Arithmetic progression Questions in English

(D) Let the second-degree polynomial be $f(x) = px^2 + qx + r$.
Given $f(1) = f(-1)$,we have $p(1)^2 + q(1) + r = p(-1)^2 + q(-1) + r$.
This simplifies to $p + q + r = p - q + r$,which implies $2q = 0$,so $q = 0$.
Thus,the function is $f(x) = px^2 + r$.
The derivative is $f'(x) = 2px$.
Since $a, b, c$ are in $A.P.$,there exists a common difference $d$ such that $b = a + d$ and $c = a + 2d$.
Then $f'(a) = 2pa$,$f'(b) = 2p(a + d) = 2pa + 2pd$,and $f'(c) = 2p(a + 2d) = 2pa + 4pd$.
Since $f'(b) - f'(a) = 2pd$ and $f'(c) - f'(b) = 2pd$,the terms $f'(a), f'(b), f'(c)$ have a common difference of $2pd$.
Therefore,$f'(a), f'(b), f'(c)$ are in $A.P.$

(A) Given $a + 2b + 3c = 6$. We want to maximize $abc^2$.
We can write this as $a + 2b + \frac{3c}{2} + \frac{3c}{2} = 6$.
By the $AM \geq GM$ inequality for four positive terms:
$\frac{a + 2b + \frac{3c}{2} + \frac{3c}{2}}{4} \geq \sqrt[4]{a \cdot 2b \cdot \frac{3c}{2} \cdot \frac{3c}{2}}$
$\frac{6}{4} \geq \sqrt[4]{a \cdot 2b \cdot \frac{9c^2}{4}}$
$\frac{3}{2} \geq \sqrt[4]{\frac{18}{4} abc^2}$
$\frac{3}{2} \geq \sqrt[4]{\frac{9}{2} abc^2}$
Raising both sides to the power of $4$:
$(\frac{3}{2})^4 \geq \frac{9}{2} abc^2$
$\frac{81}{16} \geq \frac{9}{2} abc^2$
$abc^2 \leq \frac{81}{16} \cdot \frac{2}{9} = \frac{9}{8}$.

(C) The two-digit numbers that leave a remainder of $4$ when divided by $6$ form an arithmetic progression.
These numbers are of the form $6n + 4$.
The smallest two-digit number is $10$ $(6 \times 1 + 4)$ and the largest is $94$ $(6 \times 15 + 4)$.
Thus,the sequence is $10, 16, 22, \ldots, 94$.
Here,the first term $a = 10$,the last term $l = 94$,and the common difference $d = 6$.
Using the formula for the $n^{th}$ term: $l = a + (n - 1)d$.
$94 = 10 + (n - 1)6$
$84 = (n - 1)6$
$n - 1 = 14$
$n = 15$.
The sum of the $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.
$S_{15} = \frac{15}{2}(10 + 94) = \frac{15}{2}(104) = 15 \times 52 = 780$.

(D) Let the roots of the cubic equation be $(a - d)$,$a$,and $(a + d)$.
According to the properties of roots of a cubic equation $x^3 - px^2 + qx - r = 0$,the sum of the roots is equal to the coefficient of $x^2$ with a changed sign.
$(a - d) + a + (a + d) = 9$
$3a = 9 \implies a = 3$.
Since $a = 3$ is a root of the equation,it must satisfy the equation $x^3 - 9x^2 + \alpha x - 15 = 0$.
Substituting $x = 3$ into the equation:
$(3)^3 - 9(3)^2 + \alpha(3) - 15 = 0$
$27 - 81 + 3\alpha - 15 = 0$
$3\alpha - 69 = 0$
$3\alpha = 69$
$\alpha = 23$.

(B) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant.
$a_1 + a_{16} = a_4 + a_{13} = a_7 + a_{10} = \dots = \lambda$.
Given the sum $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 147$.
This can be written as $3(a_1 + a_{16}) = 147$,so $a_1 + a_{16} = 49$.
We need to find $S = a_1 + a_6 + a_{11} + a_{16}$.
Since $a_1 + a_{16} = a_6 + a_{11} = 49$,we have $S = (a_1 + a_{16}) + (a_6 + a_{11}) = 49 + 49 = 98$.

(C) Given the ratio of the sum of $n$ terms of two $A.P.s$ is $\frac{s_n}{S_n} = \frac{3n - 13}{7n + 13}$.
Let the sum of $n$ terms be $s_n = k(3n^2 - 13n)$ and $S_n = k(7n^2 + 13n)$ for some constant $k$.
To find the sum of $2n$ terms for the second $A.P.$,we substitute $2n$ for $n$ in the expression for $S_n$:
$S_{2n} = k(7(2n)^2 + 13(2n)) = k(7(4n^2) + 26n) = k(28n^2 + 26n)$.
Therefore,the ratio $\frac{s_n}{S_{2n}} = \frac{k(3n^2 - 13n)}{k(28n^2 + 26n)} = \frac{3n^2 - 13n}{28n^2 + 26n} = \frac{n(3n - 13)}{n(28n + 26)} = \frac{3n - 13}{28n + 26}$.

(C) Given that $\log _{5} 2, \log _{5}(2^{x}-3)$,and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$
For three terms $a, b, c$ to be in $A.P.$,the condition is $2b = a + c$.
Applying this condition: $2 \log _{5}(2^{x}-3) = \log _{5} 2 + \log _{5}(\frac{17}{2}+2^{x-1})$.
Using the property $\log m + \log n = \log(mn)$,we get: $\log _{5}(2^{x}-3)^{2} = \log _{5}(2 \times (\frac{17}{2}+2^{x-1}))$.
$(2^{x}-3)^{2} = 17 + 2 \times 2^{x-1} = 17 + 2^{x}$.
Let $2^{x} = y$. Then $(y-3)^{2} = 17 + y$.
$y^{2} - 6y + 9 = 17 + y \Rightarrow y^{2} - 7y - 8 = 0$.
$(y-8)(y+1) = 0$. Since $y = 2^{x} > 0$,we have $y = 8$.
$2^{x} = 8 = 2^{3} \Rightarrow x = 3$.

(A) Let $S_n = x_n + y_n + z_n + w_n$. Since the sum of arithmetic progressions is also an arithmetic progression,$S_n$ is an $A.P.$ with first term $A$ and common difference $D$.
Given $S_4 = A + 3D = 8$ and $S_{10} = A + 9D = 20$.
Subtracting the equations: $(A + 9D) - (A + 3D) = 20 - 8$ $\Rightarrow 6D = 12$ $\Rightarrow D = 2$.
Substituting $D = 2$ into $A + 3D = 8$: $A + 6 = 8 \Rightarrow A = 2$.
Now,$S_{20} = A + 19D = 2 + 19(2) = 2 + 38 = 40$.
By the $AM-GM$ inequality,$\frac{x_{20} + y_{20} + z_{20} + w_{20}}{4} \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4}$.
$\frac{40}{4} \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4}$ $\Rightarrow 10 \geq (x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20})^{1/4}$.
Raising both sides to the power of $4$,we get $x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20} \leq 10^4$.

(C) If $a, b, c$ are in $A.P.,$ then $2b = a + c.$
$\Rightarrow 2 \log _{10}(2^x - 1) = \log _{10} 2 + \log _{10}(2^x + 3)$
$\Rightarrow \log _{10}(2^x - 1)^2 = \log _{10} [2(2^x + 3)]$
$\Rightarrow (2^x - 1)^2 = 2(2^x + 3)$
Let $2^x = y.$
$\Rightarrow (y - 1)^2 = 2(y + 3)$
$\Rightarrow y^2 - 2y + 1 = 2y + 6$
$\Rightarrow y^2 - 4y - 5 = 0$
$\Rightarrow (y - 5)(y + 1) = 0$
Since $y = 2^x > 0,$ we have $y = 5.$
$\Rightarrow 2^x = 5$
$\Rightarrow x = \log_2 5.$

(D) Given $x + y + z = 4$ where $x, y, z \in R^+$.
We want to maximize $xyz^2 = x \cdot y \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot 4$.
By the Arithmetic Mean-Geometric Mean Inequality ($AM$-$GM$),for positive real numbers $x, y, \frac{z}{2}, \frac{z}{2}$:
$\frac{x + y + \frac{z}{2} + \frac{z}{2}}{4} \geq \sqrt[4]{x \cdot y \cdot \frac{z}{2} \cdot \frac{z}{2}}$
Substituting $x + y + z = 4$:
$\frac{4}{4} \geq \sqrt[4]{\frac{xyz^2}{4}}$
$1 \geq \sqrt[4]{\frac{xyz^2}{4}}$
Raising both sides to the power of $4$:
$1 \geq \frac{xyz^2}{4}$
$xyz^2 \leq 4$
Thus,the maximum value is $4$.

(C) Let $z = {\left( {\frac{3}{a} - 1} \right)^2} + {\left( {\frac{a}{b} - 1} \right)^2} + {\left( {\frac{b}{c} - 1} \right)^2} + {\left( {3c - 1} \right)^2}$.
Expanding the terms,we get $z = \left( \frac{9}{a^2} + \frac{a^2}{b^2} + \frac{b^2}{c^2} + 9c^2 \right) + 4 - 2 \left( \frac{3}{a} + \frac{a}{b} + \frac{b}{c} + 3c \right)$.
Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality:
$\frac{1}{4} \left( \frac{9}{a^2} + \frac{a^2}{b^2} + \frac{b^2}{c^2} + 9c^2 \right) \ge \sqrt[4]{\frac{9}{a^2} \cdot \frac{a^2}{b^2} \cdot \frac{b^2}{c^2} \cdot 9c^2} = \sqrt[4]{81} = 3$.
So,$\frac{9}{a^2} + \frac{a^2}{b^2} + \frac{b^2}{c^2} + 9c^2 \ge 12$.
Similarly,$\frac{1}{4} \left( \frac{3}{a} + \frac{a}{b} + \frac{b}{c} + 3c \right) \ge \sqrt[4]{\frac{3}{a} \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot 3c} = \sqrt[4]{9} = \sqrt{3}$.
So,$2 \left( \frac{3}{a} + \frac{a}{b} + \frac{b}{c} + 3c \right) \ge 8\sqrt{3}$.
Thus,$z \ge 12 + 4 - 8\sqrt{3} = 16 - 8\sqrt{3}$.
Comparing with $p - q\sqrt{r}$,we have $p = 16$,$q = 8$,$r = 3$.
Since $q$ and $r$ are coprime,$p + q + r = 16 + 8 + 3 = 27$.

(D) Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for the four positive numbers $x, y, \frac{z}{2}, \frac{z}{2}$:
$\frac{x + y + \frac{z}{2} + \frac{z}{2}}{4} \geq \sqrt[4]{x \cdot y \cdot \frac{z}{2} \cdot \frac{z}{2}}$
Given $x + y + z = 4$,we substitute this into the inequality:
$\frac{4}{4} \geq \sqrt[4]{\frac{xyz^2}{4}}$
$1 \geq \sqrt[4]{\frac{xyz^2}{4}}$
Raising both sides to the power of $4$:
$1 \geq \frac{xyz^2}{4}$
$xyz^2 \leq 4$
Thus,the maximum value is $4$.

(A) Let the common difference of the $A.P.$ be $d$. Then $S_{n+1} - S_n = d$ for all $n$.
The given sum is $\sum_{n=1}^{100} \frac{1}{S_n S_{n+1}} = \frac{1}{d} \sum_{n=1}^{100} (\frac{1}{S_n} - \frac{1}{S_{n+1}}) = \frac{1}{d} (\frac{1}{S_1} - \frac{1}{S_{101}}) = \frac{1}{d} (\frac{S_{101} - S_1}{S_1 S_{101}}) = \frac{1}{6}$.
Since $S_{101} = S_1 + 100d$,we have $S_{101} - S_1 = 100d$.
Substituting this into the equation: $\frac{1}{d} \cdot \frac{100d}{S_1 S_{101}} = \frac{100}{S_1 S_{101}} = \frac{1}{6}$,so $S_1 S_{101} = 600$.
We are given $S_1 + S_{101} = 50$. Let $x = S_1$ and $y = S_{101}$. Then $x + y = 50$ and $xy = 600$.
The value $|x - y| = \sqrt{(x+y)^2 - 4xy} = \sqrt{50^2 - 4(600)} = \sqrt{2500 - 2400} = \sqrt{100} = 10$.

(B) Given the sum of the first $n$ terms $S_n = cn(n-1) = cn^2 - cn$.
The $n^{th}$ term $t_n$ is given by $S_n - S_{n-1}$.
$t_n = [cn^2 - cn] - [c(n-1)^2 - c(n-1)]$
$t_n = cn^2 - cn - [c(n^2 - 2n + 1) - cn + c]$
$t_n = cn^2 - cn - [cn^2 - 3cn + 2c] = 2cn - 2c = 2c(n-1)$.
We need the sum of the squares of these terms,i.e.,$\sum_{k=1}^{n} (t_k)^2$.
$t_k^2 = [2c(k-1)]^2 = 4c^2(k-1)^2 = 4c^2(k^2 - 2k + 1)$.
Sum $= \sum_{k=1}^{n} 4c^2(k^2 - 2k + 1) = 4c^2 [\sum_{k=1}^{n} k^2 - 2\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1]$.
Using standard summation formulas:
Sum $= 4c^2 [\frac{n(n+1)(2n+1)}{6} - 2\frac{n(n+1)}{2} + n]$.
Sum $= 4c^2 [\frac{n(n+1)(2n+1) - 6n(n+1) + 6n}{6}] = \frac{2c^2}{3} [n(2n^2 + 3n + 1 - 6n - 6 + 6)]$.
Sum $= \frac{2c^2}{3} [n(2n^2 - 3n + 1)] = \frac{2c^2}{3} [n(n-1)(2n-1)]$.

(A) For $S_1$,the first term $a_1 = 1$ and common difference $d_1 = 6 - 1 = 5$. The general term is $T_n = 1 + (n-1)5 = 5n - 4$.
For $S_2$,the first term $a_2 = 3$ and common difference $d_2 = 7 - 3 = 4$. The general term is $T_m = 3 + (m-1)4 = 4m - 1$.
For common terms,$5n - 4 = 4m - 1$,which implies $5n = 4m + 3$.
The first common term is $11$ (where $n=3, m=3$).
The common difference of the new sequence formed by common terms is $LCM(d_1, d_2) = LCM(5, 4) = 20$.
The sequence of common terms is $11, 31, 51, .....$ which is an $A.P.$ with first term $A = 11$ and common difference $D = 20$.
The $25^{th}$ common term is $A_{25} = A + (25-1)D = 11 + 24 \times 20 = 11 + 480 = 491$.

(D) For the sum to be maximum,we consider terms until they become non-negative. Since the common difference $d = -2$ is negative,the sum increases as long as the terms are positive.
Set $T_n \geq 0$:
$a + (n - 1)d \geq 0$
$50 + (n - 1)(-2) \geq 0$
$50 - 2n + 2 \geq 0$
$52 \geq 2n \Rightarrow n \leq 26$.
Thus,the sum of the first $26$ terms is the maximum sum.
$S_{26} = \frac{26}{2} [2(50) + (26 - 1)(-2)]$
$S_{26} = 13 [100 - 50]$
$S_{26} = 13 \times 50 = 650$.

(D) Let the four angles of the quadrilateral in $A.P.$ be $(a-3d), (a-d), (a+d), (a+3d)$.
The common difference of this sequence is $2d = 10^{\circ}$,which implies $d = 5^{\circ}$.
The sum of interior angles of a quadrilateral is $360^{\circ}$.
$(a-3d) + (a-d) + (a+d) + (a+3d) = 360^{\circ}$
$4a = 360^{\circ} \Rightarrow a = 90^{\circ}$.
The smallest angle is $a-3d = 90^{\circ} - 3(5^{\circ}) = 90^{\circ} - 15^{\circ} = 75^{\circ}$.

(B) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant,i.e.,$a_1 + a_{21} = a_3 + a_{19} = a_5 + a_{17} = 2a_{11}$.
Given $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$.
Substituting the relations,we get $(a_3 + a_{19}) + (a_5 + a_{17}) + a_{11} = 10$.
Since $a_3 + a_{19} = 2a_{11}$ and $a_5 + a_{17} = 2a_{11}$,the equation becomes $2a_{11} + 2a_{11} + a_{11} = 10$.
$5a_{11} = 10 \Rightarrow a_{11} = 2$.
We know that $a_1 + a_{21} = 2a_{11} = 2(2) = 4$.
The sum of the first $21$ terms is given by $S_{21} = \frac{21}{2}(a_1 + a_{21})$.
$S_{21} = \frac{21}{2}(4) = 21 \times 2 = 42$.

(B) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is $T_n = a + (n-1)d$.
Given $T_9 = 5T_2$ $\Rightarrow a + 8d = 5(a + d)$ $\Rightarrow a + 8d = 5a + 5d$ $\Rightarrow 4a - 3d = 0$ $\Rightarrow d = \frac{4a}{3}$.
Given $T_{13} = 2T_6 + 5$ $\Rightarrow a + 12d = 2(a + 5d) + 5$ $\Rightarrow a + 12d = 2a + 10d + 5$ $\Rightarrow 2d - a = 5$.
Substitute $d = \frac{4a}{3}$ into the second equation: $2(\frac{4a}{3}) - a = 5$.
$\frac{8a}{3} - a = 5$ $\Rightarrow \frac{5a}{3} = 5$ $\Rightarrow a = 3$.

(A) The arithmetic mean $A$ is given by:
$A = \frac{9 + 99 + 999 + \dots + 999999999}{9}$
Dividing each term by $9$,we get:
$A = 1 + 11 + 111 + \dots + 111111111$
Summing these terms:
$1 = 1$
$1 + 11 = 12$
$1 + 11 + 111 = 123$
Continuing this pattern for $9$ terms,we get:
$A = 123456789$
The number $N = 123456789$ consists of the digits $1$ through $9$. It does not contain the digit $0$.

(D) For the sum to be maximum,we consider terms until they become non-negative.
Let the $n^{th}$ term be $T_n = a + (n-1)d$.
Here,$a = 50$ and $d = -2$.
$T_n = 50 + (n-1)(-2) = 50 - 2n + 2 = 52 - 2n$.
Setting $T_n \geq 0$,we get $52 - 2n \geq 0$ $\Rightarrow 2n \leq 52$ $\Rightarrow n \leq 26$.
Thus,the sum is maximum for $n = 26$.
$S_{26} = \frac{26}{2} [2(50) + (26-1)(-2)] = 13 [100 - 50] = 13 \times 50 = 650$.

(A) Let $d_1$ be the common difference of the $A.P.$ $x_1, x_2, \dots, x_n$.
Since $x_8 - x_3 = 5d_1 = 20 - 8 = 12$,we have $d_1 = \frac{12}{5} = 2.4$.
Then $x_5 = x_3 + 2d_1 = 8 + 2(2.4) = 8 + 4.8 = 12.8$.
Let $d_2$ be the common difference of the $A.P.$ $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$.
Since $\frac{1}{h_7} - \frac{1}{h_2} = 5d_2 = \frac{1}{20} - \frac{1}{8} = \frac{2-5}{40} = -\frac{3}{40}$,we have $d_2 = -\frac{3}{200}$.
Now,$\frac{1}{h_{10}} = \frac{1}{h_7} + 3d_2 = \frac{1}{20} + 3\left(-\frac{3}{200}\right) = \frac{10-9}{200} = \frac{1}{200}$.
Thus,$h_{10} = 200$.
Finally,$x_5 \cdot h_{10} = 12.8 \times 200 = 2560$.

(C) Given that $\frac{1}{x_1}, \frac{1}{x_2}, \dots, \frac{1}{x_n}$ are in $A.P.$
Let $a = \frac{1}{x_1} = \frac{1}{4}$ and $d$ be the common difference.
We have $\frac{1}{x_{21}} = \frac{1}{20}$.
Using the $A.P.$ formula $\frac{1}{x_{21}} = a + 20d$,we get $\frac{1}{20} = \frac{1}{4} + 20d$.
$20d = \frac{1}{20} - \frac{1}{4} = \frac{1-5}{20} = -\frac{4}{20} = -\frac{1}{5}$.
So,$d = -\frac{1}{100}$.
The $n^{th}$ term of the $A.P.$ is $\frac{1}{x_n} = a + (n-1)d = \frac{1}{4} - \frac{n-1}{100} = \frac{25 - n + 1}{100} = \frac{26 - n}{100}$.
Thus,$x_n = \frac{100}{26 - n}$.
Given $x_n > 50$,we have $\frac{100}{26 - n} > 50$.
Since $x_n > 50$,$26 - n$ must be positive and $26 - n < 2$,so $n > 24$.
The least positive integer $n$ is $25$.
We need to find $\sum_{i=1}^{25} \frac{1}{x_i} = \frac{25}{2} \left[ 2a + (25-1)d \right]$.
$= \frac{25}{2} \left[ 2 \left( \frac{1}{4} \right) + 24 \left( -\frac{1}{100} \right) \right] = \frac{25}{2} \left[ \frac{1}{2} - \frac{6}{25} \right] = \frac{25}{2} \left[ \frac{25 - 12}{50} \right] = \frac{25}{2} \times \frac{13}{50} = \frac{13}{4}$.

(B) The given series is $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$
This can be written as $\sqrt{3} + 5\sqrt{3} + 9\sqrt{3} + 13\sqrt{3} + \dots$
This is an Arithmetic Progression with first term $a = \sqrt{3}$ and common difference $d = 4\sqrt{3}$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given $S_n = 435\sqrt{3}$,we have:
$\frac{n}{2}[2\sqrt{3} + (n-1)4\sqrt{3}] = 435\sqrt{3}$
Divide both sides by $\sqrt{3}$:
$\frac{n}{2}[2 + 4n - 4] = 435$
$n(2n - 2) = 870$
$2n^2 - 2n - 870 = 0$
$n^2 - n - 435 = 0$
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{1 \pm \sqrt{1 - 4(1)(-435)}}{2} = \frac{1 \pm \sqrt{1 + 1740}}{2} = \frac{1 \pm \sqrt{1741}}{2}$
Wait,checking the series calculation: $1, 5, 9, 13$ are terms of an $AP$ with $a=1, d=4$. Sum is $\frac{n}{2}[2(1) + (n-1)4] = \frac{n}{2}[4n - 2] = n(2n-1) = 435$.
$2n^2 - n - 435 = 0$.
$n = \frac{1 \pm \sqrt{1 + 4(2)(435)}}{4} = \frac{1 \pm \sqrt{1 + 3480}}{4} = \frac{1 \pm \sqrt{3481}}{4} = \frac{1 \pm 59}{4}$.
Since $n$ must be positive,$n = \frac{60}{4} = 15$.

(A) Since $a, b, c$ are in $A.P.$,we can write $a = b - d$ and $c = b + d$ for some common difference $d$.
Given $abc = 8$,we have $(b - d)b(b + d) = 8$.
$b(b^2 - d^2) = 8$,which implies $b^2 - d^2 = \frac{8}{b}$.
Since $d^2 \ge 0$,we have $b^2 - \frac{8}{b} \ge 0$.
$b^3 - 8 \ge 0$,so $b^3 \ge 8$,which means $b \ge 2$.
The minimum value of $b$ is $2$ when $d = 0$ (i.e.,$a = b = c = 2$).

(B) Given $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$.
Using the Weighted Arithmetic Mean-Geometric Mean Inequality:
$\frac{3(\frac{x}{3}) + 4(\frac{y}{4}) + 5(\frac{z}{5})}{3+4+5} \ge ((\frac{x}{3})^3 (\frac{y}{4})^4 (\frac{z}{5})^5)^{1/12}$
$\frac{x+y+z}{12} \ge (\frac{x^3 y^4 z^5}{3^3 4^4 5^5})^{1/12}$
$\frac{12}{12} \ge (\frac{x^3 y^4 z^5}{27 \times 256 \times 3125})^{1/12}$
$1 \ge \frac{x^3 y^4 z^5}{21600000}$
$x^3 y^4 z^5 \le 21600000 = 21.6 \times 10^6 = (0.1)(600)^3$.
Since the equality holds,we must have $\frac{x}{3} = \frac{y}{4} = \frac{z}{5} = k$.
Then $x = 3k, y = 4k, z = 5k$.
$3k + 4k + 5k = 12 \implies 12k = 12 \implies k = 1$.
Thus,$x = 3, y = 4, z = 5$.
Therefore,$x^3 + y^3 + z^3 = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$.

(A) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant. Specifically,$a_k + a_{n-k+1} = a_1 + a_n$.
Given $a_3 + a_7 + a_{11} + a_{15} = 72$.
We know that $a_3 + a_{15} = a_1 + a_{17}$ and $a_7 + a_{11} = a_1 + a_{17}$.
Substituting these into the given equation:
$(a_1 + a_{17}) + (a_1 + a_{17}) = 72$
$2(a_1 + a_{17}) = 72$
$a_1 + a_{17} = 36$.
The sum of the first $17$ terms is given by $S_{17} = \frac{17}{2}(a_1 + a_{17})$.
$S_{17} = \frac{17}{2} \times 36 = 17 \times 18 = 306$.

(C) Given the first term $a_1 = 10$ and the sum of the first three terms is $39.$
$a_1 + (a_1 + d) + (a_1 + 2d) = 39$
$3a_1 + 3d = 39$
$3(10) + 3d = 39$ $\Rightarrow 30 + 3d = 39$ $\Rightarrow 3d = 9$ $\Rightarrow d = 3.$
Let the number of terms be $n.$ The last four terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n.$
Their sum is $4a_1 + ( (n-4) + (n-3) + (n-2) + (n-1) )d = 178.$
$4(10) + (4n - 10)3 = 178$
$40 + 12n - 30 = 178$ $\Rightarrow 12n + 10 = 178$ $\Rightarrow 12n = 168$ $\Rightarrow n = 14.$
The median of an $A.P.$ with $n$ terms is the average of the first and last terms: $\frac{a_1 + a_n}{2}.$
$a_n = a_1 + (n-1)d = 10 + (14-1)3 = 10 + 39 = 49.$
Median $= \frac{10 + 49}{2} = \frac{59}{2} = 29.5.$

(C) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$
The second term is $a + d = 12$ .....$(1)$
The sum of the first nine terms is given by:
${S_9} = \frac{9}{2}(2a + 8d) = 9(a + 4d)$
Given that $200 < {S_9} < 220$:
$200 < 9(a + 4d) < 220$
Substitute $a = 12 - d$ from equation $(1)$ into the inequality:
$200 < 9(12 - d + 4d) < 220$
$200 < 9(12 + 3d) < 220$
$200 < 108 + 27d < 220$
Subtract $108$ from all parts:
$92 < 27d < 112$
Since the terms are positive integers,$d$ must be an integer. Testing values for $d$:
If $d = 3$,$27 \times 3 = 81$ (too small).
If $d = 4$,$27 \times 4 = 108$ (which satisfies $92 < 108 < 112$).
If $d = 5$,$27 \times 5 = 135$ (too large).
Thus,$d = 4$.
From equation $(1)$,$a + 4 = 12$,so $a = 8$.
The $4^{th}$ term is $a + 3d = 8 + 3(4) = 8 + 12 = 20$.

(B) The first series is $3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, \dots$ with common difference $d_1 = 4$.
The second series is $1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, \dots$ with common difference $d_2 = 5$.
The first common term is $11$.
The common difference of the new series formed by common terms is $LCM(d_1, d_2) = LCM(4, 5) = 20$.
Thus,the common terms form an Arithmetic Progression with first term $a = 11$ and common difference $d = 20$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 20$,$S_{20} = \frac{20}{2}[2(11) + (20 - 1)20]$.
$S_{20} = 10[22 + 19 \times 20] = 10[22 + 380] = 10[402] = 4020$.

(B) Let the total number of terms be $2n$,the first term be $a$,and the common difference be $d$.
The terms are $a, a+d, a+2d, ..., a+(2n-1)d$.
The odd-positioned terms are $a, a+2d, ..., a+(2n-2)d$. There are $n$ such terms.
Sum of odd terms $S_o = \frac{n}{2}[2a + (n-1)(2d)] = n[a + (n-1)d] = 24$ --- $(i)$
The even-positioned terms are $a+d, a+3d, ..., a+(2n-1)d$. There are $n$ such terms.
Sum of even terms $S_e = \frac{n}{2}[2(a+d) + (n-1)(2d)] = n[a+d+(n-1)d] = 30$ --- $(ii)$
Subtracting $(i)$ from $(ii)$:
$n(a+d+(n-1)d) - n(a+(n-1)d) = 30 - 24$
$nd = 6$ --- $(iii)$
Given the last term exceeds the first term by $10\frac{1}{2} = \frac{21}{2}$:
$(a+(2n-1)d) - a = \frac{21}{2}$
$(2n-1)d = \frac{21}{2}$
$2nd - d = \frac{21}{2}$
Substitute $nd = 6$ into the equation:
$2(6) - d = \frac{21}{2}$
$12 - d = 10.5$
$d = 1.5 = \frac{3}{2}$
Using $nd = 6$:
$n(\frac{3}{2}) = 6 \Rightarrow n = 4$
Total number of terms $= 2n = 2 \times 4 = 8$.

(B) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.
Given $\frac{S_p}{S_q} = \frac{p^3}{q^3}$,we have $\frac{\frac{p}{2}[2a_1 + (p-1)d]}{\frac{q}{2}[2a_1 + (q-1)d]} = \frac{p^3}{q^3}$.
Simplifying,$\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p^2}{q^2}$.
Cross-multiplying,$q^2(2a_1 + (p-1)d) = p^2(2a_1 + (q-1)d)$.
$2a_1q^2 + p q^2 d - q^2 d = 2a_1p^2 + p^2 q d - p^2 d$.
$2a_1(q^2 - p^2) = d(p^2 q - p q^2 + q^2 - p^2)$.
$2a_1(q-p)(q+p) = d(pq(p-q) - (p-q)(p+q))$.
$2a_1(q-p)(q+p) = d(p-q)(pq - p - q)$.
Since $p \neq q$,we divide by $(q-p)$: $2a_1(q+p) = -d(pq - p - q)$.
For this to hold for all $p, q$,we compare coefficients or substitute values. Using $p=1, q=2$: $\frac{a_1}{a_1+a_2} = \frac{1}{8}$ $\Rightarrow 8a_1 = a_1 + a_1 + d$ $\Rightarrow d = 6a_1$.
Then $\frac{a_6}{a_{21}} = \frac{a_1 + 5d}{a_1 + 20d} = \frac{a_1 + 5(6a_1)}{a_1 + 20(6a_1)} = \frac{31a_1}{121a_1} = \frac{31}{121}$.

(B) For the first $A.P.$,$S_n = 3n^2 + 2n$.
The first term $a = S_1 = 3(1)^2 + 2(1) = 5$.
The sum of the first two terms $S_2 = 3(2)^2 + 2(2) = 12 + 4 = 16$.
The second term $a_2 = S_2 - S_1 = 16 - 5 = 11$.
The common difference $d = a_2 - a = 11 - 5 = 6$.
For the new $A.P.$,the first term $a' = a = 5$ and the common difference $d' = 2d = 2(6) = 12$.
The sum of $n$ terms of the new $A.P.$ is $S_n' = \frac{n}{2} [2a' + (n - 1)d']$.
$S_n' = \frac{n}{2} [2(5) + (n - 1)12] = \frac{n}{2} [10 + 12n - 12] = \frac{n}{2} [12n - 2] = 6n^2 - n$.

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

(D) Given that the $A.M.$ of the $p^{th}$ and $q^{th}$ terms is equal to the $A.M.$ of the $r^{th}$ and $s^{th}$ terms of an $A.P.$:
$\frac{a_p + a_q}{2} = \frac{a_r + a_s}{2}$
Using the formula for the $n^{th}$ term of an $A.P.$,$a_n = a + (n-1)d$:
$a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d$
Simplifying the equation:
$2a + (p + q - 2)d = 2a + (r + s - 2)d$
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:
$p + q = r + s$

(A) Given $S = \sum_{i=1}^{30} {a_i}$ and $T = \sum_{i=1}^{15} {a_{2i-1}}$.
Let the $A.P.$ be defined as ${a_i} = a + (i-1)d$.
$S = {a_1} + {a_2} + {a_3} + \dots + {a_{30}}$
$T = {a_1} + {a_3} + {a_5} + \dots + {a_{29}}$
Then $2T = 2{a_1} + 2{a_3} + 2{a_5} + \dots + 2{a_{29}}$.
$S - 2T = ({a_2} - {a_1}) + ({a_4} - {a_3}) + ({a_6} - {a_5}) + \dots + ({a_{30}} - {a_{29}})$.
Since ${a_{2k}} - {a_{2k-1}} = d$,we have $S - 2T = 15d$.
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 ${a_{10}} = a + 9d$.
${a_{10}} = 7 + 9(5) = 7 + 45 = 52$.

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

(C) Let the first term 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: $t_{19} = a + 18d = 0$,which implies $a = -18d$.
We need to find the ratio $\frac{t_{49}}{t_{29}}$.
$t_{49} = a + 48d = -18d + 48d = 30d$.
$t_{29} = a + 28d = -18d + 28d = 10d$.
Therefore,the ratio is $\frac{30d}{10d} = \frac{3}{1}$,which is $3 : 1$.

(D) We need to find the sum of all $n$ such that $100 < n < 200$ and $H.C.F. (91, n) > 1$.
Since $91 = 7 \times 13$,$H.C.F. (91, n) > 1$ means $n$ 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 $k = \frac{196 - 105}{7} + 1 = 14$.
$S_A = \frac{14}{2} (105 + 196) = 7 \times 301 = 2107$.
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 $m = \frac{195 - 104}{13} + 1 = 8$.
$S_B = \frac{8}{2} (104 + 195) = 4 \times 299 = 1196$.
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 number is $182$.
$S_C = 182$.
By the Principle of Inclusion-Exclusion,the required sum is $S_A + S_B - S_C = 2107 + 1196 - 182 = 3121$.

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

(A) Let the three numbers in $A.P.$ be $a-d, a, a+d$.
Given that $(a-d) + a + (a+d) = 33$.
$3a = 33 \Rightarrow a = 11$.
Also,$(a-d)(a)(a+d) = 1155$.
$a(a^2 - d^2) = 1155$.
$11(121 - d^2) = 1155$.
$121 - d^2 = 105$.
$d^2 = 16 \Rightarrow d = \pm 4$.
If $d = 4$,the first term $A = a-d = 7$. The $11^{th}$ term $T_{11} = A + 10d = 7 + 10(4) = 47$.
If $d = -4$,the first term $A = a-d = 15$. The $11^{th}$ term $T_{11} = A + 10d = 15 + 10(-4) = -25$.

(A) The given sum is $a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114$.
This is an arithmetic progression with $6$ terms,where the first term is $a_1$ and the last term is $a_{16}$.
The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(first + last)$.
Thus,$\frac{6}{2}(a_1 + a_{16}) = 114$.
$3(a_1 + a_{16}) = 114 \Rightarrow a_1 + a_{16} = 38$.
We need to find the sum $S = a_1 + a_6 + a_{11} + a_{16}$.
This is also an arithmetic progression with $4$ terms,where the first term is $a_1$ and the last term is $a_{16}$.
$S = \frac{4}{2}(a_1 + a_{16}) = 2(38) = 76$.

(B) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
Given $a_6 = a + 5d = 2$,so $a = 2 - 5d$.
The terms are $a_1 = a = 2 - 5d$,$a_4 = a + 3d = 2 - 2d$,and $a_5 = a + 4d = 2 - d$.
Let the product be $f(d) = a_1 a_4 a_5 = (2 - 5d)(2 - 2d)(2 - d)$.
Expanding the expression: $f(d) = (4 - 4d - 10d + 10d^2)(2 - d) = (10d^2 - 14d + 4)(2 - d) = 20d^2 - 10d^3 - 28d + 14d^2 + 8 - 4d = -10d^3 + 34d^2 - 32d + 8$.
To find the maximum,find the derivative $f'(d) = -30d^2 + 68d - 32$.
Set $f'(d) = 0 \Rightarrow -2(15d^2 - 34d + 16) = 0 \Rightarrow 15d^2 - 34d + 16 = 0$.
Using the quadratic formula: $d = \frac{34 \pm \sqrt{34^2 - 4(15)(16)}}{2(15)} = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm \sqrt{196}}{30} = \frac{34 \pm 14}{30}$.
So,$d_1 = \frac{48}{30} = \frac{8}{5}$ and $d_2 = \frac{20}{30} = \frac{2}{3}$.
Find the second derivative: $f''(d) = -60d + 68$.
At $d = \frac{2}{3}$,$f''(\frac{2}{3}) = -60(\frac{2}{3}) + 68 = -40 + 68 = 28 > 0$ (local minimum).
At $d = \frac{8}{5}$,$f''(\frac{8}{5}) = -60(\frac{8}{5}) + 68 = -96 + 68 = -28 < 0$ (local maximum).
Thus,the product is maximized at $d = \frac{8}{5}$.

(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 $2a + 5d = -16$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(2a + 5d) - (2a + 3d) = -16 - 8$,so $2d = -24$,which gives $d = -12$.
Substituting $d = -12$ into Equation $1$: $2a + 3(-12) = 8$,so $2a - 36 = 8$,which gives $2a = 44$,so $a = 22$.
Now,$S_{10} = \frac{10}{2} \{2a + 9d\} = 5 \{2(22) + 9(-12)\} = 5 \{44 - 108\} = 5 \{-64\} = -320$.

(A) Let the first term be $a$ and the common difference be $d$.
Given that $a_1 + a_7 + a_{16} = 40$.
Using the formula for the $n^{th}$ term of an $A.P.$,$a_n = a + (n-1)d$,we have:
$a + (a + 6d) + (a + 15d) = 40$
$3a + 21d = 40$
$3(a + 7d) = 40$
$a + 7d = \frac{40}{3}$
We need to find the sum of the first $15$ terms,$S_{15}$.
$S_{15} = \frac{15}{2} [2a + (15-1)d]$
$S_{15} = \frac{15}{2} [2a + 14d]$
$S_{15} = 15(a + 7d)$
Substituting the value of $(a + 7d)$:
$S_{15} = 15 \times \frac{40}{3} = 5 \times 40 = 200$.

(C) Let the five numbers in $A.P.$ be $(a-2d, a-d, a, a+d, a+2d)$.
Given the sum is $25$,so $(a-2d) + (a-d) + a + (a+d) + (a+2d) = 25$,which simplifies to $5a = 25$,so $a = 5$.
The product is $(a-2d)(a-d)(a)(a+d)(a+2d) = 2520$.
Substituting $a=5$,we get $5(25-4d^2)(25-d^2) = 2520$,which simplifies to $(25-4d^2)(25-d^2) = 504$.
Expanding this,we get $625 - 25d^2 - 100d^2 + 4d^4 = 504$,or $4d^4 - 125d^2 + 121 = 0$.
Factoring the quadratic in $d^2$,we get $(4d^2 - 121)(d^2 - 1) = 0$,so $d^2 = 1$ or $d^2 = \frac{121}{4}$.
If $d^2 = 1$,the terms are $3, 4, 5, 6, 7$ or $7, 6, 5, 4, 3$. None of these is $-\frac{1}{2}$.
If $d^2 = \frac{121}{4}$,then $d = \pm \frac{11}{2}$.
For $d = \frac{11}{2}$,the terms are $5-11, 5-5.5, 5, 5+5.5, 5+11$,which are $-6, -0.5, 5, 10.5, 16$.
Since one of the numbers is $-\frac{1}{2}$,this sequence is valid. The greatest number is $16$.

(B) Let the first term be $a$ and the common difference be $d$.
Given $T_{10} = a + 9d = \frac{1}{20} \quad \dots (i)$
Given $T_{20} = a + 19d = \frac{1}{10} \quad \dots (ii)$
Subtracting $(i)$ 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 $(i)$:
$a + 9(\frac{1}{200}) = \frac{1}{20}$
$a = \frac{10}{200} - \frac{9}{200} = \frac{1}{200}$
Now,the sum of the first $200$ terms $S_{200}$ is given by:
$S_{200} = \frac{n}{2}[2a + (n-1)d]$
$S_{200} = \frac{200}{2}[2(\frac{1}{200}) + (200-1)(\frac{1}{200})]$
$S_{200} = 100[\frac{2}{200} + \frac{199}{200}]$
$S_{200} = 100[\frac{201}{200}] = \frac{201}{2} = 100 \frac{1}{2}$

(B) Given that $(2^{1+x}+2^{1-x})$,$f(x)$,and $(3^x+3^{-x})$ are in $A.P.$
By the property 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 $a^x + a^{-x} \geq 2$ for $a > 0$.
Thus,$2^x+2^{-x} \geq 2$ and $3^x+3^{-x} \geq 2$.
Therefore,$f(x) \geq 2 + \frac{1}{2}(2) = 2 + 1 = 3$.
The minimum value of $f(x)$ is $3$.

(D) The first $A$.$P$. is $A_1: 3, 7, 11, \ldots, 407$. Here,$a_1 = 3$ and $d_1 = 4$. The general term is $T_n = 3 + (n-1)4 = 4n - 1$.
The second $A$.$P$. is $A_2: 2, 9, 16, \ldots, 709$. Here,$a_2 = 2$ and $d_2 = 7$. The general term is $T_m = 2 + (m-1)7 = 7m - 5$.
For a common term,$4n - 1 = 7m - 5$,which implies $4n = 7m - 4$. This means $7m$ must be a multiple of $4$. Since $7$ is not divisible by $4$,$m$ must be a multiple of $4$. Let $m = 4k$.
Substituting $m = 4k$,we get $4n = 7(4k) - 4$,so $n = 7k - 1$.
The first common term is for $k=1$,which is $m=4$,$T_4 = 7(4)-5 = 23$.
The common difference of the new $A$.$P$. is $\text{lcm}(4, 7) = 28$.
The common terms are $23, 51, 79, \ldots$. The last term must be $\leq 407$.
$23 + (N-1)28 \leq 407$
$(N-1)28 \leq 384$
$N-1 \leq 13.71$
$N \leq 14.71$.
Thus,the number of common terms is $N = 14$.

(A) Given the sequence formula $a_{n} = 2n + 5$.
To find the first three terms,we substitute $n = 1, 2, 3$ into the formula:
For $n = 1$: $a_{1} = 2(1) + 5 = 2 + 5 = 7$
For $n = 2$: $a_{2} = 2(2) + 5 = 4 + 5 = 9$
For $n = 3$: $a_{3} = 2(3) + 5 = 6 + 5 = 11$
Therefore,the first three terms are $7, 9, 11$.