Geometric progression Questions in English

(D) Given that $G_1 = (x_1 \times x_2 \times \dots \times x_n)^{1/n}$ and $G_2 = (y_1 \times y_2 \times \dots \times y_n)^{1/n}$.
The geometric mean $G$ of the series $\frac{x_i}{y_i}$ is defined as:
$G = \left( \frac{x_1}{y_1} \times \frac{x_2}{y_2} \times \dots \times \frac{x_n}{y_n} \right)^{1/n}$
$G = \frac{(x_1 \times x_2 \times \dots \times x_n)^{1/n}}{(y_1 \times y_2 \times \dots \times y_n)^{1/n}}$
Substituting the values of $G_1$ and $G_2$:
$G = \frac{G_1}{G_2}$

(C) The geometric mean $(GM)$ of $n$ numbers $x_1, x_2, ..., x_n$ is defined as $(x_1 \times x_2 \times ... \times x_n)^{1/n}$.
For the first $n$ natural numbers,the set is ${1, 2, 3, ..., n}$.
The product of these numbers is $1 \times 2 \times 3 \times ... \times n = n!$.
Therefore,the geometric mean is $(n!)^{1/n}$.

(C) The sequence is a geometric progression: $a, ar, ar^2, \dots, ar^{n-1}$.
The geometric mean $(GM)$ of $n$ terms $x_1, x_2, \dots, x_n$ is given by $(x_1 \times x_2 \times \dots \times x_n)^{1/n}$.
Here,$GM = (a \times ar \times ar^2 \times \dots \times ar^{n-1})^{1/n}$.
$GM = (a^n \times r^{(0+1+2+\dots+(n-1))})^{1/n}$.
The sum of the first $(n-1)$ integers is $\frac{(n-1)n}{2}$.
$GM = (a^n \times r^{n(n-1)/2})^{1/n}$.
$GM = a \times r^{(n-1)/2} = ar^{(n-1)/2}$.

(D) The sequence is $1, 2, 2^2, \dots, 2^n$. The total number of terms is $n+1$.
The Geometric Mean $(G.M.)$ is given by the $(n+1)^{th}$ root of the product of all terms:
$G.M. = (1 \times 2 \times 2^2 \times \dots \times 2^n)^{\frac{1}{n+1}}$
The product of the terms is $2^{(0+1+2+\dots+n)}$. Using the sum of the first $n$ natural numbers formula $\frac{n(n+1)}{2}$,we get:
$G.M. = (2^{\frac{n(n+1)}{2}})^{\frac{1}{n+1}}$
Simplifying the exponent:
$G.M. = 2^{\frac{n(n+1)}{2(n+1)}} = 2^{n/2}$

(B) The given sequence is $1, 2, 4, 8, 16, \ldots, 2^n$.
This is a geometric progression $(G.P.)$ with the first term $a = 1$ and common ratio $r = 2$.
The number of terms in the sequence is $n+1$ (since the powers of $2$ range from $2^0$ to $2^n$).
The sum of the terms of the $G.P.$ is given by $S = \frac{a(r^{n+1} - 1)}{r - 1} = \frac{1(2^{n+1} - 1)}{2 - 1} = 2^{n+1} - 1$.
The arithmetic mean $(A.M.)$ is defined as the sum of the terms divided by the number of terms.
Therefore,$A.M. = \frac{S}{n+1} = \frac{2^{n+1} - 1}{n+1}$.

(B) The geometric mean of $n$ observations $x_1, x_2, ..., x_n$ is given by $(x_1 \cdot x_2 \cdot ... \cdot x_n)^{1/n}$.
For the given observations $2, 4, 8, 16, 32, 64$,we can write them as powers of $2$:
$2^1, 2^2, 2^3, 2^4, 2^5, 2^6$.
Geometric mean $= (2^1 \cdot 2^2 \cdot 2^3 \cdot 2^4 \cdot 2^5 \cdot 2^6)^{1/6}$.
Sum of exponents $= 1 + 2 + 3 + 4 + 5 + 6 = 21$.
Geometric mean $= (2^{21})^{1/6} = 2^{21/6} = 2^{7/2}$.

(A) Let the terms of the geometric progression be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 0$.
Given that each term equals the sum of the next two terms,we have:
$a = ar + ar^2$
Since $a \neq 0$,we can divide by $a$:
$1 = r + r^2$
$r^2 + r - 1 = 0$
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$
$r = \frac{-1 \pm \sqrt{5}}{2}$
Since the terms are positive,the common ratio $r$ must be positive.
Therefore,$r = \frac{\sqrt{5} - 1}{2}$.

(B) Let the first term be $a$ and the common ratio be $r$.
Given that the sum of the first two terms is $a + ar = a(1 + r) = 12$ $(i)$.
The sum of the third and fourth terms is $ar^2 + ar^3 = ar^2(1 + r) = 48$ $(ii)$.
Dividing equation $(ii)$ by equation $(i)$,we get:
$\frac{ar^2(1 + r)}{a(1 + r)} = \frac{48}{12}$
$r^2 = 4$
$r = \pm 2$.
Since the terms of the geometric progression are alternately positive and negative,the common ratio $r$ must be negative. Thus,$r = -2$.
Substituting $r = -2$ into equation $(i)$:
$a(1 + (-2)) = 12$
$a(-1) = 12$
$a = -12$.
Therefore,the first term is $-12$.

(A) Given that ${x_1}, {x_2}, {x_3}$ and ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio $r$,we have:
${x_2} = r{x_1}, {x_3} = {r^2}{x_1}$
${y_2} = r{y_1}, {y_3} = {r^2}{y_1}$
To check if the points are collinear,we calculate the area of the triangle formed by these points:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the values:
$\text{Area} = \frac{1}{2} |x_1(ry_1 - r^2y_1) + rx_1(r^2y_1 - y_1) + r^2x_1(y_1 - ry_1)|$
$\text{Area} = \frac{1}{2} |x_1y_1(r - r^2) + x_1y_1(r^3 - r) + x_1y_1(r^2 - r^3)|$
$\text{Area} = \frac{1}{2} |x_1y_1(r - r^2 + r^3 - r + r^2 - r^3)| = 0$
Since the area is $0$,the points lie on a straight line.

(A) Let the $G$.$P$. be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.
Given $\alpha = \sum\limits_{n = 1}^{100} {{a_{2n}}} = a_2 + a_4 + \dots + a_{200}$.
This is a $G$.$P$. with first term $ar$ and common ratio $r^2$ having $100$ terms.
So,$\alpha = ar(1 + r^2 + r^4 + \dots + r^{198})$.
Given $\beta = \sum\limits_{n = 1}^{100} {{a_{2n - 1}}} = a_1 + a_3 + \dots + a_{199}$.
This is a $G$.$P$. with first term $a$ and common ratio $r^2$ having $100$ terms.
So,$\beta = a(1 + r^2 + r^4 + \dots + r^{198})$.
Dividing $\alpha$ by $\beta$:
$\frac{\alpha}{\beta} = \frac{ar(1 + r^2 + r^4 + \dots + r^{198})}{a(1 + r^2 + r^4 + \dots + r^{198})} = r$.
Thus,the common ratio is $\frac{\alpha}{\beta}$.

(C) Given that $a, b$ are roots of $x^2 - 3x + p = 0$,we have $a + b = 3$ and $ab = p$.
Given that $c, d$ are roots of $x^2 - 12x + q = 0$,we have $c + d = 12$ and $cd = q$.
Since $a, b, c, d$ are in an increasing $G$.$P$.,let the terms be $a, ar, ar^2, ar^3$.
Then $a + b = a(1 + r) = 3$ and $c + d = ar^2(1 + r) = 12$.
Dividing the two equations: $\frac{ar^2(1 + r)}{a(1 + r)} = \frac{12}{3} \Rightarrow r^2 = 4$. Since the $G$.$P$. is increasing,$r = 2$.
Substituting $r = 2$ into $a(1 + 2) = 3$,we get $3a = 3$,so $a = 1$.
The terms are $1, 2, 4, 8$.
Thus,$p = ab = 1 \times 2 = 2$ and $q = cd = 4 \times 8 = 32$.
The ratio $(q + p) : (q - p) = (32 + 2) : (32 - 2) = 34 : 30 = 17 : 15$.

(C) Given the infinite geometric series $s_1 = 1 - \alpha + \alpha^2 - \alpha^3 + \dots = \frac{1}{1 + \alpha}$ and $s_2 = 1 - \beta + \beta^2 - \beta^3 + \dots = \frac{1}{1 + \beta}$.
From these,we have $1 + \alpha = \frac{1}{s_1} \implies \alpha = \frac{1}{s_1} - 1$ and $1 + \beta = \frac{1}{s_2} \implies \beta = \frac{1}{s_2} - 1$.
Let $s = 1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots = \frac{1}{1 + \alpha\beta}$.
Substituting the values of $\alpha$ and $\beta$:
$s = \frac{1}{1 + (\frac{1}{s_1} - 1)(\frac{1}{s_2} - 1)} = \frac{1}{1 + (\frac{1 - s_1}{s_1})(\frac{1 - s_2}{s_2})} = \frac{s_1s_2}{s_1s_2 + (1 - s_1)(1 - s_2)}$.
$s = \frac{s_1s_2}{s_1s_2 + 1 - s_1 - s_2 + s_1s_2} = \frac{s_1s_2}{1 - s_1 - s_2 + 2s_1s_2}$.

(B) Given the inequality: $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$ $(i)$
We can rewrite the expression on the left-hand side as:
$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0$
This simplifies to:
$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$ $(ii)$
Since the sum of squares of real numbers is always non-negative,the only way the sum can be less than or equal to $0$ is if each square term is exactly $0$:
$(ap - b)^2 = 0, (bp - c)^2 = 0, (cp - d)^2 = 0$
This implies:
$ap = b, bp = c, cp = d$
Therefore:
$\frac{b}{a} = p, \frac{c}{b} = p, \frac{d}{c} = p$
Since the ratio between consecutive terms is constant $(p)$,the sequence $a, b, c, d$ is in $G.P.$

(C) Let $r > 1$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$.
Then $\beta = r\alpha, \gamma = r^2\alpha, \delta = r^3\alpha$.
From the sum of roots for the first equation: $\alpha + \beta = \alpha(1 + r) = 3$ $(i)$.
From the product of roots for the first equation: $\alpha\beta = \alpha^2r = a$ $(ii)$.
From the sum of roots for the second equation: $\gamma + \delta = \alpha r^2(1 + r) = 12$ $(iii)$.
From the product of roots for the second equation: $\gamma\delta = \alpha^2r^5 = b$ $(iv)$.
Dividing $(iii)$ by $(i)$: $\frac{\alpha r^2(1 + r)}{\alpha(1 + r)} = \frac{12}{3} \Rightarrow r^2 = 4$. Since the $G.P.$ is increasing,$r = 2$.
Substituting $r = 2$ into $(i)$: $\alpha(1 + 2) = 3$ $\Rightarrow 3\alpha = 3$ $\Rightarrow \alpha = 1$.
Now,$a = \alpha^2r = (1)^2(2) = 2$.
And $b = \alpha^2r^5 = (1)^2(2^5) = 32$.
Thus,$a = 2$ and $b = 32$.

(B) Given $x = 1 + \cos^2\phi + \cos^4\phi + \dots = \frac{1}{1 - \cos^2\phi} = \frac{1}{\sin^2\phi}$.
Given $y = 1 + \sin^2\phi + \sin^4\phi + \dots = \frac{1}{1 - \sin^2\phi} = \frac{1}{\cos^2\phi}$.
Given $z = 1 + \cos^2\phi \sin^2\phi + \cos^4\phi \sin^4\phi + \dots = \frac{1}{1 - \cos^2\phi \sin^2\phi}$.
Now,$xy = \frac{1}{\sin^2\phi \cos^2\phi}$.
Then $xy + z = \frac{1}{\sin^2\phi \cos^2\phi} + \frac{1}{1 - \cos^2\phi \sin^2\phi} = \frac{1 - \cos^2\phi \sin^2\phi + \sin^2\phi \cos^2\phi}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)} = \frac{1}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)}$.
Also,$xyz = \left(\frac{1}{\sin^2\phi}\right) \left(\frac{1}{\cos^2\phi}\right) \left(\frac{1}{1 - \cos^2\phi \sin^2\phi}\right) = \frac{1}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)}$.
Thus,$xyz = xy + z$.

(C) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = -x$.
For the series to converge,$|r| < 1$,i.e.,$|-x| < 1$ or $|x| < 1$.
The sum of the infinite geometric series is given by $S = \frac{a}{1 - r}$.
Substituting the values,we get $x = \frac{1}{1 - (-x)} = \frac{1}{1 + x}$.
This simplifies to $x(1 + x) = 1$,which is $x^2 + x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Since $|x| < 1$,we reject $x = \frac{-1 - \sqrt{5}}{2} \approx -1.618$.
Thus,$x = \frac{\sqrt{5} - 1}{2}$.
We know that $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$.
Therefore,$x = 2 \sin 18^\circ$.

(A) Let $P(x) = (x - 1)(x - \frac{1}{2})(x - \frac{1}{2^2}) \dots (x - \frac{1}{2^{49}})$.
This is a polynomial of degree $50$.
The coefficient of $x^{50}$ is $1$.
The coefficient of $x^{49}$ is the sum of the constant terms of each factor multiplied by $-1$,which is $-(1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{49}})$.
This is a geometric progression with $a = 1$,$r = \frac{1}{2}$,and $n = 50$ terms.
The sum $S_{50} = \frac{1(1 - (1/2)^{50})}{1 - 1/2} = 2(1 - \frac{1}{2^{50}})$.
Therefore,the coefficient of $x^{49}$ is $-2(1 - \frac{1}{2^{50}})$.

(C) Given $\log_8(a_1 a_2 \dots a_{12}) = 2014$.
Since $a_k = a_1 r^{k-1}$,the product is $a_1^{12} r^{0+1+2+\dots+11} = a_1^{12} r^{66}$.
Thus,$\log_8(a_1^{12} r^{66}) = 2014 \Rightarrow a_1^{12} r^{66} = 8^{2014} = 2^{6042}$.
Let $a_1 = 2^m$ and $r = 2^n$ where $m, n$ are integers.
Then $(2^m)^{12} (2^n)^{66} = 2^{12m + 66n} = 2^{6042}$.
This implies $12m + 66n = 6042$,which simplifies to $2m + 11n = 1007$.
Since $r > 1$ and the progression is increasing,$n \ge 1$. Also $a_1 \ge 1$,so $m \ge 0$.
$2m = 1007 - 11n$. For $m$ to be an integer,$1007 - 11n$ must be even,so $n$ must be odd.
Also $m > 0$ $\Rightarrow 1007 - 11n > 0$ $\Rightarrow n < \frac{1007}{11} \approx 91.54$.
Thus $n \in \{1, 3, 5, \dots, 91\}$.
The number of such odd values is $\frac{91 - 1}{2} + 1 = 46$.

(B) Given $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} = k$.
Applying Componendo and Dividendo to each term:
$\frac{(a + bx) + (a - bx)}{(a + bx) - (a - bx)} = \frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)} = \frac{(c + dx) + (c - dx)}{(c + dx) - (c - dx)}$.
This simplifies to:
$\frac{2a}{2bx} = \frac{2b}{2cx} = \frac{2c}{2dx}$.
Dividing by $\frac{2}{x}$ (since $x \neq 0$):
$\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$.
This implies that $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.
Therefore,$a, b, c, d$ are in $G.P.$

(A) Let the first term $a_1 = 1$ and the common ratio be $r$.
Then the terms are $a_1 = 1, a_2 = r, a_3 = r^2$.
We want to minimize the expression $f(r) = 4a_2 + 5a_3 = 4r + 5r^2$.
To find the minimum,we take the derivative with respect to $r$ and set it to zero:
$f'(r) = \frac{d}{dr}(4r + 5r^2) = 4 + 10r = 0$.
Solving for $r$:
$10r = -4 \Rightarrow r = -\frac{4}{10} = -0.4$.
Since $f''(r) = 10 > 0$,the function has a minimum at $r = -0.4$.

(B) Let the three numbers be $3^a, 3^b, 3^c$ where $1 \le a < b < c \le 20$.
For these to form a $G.P.$,the condition is $(3^b)^2 = 3^a \times 3^c$,which implies $2b = a + c$.
This means $a$ and $c$ must have the same parity (both even or both odd) so that their sum is even.
Case $1$: $a$ and $c$ are both odd.
There are $10$ odd numbers in the set $\{1, 2, \dots, 20\}$,which are $\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}$.
We need to choose $2$ distinct numbers from these $10$ to be $a$ and $c$,which can be done in $^{10}C_2 = 45$ ways.
Once $a$ and $c$ are chosen,$b = (a+c)/2$ is uniquely determined and satisfies $a < b < c$.
Case $2$: $a$ and $c$ are both even.
There are $10$ even numbers in the set $\{1, 2, \dots, 20\}$,which are $\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$.
We need to choose $2$ distinct numbers from these $10$ to be $a$ and $c$,which can be done in $^{10}C_2 = 45$ ways.
Once $a$ and $c$ are chosen,$b = (a+c)/2$ is uniquely determined and satisfies $a < b < c$.
Total ways = $45 + 45 = 90$.

(A) The sum of interior angles of an $n$-sided polygon is given by $(n-2) \times 180^\circ$.
The angles are in $G.P.$ with first term $a = 1^\circ$ and common ratio $r = 2$.
The sum of $n$ terms of a $G.P.$ is $S_n = a \frac{r^n - 1}{r - 1}$.
Substituting the values: $S_n = 1 \times \frac{2^n - 1}{2 - 1} = 2^n - 1$.
Equating the two expressions for the sum: $2^n - 1 = (n - 2) \times 180$.
$2^n - 1 = 180n - 360$.
$2^n = 180n - 359$.
For any integer $n \ge 3$,the $L.H.S.$ $(2^n)$ is always even,while the $R.H.S.$ $(180n - 359)$ is always odd.
Since an even number cannot equal an odd number,there are no possible values for $n$.

(C) Since $2a+b, a-b, a+3b$ are in $G.P.$,we have $(a-b)^2 = (2a+b)(a+3b)$.
Expanding both sides: $a^2 - 2ab + b^2 = 2a^2 + 6ab + ab + 3b^2$.
Rearranging the terms: $a^2 + 9ab + 2b^2 = 0$.
Dividing by $a^2$ (since $a \neq 0$): $2(\frac{b}{a})^2 + 9(\frac{b}{a}) + 1 = 0$.
Let $x = \frac{b}{a}$. Then $2x^2 + 9x + 1 = 0$. The roots are $x_1 = \frac{b_1}{a_1}$ and $x_2 = \frac{b_2}{a_2}$.
From the quadratic equation,$x_1 + x_2 = -\frac{9}{2}$ and $x_1x_2 = \frac{1}{2}$.
We need to find $2(a_1b_2 + a_2b_1) + 9a_1a_2$.
Divide the expression by $a_1a_2$: $\frac{2(a_1b_2 + a_2b_1) + 9a_1a_2}{a_1a_2} = 2(\frac{b_2}{a_2} + \frac{b_1}{a_1}) + 9 = 2(x_1 + x_2) + 9$.
Substituting the sum of roots: $2(-\frac{9}{2}) + 9 = -9 + 9 = 0$.

(A) The given series is an infinite geometric progression with first term $a = \frac{4}{3}$ and common ratio $r = -\frac{x}{3}$.
For the sum to exist,the condition $|r| < 1$ must be satisfied,which implies $|-\frac{x}{3}| < 1$,or $|x| < 3$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
Substituting the values,we get $x = \frac{4/3}{1 - (-x/3)} = \frac{4/3}{1 + x/3} = \frac{4}{3+x}$.
Cross-multiplying gives $x(3+x) = 4$,which simplifies to $x^2 + 3x - 4 = 0$.
Factoring the quadratic equation,we get $(x+4)(x-1) = 0$,so $x = 1$ or $x = -4$.
Since the condition for convergence is $|x| < 3$,we must have $x = 1$ because $|-4| > 3$ is false.
Therefore,$x = 1$ is the only valid solution.

(D) Let the roots be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$.
From the sum of roots,$\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 40$.
From the sum of reciprocals,$\frac{r^2}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} = 10$.
This can be written as $\frac{1}{a} (r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2}) = 10$.
Also,the sum of roots is $a(\frac{1}{r^2} + \frac{1}{r} + 1 + r + r^2) = 40$.
Dividing the two equations: $\frac{a(r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2})}{\frac{1}{a}(r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2})} = \frac{40}{10}$ $\Rightarrow a^2 = 4$ $\Rightarrow a = 2$ (assuming positive roots).
The product of the roots is $s = -(\frac{a}{r^2} \cdot \frac{a}{r} \cdot a \cdot ar \cdot ar^2) = -a^5$.
Since $a = 2$,$s = -(2)^5 = -32$.
Therefore,$|s| = |-32| = 32$.

(C) The given series is a geometric progression with first term $a = 1$ and common ratio $r = \frac{1}{2}$.
The sum of the first $n$ terms is given by $S_n = \frac{a(1 - r^n)}{1 - r} = \frac{1(1 - (1/2)^n)}{1 - 1/2} = 2(1 - \frac{1}{2^n}) = 2 - \frac{2}{2^n} = 2 - \frac{1}{2^{n-1}}$.
We are given the condition $2 - S_n < \frac{1}{100}$.
Substituting $S_n$,we get $2 - (2 - \frac{1}{2^{n-1}}) < \frac{1}{100}$,which simplifies to $\frac{1}{2^{n-1}} < \frac{1}{100}$.
This implies $2^{n-1} > 100$.
We know that $2^6 = 64$ and $2^7 = 128$.
Therefore,$n - 1 \geq 7$,which gives $n \geq 8$.
The least integral value of $n$ is $8$.

(B) The sum of the infinite geometric series is given by $S = \frac{a}{1-r} = 7 \quad \dots(1)$
The terms with odd powers of $r$ are $ar, ar^3, ar^5, \dots$,which form a geometric series with first term $ar$ and common ratio $r^2$.
The sum of this series is $\frac{ar}{1-r^2} = 3 \quad \dots(2)$
Dividing $(2)$ by $(1)$:
$\frac{ar}{1-r^2} \div \frac{a}{1-r} = \frac{3}{7}$
$\frac{ar}{(1-r)(1+r)} \times \frac{1-r}{a} = \frac{3}{7}$
$\frac{r}{1+r} = \frac{3}{7}$
$7r = 3 + 3r$ $\Rightarrow 4r = 3$ $\Rightarrow r = \frac{3}{4}$
Substitute $r = \frac{3}{4}$ into $(1)$:
$\frac{a}{1 - 3/4} = 7$ $\Rightarrow \frac{a}{1/4} = 7$ $\Rightarrow a = \frac{7}{4}$
Now,calculate $(a^2 - r^2)$:
$a^2 - r^2 = (\frac{7}{4})^2 - (\frac{3}{4})^2 = \frac{49}{16} - \frac{9}{16} = \frac{40}{16} = \frac{5}{2}$

(B) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin x$.
For the sum to exist,$|\sin x| < 1$.
The sum of the infinite series is given by $S = \frac{a}{1 - r} = \frac{1}{1 - \sin x}$.
Given $\frac{1}{1 - \sin x} = 4 + 2\sqrt{3}$.
$1 - \sin x = \frac{1}{4 + 2\sqrt{3}} = \frac{4 - 2\sqrt{3}}{(4 + 2\sqrt{3})(4 - 2\sqrt{3})} = \frac{4 - 2\sqrt{3}}{16 - 12} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$.
Therefore,$\sin x = \frac{\sqrt{3}}{2}$.
Since $0 < x < \pi$,the possible values for $x$ are $x = \frac{\pi}{3}$ or $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
Comparing with the given options,$x = \frac{\pi}{3}$ is a valid solution.

(B) $A_n$ is a geometric progression with first term $a = \frac{3}{4}$ and common ratio $r = -\frac{3}{4}$.
The sum of $n$ terms is $A_n = \frac{a(1-r^n)}{1-r} = \frac{\frac{3}{4}(1 - (-\frac{3}{4})^n)}{1 - (-\frac{3}{4})} = \frac{\frac{3}{4}(1 - (-\frac{3}{4})^n)}{\frac{7}{4}} = \frac{3}{7} \left[ 1 - \left( -\frac{3}{4} \right)^n \right]$.
Given $B_n = 1 - A_n$,the condition $B_n > A_n$ implies $1 - A_n > A_n$,which simplifies to $1 > 2A_n$,or $A_n < \frac{1}{2}$.
Substituting $A_n$: $\frac{3}{7} \left[ 1 - \left( -\frac{3}{4} \right)^n \right] < \frac{1}{2} \implies 1 - \left( -\frac{3}{4} \right)^n < \frac{7}{6}$.
This gives $-\left( -\frac{3}{4} \right)^n < \frac{1}{6}$,or $\left( -\frac{3}{4} \right)^n > -\frac{1}{6}$.
For odd $n$,$\left( -\frac{3}{4} \right)^n = -\left( \frac{3}{4} \right)^n$. So,$-\left( \frac{3}{4} \right)^n > -\frac{1}{6}$,which means $\left( \frac{3}{4} \right)^n < \frac{1}{6}$.
Taking logarithms: $n \log(\frac{3}{4}) < \log(\frac{1}{6}) \implies n(\log 3 - \log 4) < -\log 6$.
$n(0.4771 - 0.6020) < -0.7781 \implies n(-0.1249) < -0.7781 \implies n > \frac{0.7781}{0.1249} \approx 6.23$.
The least odd natural number $p$ satisfying $n > 6.23$ is $p = 7$.

(D) Let the first three terms of the $G.P.$ be $a, ar, ar^2$.
According to the problem,the product of the first three terms is $1000$:
$a(ar)(ar^2) = 1000$ $\Rightarrow (ar)^3 = 1000$ $\Rightarrow ar = 10$.
The sum of the $3^{rd}$ and $4^{th}$ terms is $60$:
$ar^2 + ar^3 = 60 \Rightarrow ar(r + r^2) = 60$.
Substituting $ar = 10$ into the equation:
$10(r + r^2) = 60 \Rightarrow r^2 + r - 6 = 0$.
Factoring the quadratic equation:
$(r + 3)(r - 2) = 0 \Rightarrow r = 2$ or $r = -3$.
Case $1$: If $r = 2$,then $a(2) = 10 \Rightarrow a = 5$.
Case $2$: If $r = -3$,then $a(-3) = 10 \Rightarrow a = -10/3$.
Since the first term $a$ must be positive,we choose $a = 5$ and $r = 2$.
The $7^{th}$ term is given by $T_7 = ar^6 = 5(2)^6 = 5 \times 64 = 320$.

(C) Let the geometric progression be $a, ar, ar^2, ar^3, ar^4$.
The sum of the first $5$ terms is $S_5 = a + ar + ar^2 + ar^3 + ar^4 = \frac{a(r^5 - 1)}{r - 1}$.
The reciprocals are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}, \frac{1}{ar^4}$.
The sum of these reciprocals is $S'_5 = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4} = \frac{1}{a} \left( \frac{1 - (1/r)^5}{1 - 1/r} \right) = \frac{1}{a} \left( \frac{(r^5 - 1)/r^5}{(r - 1)/r} \right) = \frac{1}{a} \cdot \frac{r^5 - 1}{r^5} \cdot \frac{r}{r - 1} = \frac{r^5 - 1}{a r^4 (r - 1)}$.
Given the ratio $\frac{S_5}{S'_5} = 49$:
$\frac{\frac{a(r^5 - 1)}{r - 1}}{\frac{r^5 - 1}{a r^4 (r - 1)}} = 49$ $\Rightarrow a^2 r^4 = 49$ $\Rightarrow ar^2 = 7$ (since $a$ and $r$ are real).
Given the sum of the first and third term is $35$:
$a + ar^2 = 35$.
Substituting $ar^2 = 7$ into the equation:
$a + 7 = 35 \Rightarrow a = 28$.

(C) The given expression is $1 - \sum_{k=1}^{n-1} \frac{2}{3^k} < \frac{1}{100}$.
This can be written as $1 - 2 \left( \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^{n-1}} \right) < \frac{1}{100}$.
The sum inside the bracket is a geometric progression with $a = \frac{1}{3}$,$r = \frac{1}{3}$,and $n-1$ terms.
The sum $S_{n-1} = \frac{a(1-r^{n-1})}{1-r} = \frac{\frac{1}{3}(1 - (\frac{1}{3})^{n-1})}{1 - \frac{1}{3}} = \frac{\frac{1}{3}(1 - \frac{1}{3^{n-1}})}{\frac{2}{3}} = \frac{1}{2} (1 - \frac{1}{3^{n-1}}) = \frac{1}{2} - \frac{1}{2 \cdot 3^{n-1}}$.
Substituting this back: $1 - 2 \left( \frac{1}{2} - \frac{1}{2 \cdot 3^{n-1}} \right) < \frac{1}{100}$.
$1 - 1 + \frac{1}{3^{n-1}} < \frac{1}{100}$.
$\frac{1}{3^{n-1}} < \frac{1}{100} \Rightarrow 3^{n-1} > 100$.
For $n=5$,$3^{5-1} = 3^4 = 81$ (which is not $> 100$).
For $n=6$,$3^{6-1} = 3^5 = 243$ (which is $> 100$).
Thus,the least positive integer $n$ is $6$.

(C) Let the terms of the $G.P.$ 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:
$ar^3 - a = 52 \Rightarrow a(r^3 - 1) = 52 \quad ......(1)$
$a + ar + ar^2 = 26 \Rightarrow a(1 + r + r^2) = 26 \quad ......(2)$
We know that $r^3 - 1 = (r - 1)(r^2 + r + 1)$.
Dividing equation $(1)$ by equation $(2)$:
$\frac{a(r - 1)(r^2 + r + 1)}{a(1 + r + r^2)} = \frac{52}{26}$
$r - 1 = 2 \Rightarrow r = 3$.
Substituting $r = 3$ in equation $(2)$:
$a(1 + 3 + 9) = 26$ $\Rightarrow 13a = 26$ $\Rightarrow a = 2$.
The sum of the first six terms is $S_6 = a(1 + r + r^2 + r^3 + r^4 + r^5) = a(1 + r + r^2)(1 + r^3)$.
$S_6 = 26 \times (1 + 3^3) = 26 \times (1 + 27) = 26 \times 28 = 728$.

(D) The given inequality is $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$.
This can be rearranged as:
$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bpc + c^2) + (c^2p^2 - 2pcd + d^2) \le 0$.
This simplifies to:
$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$.
Since the sum of squares of real numbers is non-negative,the only way for the sum to be $\le 0$ is if each term is exactly zero:
$ap - b = 0$,$bp - c = 0$,and $cp - d = 0$.
This implies $p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.
Therefore,$a, b, c, d$ are in $G.P.$

(D) Let the three numbers in $G.P.$ be $a, ar, ar^2$ where $r \neq 1$ (since the numbers are distinct).
Given $a + ar + ar^2 = x(ar)$.
Since $a \neq 0$,we can divide by $a$:
$1 + r + r^2 = xr$
$x = \frac{1 + r + r^2}{r} = r + 1 + \frac{1}{r} = (r + \frac{1}{r}) + 1$.
We know that for $r > 0$,$r + \frac{1}{r} \geq 2$,so $x \geq 2 + 1 = 3$.
For $r < 0$,let $r = -k$ where $k > 0$. Then $r + \frac{1}{r} = -(k + \frac{1}{k}) \leq -2$.
So $x \leq -2 + 1 = -1$.
Thus,$x \in (-\infty, -1] \cup [3, \infty)$.
Since the numbers are distinct,$r \neq 1$,so $x \neq 1 + 1 + 1 = 3$.
Therefore,$x$ cannot be any value in the interval $(-1, 3)$.
Among the given options,$-2$ is in the range $(-\infty, -1]$,$-3$ is in $(-\infty, -1]$,and $4$ is in $[3, \infty)$.
However,$2$ lies in the interval $(-1, 3)$,so $x$ cannot be $2$.

(C) For $5, 5r, 5r^2$ to be the sides of a triangle,the sum of any two sides must be greater than the third side.
$1) 5 + 5r > 5r^2 \Rightarrow r^2 - r - 1 < 0$. The roots of $r^2 - r - 1 = 0$ are $r = \frac{1 \pm \sqrt{5}}{2}$. Since $r > 0$,we have $0 < r < \frac{1 + \sqrt{5}}{2} \approx 1.618$.
$2) 5 + 5r^2 > 5r \Rightarrow r^2 - r + 1 > 0$. This is true for all $r \in \mathbb{R}$ as the discriminant $D = (-1)^2 - 4(1)(1) = -3 < 0$.
$3) 5r + 5r^2 > 5 \Rightarrow r^2 + r - 1 > 0$. The roots of $r^2 + r - 1 = 0$ are $r = \frac{-1 \pm \sqrt{5}}{2}$. Since $r > 0$,we have $r > \frac{-1 + \sqrt{5}}{2} \approx 0.618$.
Combining these,we get $\frac{\sqrt{5} - 1}{2} < r < \frac{\sqrt{5} + 1}{2}$,which is approximately $0.618 < r < 1.618$.
Checking the options:
$A) \frac{3}{4} = 0.75$ (in range)
$B) \frac{5}{4} = 1.25$ (in range)
$C) \frac{7}{4} = 1.75$ (outside range)
$D) \frac{3}{2} = 1.5$ (in range)
Thus,$r$ cannot be $\frac{7}{4}$.

(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r} = 3$.
Cubing both sides,we get $\frac{a^3}{(1-r)^3} = 27 \quad (1)$.
The series of the cubes of the terms is $a^3, a^3r^3, a^3r^6, \dots$,which is also a geometric series with first term $a^3$ and common ratio $r^3$.
The sum of this series is $\frac{a^3}{1-r^3} = \frac{27}{19} \quad (2)$.
Dividing equation $(1)$ by equation $(2)$,we get $\frac{1-r^3}{(1-r)^3} = \frac{27}{27/19} = 19$.
Using the identity $1-r^3 = (1-r)(1+r+r^2)$,we have $\frac{(1-r)(1+r+r^2)}{(1-r)^3} = 19$,which simplifies to $\frac{1+r+r^2}{(1-r)^2} = 19$.
$1+r+r^2 = 19(1-2r+r^2) \implies 1+r+r^2 = 19-38r+19r^2$.
$18r^2 - 39r + 18 = 0$. Dividing by $3$,we get $6r^2 - 13r + 6 = 0$.
$(2r-3)(3r-2) = 0$. Since $|r| < 1$ for an infinite geometric series,$r = \frac{2}{3}$.

(A) Given that $a_1, a_2, ..., a_{10}$ is a $G.P.$ with common ratio $r$.
We know that $a_n = a_1 r^{n-1}$.
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 $\frac{a_9}{a_5}$.
$\frac{a_9}{a_5} = \frac{a_1 r^8}{a_1 r^4} = r^4$.
Since $r^2 = 25$,then $r^4 = (r^2)^2 = 25^2 = (5^2)^2 = 5^4$.
Thus,the value is $5^4$.

(A) Given $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$.
Since it is a $G$.$P$.,$a_{2} = a_{1}r$ and $a_{3} = a_{1}r^{2}$,$a_{4} = a_{1}r^{3}$.
$a_{1}(1 + r) = 4$ --- $(1)$
$a_{1}r^{2}(1 + r) = 16$ --- $(2)$
Dividing $(2)$ by $(1)$,we get $r^{2} = 4$,so $r = 2$ or $r = -2$.
If $r = 2$,$a_{1}(1 + 2) = 4 \Rightarrow a_{1} = 4/3 > 0$,which contradicts $a_{1} < 0$.
Thus,$r = -2$. Substituting in $(1)$,$a_{1}(1 - 2) = 4$ $\Rightarrow -a_{1} = 4$ $\Rightarrow a_{1} = -4$.
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 $4 \lambda = -684 \Rightarrow \lambda = -171$.

(C) The given sum is a geometric series: $S = 1 + 49 + 49^2 + \ldots + 49^{125}$.
Using the formula for the sum of a geometric progression,$S = \frac{49^{126}-1}{49-1} = \frac{49^{126}-1}{48}$.
We can write $49^{126}-1$ as $(49^{63})^2 - 1^2 = (49^{63}-1)(49^{63}+1)$.
Thus,$S = \frac{(49^{63}-1)(49^{63}+1)}{48}$.
For $49^k+1$ to be a factor of $S$,we observe that $49^{63}+1$ is a factor of $S$ because $48$ divides $(49^{63}-1)$ (since $49 \equiv 1 \pmod{48}$,so $49^{63} \equiv 1^{63} \equiv 1 \pmod{48}$,implying $49^{63}-1$ is a multiple of $48$).
Therefore,the greatest positive integer $k$ is $63$.

(D) Given that $a_n$ is a $G$.$P$. with common ratio $r$.
$\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + \dots + a_{201} = 200$
This is a $G$.$P$. with first term $a_3 = ar^2$ and common ratio $r^2$ with $100$ terms.
So,$ar^2 \frac{(r^2)^{100} - 1}{r^2 - 1} = ar^2 \frac{r^{200} - 1}{r^2 - 1} = 200$
$\sum_{n=1}^{100} a_{2n} = a_2 + a_4 + \dots + a_{200} = 100$
This is a $G$.$P$. with first term $a_2 = ar$ and common ratio $r^2$ with $100$ terms.
So,$ar \frac{(r^2)^{100} - 1}{r^2 - 1} = ar \frac{r^{200} - 1}{r^2 - 1} = 100$
Dividing the two equations: $\frac{ar^2 \frac{r^{200} - 1}{r^2 - 1}}{ar \frac{r^{200} - 1}{r^2 - 1}} = \frac{200}{100} \Rightarrow r = 2$
We need to find $S = \sum_{n=1}^{200} a_n = a_1 + a_2 + \dots + a_{200}$.
Note that $\sum_{n=1}^{100} a_{2n+1} + \sum_{n=1}^{100} a_{2n} = a_2 + a_3 + a_4 + \dots + a_{201} = 300$.
Since $a_{k+1} = r a_k$,we have $a_2 + a_3 + \dots + a_{201} = r(a_1 + a_2 + \dots + a_{200}) = 300$.
Substituting $r = 2$: $2 \sum_{n=1}^{200} a_n = 300 \Rightarrow \sum_{n=1}^{200} a_n = 150$.

(A) The given expression is $P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \dots \infty$.
Expressing all terms with base $2$:
$P = 2^{\frac{1}{4}} \cdot (2^2)^{\frac{1}{16}} \cdot (2^3)^{\frac{1}{48}} \cdot (2^4)^{\frac{1}{128}} \cdot \dots$
$P = 2^{\frac{1}{4}} \cdot 2^{\frac{2}{16}} \cdot 2^{\frac{3}{48}} \cdot 2^{\frac{4}{128}} \cdot \dots$
$P = 2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 2^{\frac{1}{16}} \cdot 2^{\frac{1}{32}} \cdot \dots$
Using the property $a^m \cdot a^n = a^{m+n}$,we get:
$P = 2^{(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots)}$
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 $n^{th}$ term of the sequence is $a_{n} = 2^{n}$.
To find the first five terms,we substitute $n = 1, 2, 3, 4, 5$ into the formula:
For $n = 1$: $a_{1} = 2^{1} = 2$
For $n = 2$: $a_{2} = 2^{2} = 4$
For $n = 3$: $a_{3} = 2^{3} = 8$
For $n = 4$: $a_{4} = 2^{4} = 16$
For $n = 5$: $a_{5} = 2^{5} = 32$
Thus,the first five terms are $2, 4, 8, 16, 32$.

(A) The given $G.P.$ is $5, 25, 125, \ldots$
Here,the first term $a = 5$ and the common ratio $r = \frac{25}{5} = 5$.
The $n^{\text{th}}$ term of a $G.P.$ is given by $a_n = a \cdot r^{n-1}$.
Substituting the values,$a_n = 5 \cdot 5^{n-1} = 5^{1 + n - 1} = 5^n$.
For the $10^{\text{th}}$ term,we set $n = 10$:
$a_{10} = 5^{10}$.

(B) The given $GP$ is $2, 8, 32, \ldots$ where the first term $a = 2$ and the common ratio $r = \frac{8}{2} = 4$.
Let $131072$ be the $n^{\text{th}}$ term of the $GP$.
The formula for the $n^{\text{th}}$ term is $a_n = a \cdot r^{n-1}$.
Substituting the values,we get $131072 = 2 \cdot 4^{n-1}$.
Dividing both sides by $2$,we get $65536 = 4^{n-1}$.
Since $65536 = 4^8$,we have $4^8 = 4^{n-1}$.
Equating the exponents,$n - 1 = 8$,which gives $n = 9$.
Therefore,$131072$ is the $9^{\text{th}}$ term of the $GP$.

(A) Let the first term be $a$ and the common ratio be $r$.
In a $G.P.$,the $n^{th}$ term is given by $a_n = ar^{n-1}$.
Given $a_3 = ar^2 = 24$ $(1)$.
Given $a_6 = ar^5 = 192$ $(2)$.
Dividing $(2)$ by $(1)$,we get $\frac{ar^5}{ar^2} = \frac{192}{24}$.
$r^3 = 8$,which implies $r = 2$.
Substituting $r = 2$ in $(1)$,we get $a(2)^2 = 24$,so $4a = 24$,which gives $a = 6$.
The $10^{th}$ term is $a_{10} = ar^9 = 6 \times (2)^9$.
$a_{10} = 6 \times 512 = 3072$.

(A) Given the geometric series $1 + \frac{2}{3} + \frac{4}{9} + \dots$
Here,the first term $a = 1$ and the common ratio $r = \frac{2/3}{1} = \frac{2}{3}$.
Since $|r| < 1$,the sum of the first $n$ terms is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.
Substituting the values,$S_n = \frac{1(1 - (2/3)^n)}{1 - 2/3} = \frac{1 - (2/3)^n}{1/3} = 3[1 - (\frac{2}{3})^n]$.
For the sum of the first $5$ terms,$n = 5$:
$S_5 = 3[1 - (\frac{2}{3})^5] = 3[1 - \frac{32}{243}] = 3[\frac{243 - 32}{243}] = 3[\frac{211}{243}] = \frac{211}{81}$.

(C) Let $n$ be the number of terms needed. Given that the first term $a = 3$ and the common ratio $r = \frac{1}{2}$.
The sum of the first $n$ terms of a $G.P.$ is given by $S_{n} = \frac{a(1 - r^{n})}{1 - r}$.
Substituting the given values,we have $\frac{3069}{512} = \frac{3(1 - (\frac{1}{2})^{n})}{1 - \frac{1}{2}}$.
$\frac{3069}{512} = \frac{3(1 - \frac{1}{2^{n}})}{\frac{1}{2}} = 6(1 - \frac{1}{2^{n}})$.
Dividing both sides by $6$,we get $\frac{3069}{3072} = 1 - \frac{1}{2^{n}}$.
$\frac{1}{2^{n}} = 1 - \frac{3069}{3072} = \frac{3072 - 3069}{3072} = \frac{3}{3072} = \frac{1}{1024}$.
Since $1024 = 2^{10}$,we have $2^{n} = 2^{10}$,which implies $n = 10$.

Let the first three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given the product is $-1$:
$\left(\frac{a}{r}\right)(a)(ar) = -1$
$a^3 = -1 \implies a = -1$.
Given the sum is $\frac{13}{12}$:
$\frac{a}{r} + a + ar = \frac{13}{12}$
Substituting $a = -1$:
$-\frac{1}{r} - 1 - r = \frac{13}{12}$
$-\frac{1+r+r^2}{r} = \frac{13}{12}$
$-12 - 12r - 12r^2 = 13r$
$12r^2 + 25r + 12 = 0$
Solving the quadratic equation:
$12r^2 + 16r + 9r + 12 = 0$
$4r(3r + 4) + 3(3r + 4) = 0$
$(4r + 3)(3r + 4) = 0$
$r = -\frac{3}{4}$ or $r = -\frac{4}{3}$.
For $r = -\frac{3}{4}$,the terms are $\frac{4}{3}, -1, \frac{3}{4}$.
For $r = -\frac{4}{3}$,the terms are $\frac{3}{4}, -1, \frac{4}{3}$.