a \frac{(r^{200}-1)}{(2^2-1)} = 100 \Rightarrow 2a \frac{(r^{200}-1)}{3} = 100 \Rightarrow a \frac{(r^{200}-1)}{3} = 50 \Rightarrow a(r^{200}-1) = 150$.
Substituting this into the sum formula: $\sum_{n=1}^{200} a_n = \frac{150}{2-1} = 150$.

(D) For the first $A$.$P$.: $a_1 = 3$,$d_1 = 4$. The general term is $T_n = 3 + (n-1)4 = 4n - 1$.
For the second $A$.$P$.: $a_2 = 2$,$d_2 = 7$. The general term is $T_m = 2 + (m-1)7 = 7m - 5$.
Equating the terms: $4n - 1 = 7m - 5 \Rightarrow 4n = 7m - 4$.
This implies $7m$ must be a multiple of $4$. Since $7$ is not divisible by $4$,$m$ must be a multiple of $4$. Let $m = 4k$.
Then $4n = 7(4k) - 4 \Rightarrow n = 7k - 1$.
For the first sequence,$T_n \leq 407 \Rightarrow 4n - 1 \leq 407 \Rightarrow 4n \leq 408 \Rightarrow n \leq 102$.
Substituting $n = 7k - 1$: $7k - 1 \leq 102 \Rightarrow 7k \leq 103 \Rightarrow k \leq 14.71$.
Since $k$ must be a positive integer,the possible values for $k$ are $1, 2, \ldots, 14$.
Thus,there are $14$ common terms.

(A) Let the given expression be $P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots \infty$.
Expressing all bases as powers of $2$:
$P = 2^{\frac{1}{4}} \cdot (2^2)^{\frac{1}{16}} \cdot (2^3)^{\frac{1}{48}} \cdot (2^4)^{\frac{1}{128}} \cdot \ldots \infty$
$P = 2^{\frac{1}{4}} \cdot 2^{\frac{2}{16}} \cdot 2^{\frac{3}{48}} \cdot 2^{\frac{4}{128}} \cdot \ldots \infty$
Simplifying the exponents:
$P = 2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 2^{\frac{1}{16}} \cdot 2^{\frac{1}{32}} \cdot \ldots \infty$
Using the property $a^m \cdot a^n = a^{m+n}$,we sum the exponents:
$P = 2^{\left(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots \infty\right)}$
The exponent is an infinite geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{2}$.
The sum of an infinite geometric series is $S = \frac{a}{1-r}$.
$S = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.
Therefore,$P = 2^{\frac{1}{2}}$.

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