Progression and Sequence Questions in English

(D) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given that their product is $27$,we have $\frac{a}{r} \cdot a \cdot ar = 27$,which implies $a^3 = 27$,so $a = 3$.
The sum of these terms is $S = \frac{3}{r} + 3 + 3r = 3(\frac{1}{r} + r + 1)$.
Case $1$: If $r > 0$,by $AM \geq GM$,we have $\frac{r + \frac{1}{r}}{2} \geq \sqrt{r \cdot \frac{1}{r}} = 1$,so $r + \frac{1}{r} \geq 2$.
Thus,$S = 3(r + \frac{1}{r} + 1) \geq 3(2 + 1) = 9$.
Case $2$: If $r < 0$,let $r = -k$ where $k > 0$. Then $r + \frac{1}{r} = -(k + \frac{1}{k}) \leq -2$.
Thus,$S = 3(r + \frac{1}{r} + 1) \leq 3(-2 + 1) = -3$.
Therefore,$S \in (-\infty, -3] \cup [9, \infty)$.

(B) Given the sum of the first $11$ terms of an $A.P.$ is $0$:
$S_{11} = \frac{11}{2} (2a_{1} + 10d) = 0$
$11(a_{1} + 5d) = 0 \Rightarrow a_{1} + 5d = 0 \Rightarrow d = -\frac{a_{1}}{5}$.
We need to find the sum of the series $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$.
This is an $A.P.$ with $12$ terms,first term $A = a_{1}$ and common difference $D = 2d$.
The sum $S'$ is given by:
$S' = \frac{12}{2} [2A + (12-1)D] = 6 [2a_{1} + 11(2d)] = 6 [2a_{1} + 22d]$.
Substituting $d = -\frac{a_{1}}{5}$:
$S' = 6 [2a_{1} + 22(-\frac{a_{1}}{5})] = 6 [2a_{1} - \frac{22a_{1}}{5}] = 6 [\frac{10a_{1} - 22a_{1}}{5}] = 6 [-\frac{12a_{1}}{5}] = -\frac{72}{5} a_{1}$.
Thus,$k = -\frac{72}{5}$.

(C) The given series is $S = (x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + \dots$ up to $9$ terms.
We can split the sum into three parts:
$S = (x + x^2 + x^3 + \dots + x^9) + (ka + ka + \dots + ka \text{ [9 times]}) + (0 + 2a + 4a + \dots + 16a)$.
The first part is a geometric progression: $x \frac{x^9-1}{x-1} = \frac{x^{10}-x}{x-1}$.
The second part is $9ka$.
The third part is an arithmetic progression with $n=9$,first term $0$,and common difference $2a$: $\frac{9}{2} [2(0) + (9-1)2a] = \frac{9}{2} [16a] = 72a$.
Thus,$S = \frac{x^{10}-x}{x-1} + 9ka + 72a = \frac{x^{10}-x + (9k+72)a(x-1)}{x-1}$.
Comparing this with the given expression $S = \frac{x^{10}-x+45a(x-1)}{x-1}$,we get $9k + 72 = 45$.
$9k = 45 - 72 = -27$.
$k = -3$.

(A) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$,where $d > 0$.
The terms are $a, a+d, a+2d, \ldots, a+10d$.
The mean $\bar{X} = \frac{1}{11} \sum_{i=0}^{10} (a + id) = a + \frac{d}{11} \times \frac{10 \times 11}{2} = a + 5d$.
The variance is given by $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{X})^2$.
$\sigma^2 = \frac{1}{11} \sum_{i=0}^{10} (a + id - (a + 5d))^2 = \frac{1}{11} \sum_{i=0}^{10} (i-5)^2 d^2$.
$\sigma^2 = \frac{d^2}{11} [(-5)^2 + (-4)^2 + (-3)^2 + (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2]$.
$\sigma^2 = \frac{d^2}{11} [25 + 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 + 25] = \frac{d^2}{11} [110] = 10d^2$.
Given variance is $90$,so $10d^2 = 90 \Rightarrow d^2 = 9$.
Since the $A.P.$ is increasing,$d > 0$,therefore $d = 3$.

(C) The given series is an Arithmetic Progression $(AP)$ where the first term $a = 20$ and the common difference $d = 19 \frac{3}{5} - 20 = \frac{98}{5} - \frac{100}{5} = -\frac{2}{5}$.
The sum of $n$ terms is given by $S_n = \frac{n}{2} [2a + (n - 1)d] = 488$.
Substituting the values: $\frac{n}{2} [2(20) + (n - 1)(-\frac{2}{5})] = 488$.
$\frac{n}{2} [40 - \frac{2n}{5} + \frac{2}{5}] = 488$.
$n [20 - \frac{n}{5} + \frac{1}{5}] = 488$.
$n [\frac{100 - n + 1}{5}] = 488$.
$n(101 - n) = 2440$.
$n^2 - 101n + 2440 = 0$.
Solving the quadratic equation: $(n - 40)(n - 61) = 0$.
So,$n = 40$ or $n = 61$.
If $n = 61$,the $n^{th}$ term $T_n = a + (n - 1)d = 20 + (60)(-\frac{2}{5}) = 20 - 24 = -4$.
If $n = 40$,the $n^{th}$ term $T_n = 20 + (39)(-\frac{2}{5}) = 20 - 15.6 = 4.4$ (which is not negative).
Thus,for the $n^{th}$ term to be negative,$n = 61$ and the $n^{th}$ term is $-4$.

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

(D) Let the first term be $a = 3$ and the common difference be $d$.
The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
The sum of the first $25$ terms is $S_{25} = \frac{25}{2}[2(3) + 24d] = \frac{25}{2}[6 + 24d] = 25(3 + 12d) = 75 + 300d$.
The sum of the next $15$ terms is $S_{26-40} = S_{40} - S_{25}$.
$S_{40} = \frac{40}{2}[2(3) + 39d] = 20[6 + 39d] = 120 + 780d$.
Given $S_{25} = S_{40} - S_{25}$,which implies $2S_{25} = S_{40}$.
$2(75 + 300d) = 120 + 780d$.
$150 + 600d = 120 + 780d$.
$150 - 120 = 780d - 600d$.
$30 = 180d$.
$d = \frac{30}{180} = \frac{1}{6}$.

(C) Let the given expression be $S = S_1 + S_2$,where $S_1 = (2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots$ up to $51$ terms) and $S_2 = (1! - 2! + 3! - \dots$ up to $51$ terms).
We know that ${}^{n}P_{n-1} = n!$.
Thus,$S_1 = (2 \cdot 1! - 3 \cdot 2! + 4 \cdot 3! - \dots + 52 \cdot 51!)$.
Since $(n+1) \cdot n! = (n+1)!$,we have $S_1 = (2! - 3! + 4! - \dots + 52!)$.
Now,$S_2 = (1! - 2! + 3! - 4! + \dots + 51!)$.
Adding $S_1$ and $S_2$:
$S = (2! - 3! + 4! - \dots + 52!) + (1! - 2! + 3! - 4! + \dots + 51!)$.
All intermediate terms cancel out:
$S = 1! + 52! = 1 + 52!$.

(C) The $n$-th term of an $A.P.$ is given by $a_{n} = a_{1} + (n-1)d$.
Given $a_{1} = 1$ and $a_{n} = 300$,we have $300 = 1 + (n-1)d$,which implies $(n-1)d = 299$.
The prime factorization of $299$ is $13 \times 23$.
Since $15 \leq n \leq 50$,we have $14 \leq n-1 \leq 49$.
The possible values for $(n-1)$ are $23$ or $13$ (not possible as $n-1 \geq 14$).
Thus,$n-1 = 23 \Rightarrow n = 24$ and $d = 13$.
We need to find $(S_{n-4}, a_{n-4}) = (S_{20}, a_{20})$.
$a_{20} = a_{1} + 19d = 1 + 19(13) = 1 + 247 = 248$.
$S_{20} = \frac{20}{2}(a_{1} + a_{20}) = 10(1 + 248) = 10(249) = 2490$.
Therefore,the ordered pair is $(2490, 248)$.

(A) Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for two positive numbers $2^{\sin x}$ and $2^{\cos x}$:
$\frac{2^{\sin x} + 2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$
$\Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x + \cos x}{2}}$
$\Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2^{1 + \frac{\sin x + \cos x}{2}}$
We know that the minimum value of $\sin x + \cos x$ is $-\sqrt{2}$.
Substituting this value:
$\min(2^{\sin x} + 2^{\cos x}) = 2^{1 + \frac{-\sqrt{2}}{2}} = 2^{1 - \frac{\sqrt{2}}{2}} = 2^{1 - \frac{1}{\sqrt{2}}}$

(C) Let the roots be $\alpha, \alpha r, \alpha r^2, \alpha r^3$ in geometric progression.
From the first equation $x^2 - 3x + p = 0$,the sum of roots is $\alpha + \alpha r = 3 \implies \alpha(1+r) = 3$ (Equation $1$).
The product of roots is $p = \alpha^2 r$.
From the second equation $x^2 - 6x + q = 0$,the sum of roots is $\alpha r^2 + \alpha r^3 = 6 \implies \alpha r^2(1+r) = 6$ (Equation $2$).
The product of roots is $q = \alpha^2 r^5$.
Dividing Equation $2$ by Equation $1$: $\frac{\alpha r^2(1+r)}{\alpha(1+r)} = \frac{6}{3} \implies r^2 = 2$.
Now,calculate $p$ and $q$ in terms of $\alpha$ and $r$: $p = \alpha^2 r$ and $q = \alpha^2 r^5 = p r^4$.
Since $r^2 = 2$,then $r^4 = 4$.
Thus,$q = p(4) = 4p$.
The ratio $\frac{2q+p}{2q-p} = \frac{2(4p)+p}{2(4p)-p} = \frac{8p+p}{8p-p} = \frac{9p}{7p} = \frac{9}{7}$.

(B) The given expression is $S = 1 + \sum_{n=1}^{10} (1 - (2n)^2(2n-1))$.
There are $10$ terms in the summation plus the initial $1$,making a total of $11$ terms.
$S = 11 - \sum_{n=1}^{10} (4n^2)(2n-1) = 11 - 4 \sum_{n=1}^{10} (2n^3 - n^2)$.
Using the summation formulas $\sum n^3 = [\frac{n(n+1)}{2}]^2$ and $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=10$:
$\sum_{n=1}^{10} n^3 = (55)^2 = 3025$.
$\sum_{n=1}^{10} n^2 = \frac{10 \times 11 \times 21}{6} = 385$.
Substituting these values:
$S = 11 - 4 [2(3025) - 385] = 11 - 4 [6050 - 385] = 11 - 4(5665) = 11 - 22660$.
We need the form $\alpha - 220 \beta$. Since $22660 = 220 \times 103$,we have $S = 11 - 220(103)$.
Thus,$(\alpha, \beta) = (11, 103)$.

(D) The given series is $\log_{(7^{1/2})} x + \log_{(7^{1/3})} x + \log_{(7^{1/4})} x + \dots + \log_{(7^{1/21})} x = 460$.
Using the property $\log_{(a^n)} x = \frac{1}{n} \log_a x$,we can rewrite the terms as:
$2 \log_7 x + 3 \log_7 x + 4 \log_7 x + \dots + 21 \log_7 x = 460$.
Factoring out $\log_7 x$,we get:
$\log_7 x \cdot (2 + 3 + 4 + \dots + 21) = 460$.
The sum of the arithmetic progression $2 + 3 + \dots + 21$ is given by $\frac{n}{2}(a + l)$,where $n = 20$,$a = 2$,and $l = 21$.
Sum $= \frac{20}{2}(2 + 21) = 10 \times 23 = 230$.
So,$230 \cdot \log_7 x = 460$.
$\log_7 x = \frac{460}{230} = 2$.
Therefore,$x = 7^2 = 49$.

(C) The given expression is a Geometric Progression ($G$.$P$.) with $n = 11$ terms.
Here,the first term $a = 2^{10}$ and the common ratio $r = \frac{3}{2}$.
The sum of a $G$.$P$. is given by $S_n = a \frac{r^n - 1}{r - 1}$.
Substituting the values: $S' = 2^{10} \cdot \frac{(\frac{3}{2})^{11} - 1}{\frac{3}{2} - 1}$.
$S' = 2^{10} \cdot \frac{\frac{3^{11}}{2^{11}} - 1}{\frac{1}{2}} = 2^{11} \cdot (\frac{3^{11} - 2^{11}}{2^{11}})$.
$S' = 3^{11} - 2^{11}$.
Given that $S' = S - 2^{11}$,we have $3^{11} - 2^{11} = S - 2^{11}$.
Therefore,$S = 3^{11}$.

(B) Let the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$ be $d.$ Then,the common difference of $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $d + 2.$
For the first $A.P.:$
$a_{40} = a_{1} + 39d = -159$ $(1)$
$a_{100} = a_{1} + 99d = -399$ $(2)$
Subtracting $(1)$ from $(2):$
$60d = -240 \Rightarrow d = -4.$
Substituting $d = -4$ in $(1):$
$a_{1} + 39(-4) = -159 \Rightarrow a_{1} - 156 = -159 \Rightarrow a_{1} = -3.$
Now,find $a_{70}:$
$a_{70} = a_{1} + 69d = -3 + 69(-4) = -3 - 276 = -279.$
Given $b_{100} = a_{70} = -279.$
For the second $A.P.:$
$b_{100} = b_{1} + 99(d + 2) = -279.$
Since $d = -4,$ then $d + 2 = -2.$
$b_{1} + 99(-2) = -279$
$b_{1} - 198 = -279$
$b_{1} = -279 + 198 = -81.$

(B) Given the functional equation $f(x+y) = f(x)f(y)$.
For $x=1, y=1$,we have $f(2) = f(1)f(1) = 3^2 = 9$.
For $x=2, y=1$,we have $f(3) = f(2)f(1) = 3^2 \cdot 3 = 3^3 = 27$.
By induction,it follows that $f(x) = 3^x$ for all $x \in R$.
We are given the sum $\sum_{i=1}^{n} f(i) = 363$,which implies $\sum_{i=1}^{n} 3^i = 363$.
This is a geometric progression with the first term $a = 3$,common ratio $r = 3$,and $n$ terms.
The sum of a geometric series is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values: $\frac{3(3^n - 1)}{3 - 1} = 363$.
$\frac{3(3^n - 1)}{2} = 363$.
$3(3^n - 1) = 726$.
$3^n - 1 = 242$.
$3^n = 243$.
Since $243 = 3^5$,we have $n = 5$.

(C) The given equation is $(a^{2}+b^{2}+c^{2}) p^{2}-2(ab+bc+cd) p+(b^{2}+c^{2}+d^{2})=0$.
This can be rewritten as $(a^{2}p^{2}-2abp+b^{2})+(b^{2}p^{2}-2bcp+c^{2})+(c^{2}p^{2}-2cdp+d^{2})=0$.
This simplifies to $(ap-b)^{2}+(bp-c)^{2}+(cp-d)^{2}=0$.
Since $a, b, c, d, p$ are real numbers,the sum of squares is zero only if each term is zero:
$ap-b=0, bp-c=0, cp-d=0$.
This implies $p = b/a = c/b = d/c$.
Therefore,the ratio of consecutive terms is constant,which means $a, b, c, d$ are in $G.P.$

(B) Let the series be $a_n = 1, 7, 17, 31, 49, \ldots$
The differences between consecutive terms are:
$7 - 1 = 6$
$17 - 7 = 10$
$31 - 17 = 14$
$49 - 31 = 18$
The sequence of differences is $6, 10, 14, 18, \ldots$,which is an Arithmetic Progression with first term $a = 6$ and common difference $d = 4$.
The $n^{th}$ term of the original series can be expressed as a quadratic form $a_n = An^2 + Bn + C$.
For $n=1: A + B + C = 1$
For $n=2: 4A + 2B + C = 7$
For $n=3: 9A + 3B + C = 17$
Subtracting the first from the second: $3A + B = 6$
Subtracting the second from the third: $5A + B = 10$
Subtracting these results: $2A = 4 \implies A = 2$.
Substituting $A=2$ into $3A + B = 6$,we get $6 + B = 6 \implies B = 0$.
Substituting $A=2, B=0$ into $A + B + C = 1$,we get $2 + 0 + C = 1 \implies C = -1$.
Thus,the $n^{th}$ term is $a_n = 2n^2 - 1$.

(B) The given series is $12, 72, 432, 2592, \dots$
This is a Geometric Progression $(GP)$ where the first term $a = 12$ and the common ratio $r = \frac{72}{12} = 6$.
The formula for the $n^{th}$ term of a $GP$ is $T_n = a \times r^{n-1}$.
Substituting the values,we get $T_n = 12 \times 6^{n-1}$.
We can rewrite $12$ as $2 \times 6$,so $T_n = (2 \times 6) \times 6^{n-1} = 2 \times 6^{1 + n - 1} = 2 \times 6^{n}$.
Thus,the $n^{th}$ term is $2 \times 6^{n}$.

(C) Let the series be $a_n = 5, 14, 29, 50, 77, \ldots$
Calculate the first differences: $14-5=9, 29-14=15, 50-29=21, 77-50=27$.
The differences are $9, 15, 21, 27, \ldots$,which form an arithmetic progression with a common difference of $6$.
Since the second difference is constant $(6)$,the $n^{th}$ term is a quadratic expression of the form $a_n = An^2 + Bn + C$.
For $n=1: A + B + C = 5$
For $n=2: 4A + 2B + C = 14$
For $n=3: 9A + 3B + C = 29$
Subtracting equations: $(4A+2B+C) - (A+B+C) = 14-5 \implies 3A + B = 9$.
$(9A+3B+C) - (4A+2B+C) = 29-14 \implies 5A + B = 15$.
Subtracting these results: $(5A+B) - (3A+B) = 15-9 \implies 2A = 6 \implies A = 3$.
Substituting $A=3$ into $3A+B=9$: $3(3) + B = 9 \implies B = 0$.
Substituting $A=3, B=0$ into $A+B+C=5$: $3 + 0 + C = 5 \implies C = 2$.
Thus,the $n^{th}$ term is $a_n = 3n^2 + 2$.

(C) Let the six arithmetic means be $A_1, A_2, A_3, A_4, A_5, A_6$ between $20$ and $76$.
Then,the sequence $20, A_1, A_2, A_3, A_4, A_5, A_6, 76$ forms an arithmetic progression $(AP)$ with $n = 8$ terms.
The first term $a = 20$ and the last term $a_8 = 76$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values,$76 = 20 + (8 - 1)d$.
$76 - 20 = 7d \Rightarrow 56 = 7d \Rightarrow d = 8$.
The arithmetic means are:
$A_1 = 20 + 8 = 28$
$A_2 = 28 + 8 = 36$
$A_3 = 36 + 8 = 44$
$A_4 = 44 + 8 = 52$
$A_5 = 52 + 8 = 60$
$A_6 = 60 + 8 = 68$
Thus,the arithmetic means are $28, 36, 44, 52, 60, 68$.

(B) Let the number of terms be $n$ and the common difference be $d$.
Given,first term $a = 75$ and last term $l = a_n = 375$.
The sum of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$.
Substituting the values: $3600 = \frac{n}{2}(75 + 375)$.
$3600 = \frac{n}{2}(450)$.
$3600 = 225n$.
$n = \frac{3600}{225} = 16$.
Now,use the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.
$375 = 75 + (16 - 1)d$.
$375 - 75 = 15d$.
$300 = 15d$.
$d = \frac{300}{15} = 20$.
Thus,the number of terms is $16$ and the common difference is $20$.

(C) The sum of an $AP$ is given by the formula $S_n = \frac{n}{2}(a_1 + a_n)$,where $n$ is the number of terms,$a_1$ is the first term,and $a_n$ is the last term.
Given:
Number of terms $n = 40$
First term $a_1 = 80$
Last term $a_{40} = 275$
Substituting these values into the formula:
$S_{40} = \frac{40}{2}(80 + 275)$
$S_{40} = 20(355)$
$S_{40} = 7100$
Thus,the sum of the $40$ terms is $7100$.

(C) The given sequence is a Geometric Progression $(GP)$ where the first term $a = 6$.
The common ratio $r$ is calculated as $r = \frac{12}{6} = 2$.
The formula for the $n^{th}$ term of a $GP$ is $T_n = a \times r^{n-1}$.
To find the $7^{th}$ term $(n = 7)$:
$T_7 = 6 \times 2^{7-1} = 6 \times 2^6$.
Since $2^6 = 64$,we have:
$T_7 = 6 \times 64 = 384$.

(B) The given sequence is a Geometric Progression $(GP)$.
First term $(a)$ = $39366$.
Common ratio $(r)$ = $\frac{13122}{39366} = \frac{1}{3}$.
The $n^{th}$ term of a $GP$ is given by the formula $a_n = a \cdot r^{n-1}$.
To find the seventh term $(a_7)$,we substitute $n = 7$,$a = 39366$,and $r = \frac{1}{3}$:
$a_7 = 39366 \cdot \left(\frac{1}{3}\right)^{7-1}$
$a_7 = 39366 \cdot \left(\frac{1}{3}\right)^6$
$a_7 = 39366 \cdot \frac{1}{729}$
$a_7 = 54$.

(A) The given sequence is a $GP$ where the first term $a_1 = 21a$.
To find the common ratio $r$,we divide the second term by the first term: $r = \frac{84a^3}{21a} = 4a^2$.
The $n^{th}$ term of a $GP$ is given by the formula $a_n = a_1 \cdot r^{n-1}$.
For the seventh term $(n = 7)$:
$a_7 = (21a) \cdot (4a^2)^{7-1}$
$a_7 = 21a \cdot (4a^2)^6$
$a_7 = 21a \cdot 4^6 \cdot a^{12}$
$a_7 = 21 \cdot 4096 \cdot a^{13}$
$a_7 = 86016a^{13}$.

(C) Given: First term $a = 1024$ and common ratio $r = \frac{1}{2}$.
The formula for the sum of the first $n$ terms of a $GP$ is $S_n = \frac{a(1 - r^n)}{1 - r}$ for $r < 1$.
Substituting the values for $n = 7$:
$S_7 = \frac{1024(1 - (\frac{1}{2})^7)}{1 - \frac{1}{2}}$
$S_7 = \frac{1024(1 - \frac{1}{128})}{\frac{1}{2}}$
$S_7 = 1024 \times (\frac{127}{128}) \times 2$
$S_7 = 8 \times 127 \times 2 = 16 \times 127 = 2032$.

(B) Let the three numbers in $GP$ be $\frac{a}{r}, a, ar$.
Given that their product is $8000$,so $\frac{a}{r} \times a \times ar = 8000$.
This simplifies to $a^3 = 8000$,which gives $a = 20$.
Given that their sum is $105$,so $\frac{a}{r} + a + ar = 105$.
Substituting $a = 20$,we get $\frac{20}{r} + 20 + 20r = 105$.
Dividing by $5$,we get $\frac{4}{r} + 4 + 4r = 21$,which simplifies to $4r^2 - 17r + 4 = 0$.
Solving the quadratic equation $4r^2 - 16r - r + 4 = 0$,we get $4r(r - 4) - 1(r - 4) = 0$.
Thus,$(4r - 1)(r - 4) = 0$,giving $r = 4$ or $r = \frac{1}{4}$.
If $r = 4$,the numbers are $\frac{20}{4}, 20, 20 \times 4$,which are $5, 20, 80$.
If $r = \frac{1}{4}$,the numbers are $\frac{20}{1/4}, 20, 20 \times \frac{1}{4}$,which are $80, 20, 5$.
In both cases,the set of numbers is ${5, 20, 80}$.

(C) Given $AM = 17$ and $GM = 8$.
Let the two numbers be $a$ and $b$.
We know that $AM = \frac{a+b}{2} = 17 \implies a+b = 34$ $(1)$.
We know that $GM = \sqrt{ab} = 8 \implies ab = 64$ $(2)$.
Using the relation $(a-b)^2 = (a+b)^2 - 4ab$,we get:
$(a-b)^2 = (34)^2 - 4(64) = 1156 - 256 = 900$.
Thus,$a-b = 30$ $(3)$.
Solving $(1)$ and $(3)$,we get $2a = 64 \implies a = 32$ and $b = 2$.
The Harmonic Mean $(HM)$ is given by $HM = \frac{2ab}{a+b} = \frac{2(64)}{34} = \frac{128}{34} = \frac{64}{17}$.
The $AP$ is $2, 17, 32$.
The $GP$ is $2, 8, 32$.
The $HP$ is $2, \frac{64}{17}, 32$.

(C) The given sequence is an arithmetic progression $(AP)$ where the first term $a = \frac{n-1}{n}$.
The common difference $d$ is calculated as $d = \frac{n+1}{n} - \frac{n-1}{n} = \frac{2}{n}$.
Wait,let us re-examine the sequence: $\frac{n-1}{n}, \frac{n+1}{n}, \frac{n+2}{n}, \dots$. This sequence is not a standard $AP$ as written. Assuming the sequence is $\frac{n-1}{n}, \frac{n}{n}, \frac{n+1}{n}, \dots$ (where $a = \frac{n-1}{n}$ and $d = \frac{1}{n}$):
The sum of $n$ terms $S_n = \frac{n}{2} [2a + (n-1)d]$.
Substituting the values: $S_n = \frac{n}{2} [2(\frac{n-1}{n}) + (n-1)(\frac{1}{n})]$.
$S_n = \frac{n}{2} [\frac{2n-2+n-1}{n}] = \frac{n}{2} [\frac{3n-3}{n}] = \frac{3(n-1)}{2}$.

(C) For the first arithmetic progression $(AP_1)$: $3, 7, 11, 15, \ldots$
Here,first term $a_1 = 3$ and common difference $d_1 = 7 - 3 = 4$.
The sum of $n$ terms is $S_{n_1} = \frac{n}{2} \{2a_1 + (n - 1)d_1\} = \frac{n}{2} \{2(3) + (n - 1)4\} = \frac{n}{2} \{6 + 4n - 4\} = \frac{n}{2} \{4n + 2\} = n(2n + 1)$.
For the second arithmetic progression $(AP_2)$: $30, 33, 36, 39, \ldots$
Here,first term $a_2 = 30$ and common difference $d_2 = 33 - 30 = 3$.
The sum of $n$ terms is $S_{n_2} = \frac{n}{2} \{2a_2 + (n - 1)d_2\} = \frac{n}{2} \{2(30) + (n - 1)3\} = \frac{n}{2} \{60 + 3n - 3\} = \frac{n}{2} \{57 + 3n\} = \frac{3n(19 + n)}{2}$.
Given $S_{n_1} = S_{n_2}$:
$n(2n + 1) = \frac{3n(19 + n)}{2}$
Since $n \neq 0$,we divide by $n$:
$2n + 1 = \frac{3(19 + n)}{2}$
$2(2n + 1) = 3(19 + n)$
$4n + 2 = 57 + 3n$
$4n - 3n = 57 - 2$
$n = 55$.

(B) The sum of the first $n$ terms of a $GP$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Given,$S_{20} = 1025 \times S_{10}$.
Substituting the formula: $\frac{a(r^{20} - 1)}{r - 1} = 1025 \times \frac{a(r^{10} - 1)}{r - 1}$.
Assuming $r \neq 1$ and $a \neq 0$,we simplify: $r^{20} - 1 = 1025(r^{10} - 1)$.
Since $r^{20} - 1 = (r^{10})^2 - 1^2 = (r^{10} - 1)(r^{10} + 1)$,we have:
$(r^{10} - 1)(r^{10} + 1) = 1025(r^{10} - 1)$.
If $r^{10} - 1 \neq 0$,then $r^{10} + 1 = 1025$.
$r^{10} = 1024$.
$r^{10} = 2^{10}$.
Therefore,$r = \pm 2$.

(C) The given series is $1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \cdots$.
This is an infinite geometric progression $(GP)$ where the first term $a = 1$ and the common ratio $r = \frac{1}{2}$.
The sum of an infinite $GP$ is given by the formula $S_{\infty} = \frac{a}{1-r}$,provided $|r| < 1$.
Here,$|r| = |\frac{1}{2}| < 1$,so the sum exists.
Substituting the values,we get $S_{\infty} = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$.

(D) Let the sum be $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \frac{5}{3^4} + \cdots$
This is an arithmetico-geometric series of the form $S = \sum_{n=1}^{\infty} n x^{n-1}$ where $x = \frac{1}{3}$.
The sum of this series is given by the formula $S = \frac{1}{(1-x)^2}$ for $|x| < 1$.
Substituting $x = \frac{1}{3}$ into the formula:
$S = \frac{1}{(1 - \frac{1}{3})^2}$
$S = \frac{1}{(\frac{2}{3})^2}$
$S = \frac{1}{\frac{4}{9}}$
$S = \frac{9}{4}$

(B) The geometric mean $(GM)$ between two numbers $a$ and $b$ is calculated using the formula $GM = \sqrt{a \times b}$.
Given $a = \frac{3}{2}$ and $b = \frac{27}{2}$.
Substituting these values into the formula:
$GM = \sqrt{\frac{3}{2} \times \frac{27}{2}}$
$GM = \sqrt{\frac{81}{4}}$
$GM = \frac{9}{2}$.

(C) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Given that the $9^{th}$ term $a_9 = -6$ and the common difference $d = \frac{5}{4}$.
Substituting these values into the formula:
$a_9 = a + (9 - 1)d$
$-6 = a + 8 \times \left(\frac{5}{4}\right)$
$-6 = a + 2 \times 5$
$-6 = a + 10$
$a = -6 - 10 = -16$.
Now,we need to find the $25^{th}$ term $(a_{25})$:
$a_{25} = a + (25 - 1)d$
$a_{25} = -16 + 24 \times \left(\frac{5}{4}\right)$
$a_{25} = -16 + 6 \times 5$
$a_{25} = -16 + 30 = 14$.
Thus,the $25^{th}$ term is $14$.

(A) The given arithmetic progression $(AP)$ is $5, 13, 21, \ldots$
Here,the first term $a = 5$.
The common difference $d = 13 - 5 = 8$.
Let the $n^{th}$ term of the $AP$ be $181$,i.e.,$a_n = 181$.
The formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n - 1)d$.
Substituting the values: $181 = 5 + (n - 1)8$.
Subtracting $5$ from both sides: $176 = (n - 1)8$.
Dividing by $8$: $n - 1 = 176 / 8 = 22$.
Adding $1$ to both sides: $n = 23$.
Therefore,$181$ is the $23^{rd}$ term of the $AP$.

(B) The given series is an Arithmetic Progression $(AP)$ where the first term $a = \frac{1}{n}$.
The common difference $d$ is calculated as:
$d = \frac{n+1}{n} - \frac{1}{n} = \frac{n+1-1}{n} = \frac{n}{n} = 1$.
The formula for the $n^{th}$ term of an $AP$ is $a_n = a + (n-1)d$.
Substituting the values:
$a_n = \frac{1}{n} + (n-1)(1)$
$a_n = \frac{1}{n} + n - 1$
$a_n = \frac{1 + n(n-1)}{n}$
$a_n = \frac{1 + n^2 - n}{n}$.

(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
Given $a_p = q$,we have $a + (p-1)d = q$ $....(1)$
Given $a_q = p$,we have $a + (q-1)d = p$ $....(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(a + (p-1)d) - (a + (q-1)d) = q - p$
$(p - 1 - q + 1)d = -(p - q)$
$(p - q)d = -(p - q)$
$d = -1$
Substituting $d = -1$ in equation $(1)$:
$a + (p-1)(-1) = q$
$a - p + 1 = q$
$a = p + q - 1$
Now,the $r^{th}$ term is $a_r = a + (r-1)d$.
$a_r = (p + q - 1) + (r - 1)(-1)$
$a_r = p + q - 1 - r + 1$
$a_r = p + q - r$

(A) Since $\frac{2}{3}, k, \frac{5}{8} k$ are in $A.P.$,the common difference between consecutive terms must be equal.
Therefore,$k - \frac{2}{3} = \frac{5}{8} k - k$.
Rearranging the terms to isolate $k$ on one side:
$k - \frac{5}{8} k + k = \frac{2}{3}$
$2k - \frac{5}{8} k = \frac{2}{3}$
Finding a common denominator for the left side:
$\frac{16k - 5k}{8} = \frac{2}{3}$
$\frac{11k}{8} = \frac{2}{3}$
Solving for $k$:
$k = \frac{2}{3} \times \frac{8}{11} = \frac{16}{33}$.

(D) Since $k+2, 4k-6$ and $3k-2$ are in $A.P.$,the common difference between consecutive terms must be equal.
Therefore,$(4k-6) - (k+2) = (3k-2) - (4k-6)$.
Simplifying the left side: $4k - 6 - k - 2 = 3k - 8$.
Simplifying the right side: $3k - 2 - 4k + 6 = -k + 4$.
Equating both sides: $3k - 8 = -k + 4$.
Adding $k$ to both sides: $4k - 8 = 4$.
Adding $8$ to both sides: $4k = 12$.
Dividing by $4$: $k = 3$.

(D) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
Given that the ratio of the $7^{th}$ term to the $3^{rd}$ term is $12:5$:
$\frac{a + 6d}{a + 2d} = \frac{12}{5}$
Cross-multiplying gives:
$5(a + 6d) = 12(a + 2d)$
$5a + 30d = 12a + 24d$
$30d - 24d = 12a - 5a$
$6d = 7a$
$a = \frac{6}{7}d$
Now,we need to find the ratio of the $13^{th}$ term to the $4^{th}$ term:
$\frac{a_{13}}{a_4} = \frac{a + 12d}{a + 3d}$
Substituting $a = \frac{6}{7}d$ into the expression:
$\frac{\frac{6}{7}d + 12d}{\frac{6}{7}d + 3d} = \frac{\frac{6d + 84d}{7}}{\frac{6d + 21d}{7}} = \frac{90d}{27d} = \frac{90}{27}$
Simplifying the fraction by dividing by $9$:
$\frac{90}{27} = \frac{10}{3}$
Thus,the ratio is $10:3$.

(C) Let $a$ be the first term and $d$ be the common difference of an $A.P.$
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
Therefore,$a_7 = a + 6d$ and $a_{11} = a + 10d$.
According to the problem,$7a_7 = 11a_{11}$.
Substituting the values,we get $7(a + 6d) = 11(a + 10d)$.
Expanding both sides: $7a + 42d = 11a + 110d$.
Rearranging the terms: $7a - 11a = 110d - 42d$.
$-4a = 68d$,which simplifies to $a = -17d$ (Equation $1$).
We need to find the $18^{th}$ term,$a_{18} = a + 17d$.
Substituting $a = -17d$ from Equation $1$ into the expression for $a_{18}$:
$a_{18} = -17d + 17d = 0$.
Thus,the $18^{th}$ term of the $A.P.$ is $0$.

(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
The $4$th term is $a_4 = a + 3d$. According to the problem,$a_4 = 3a$.
So,$a + 3d = 3a \Rightarrow 3d = 2a \Rightarrow a = \frac{3}{2}d$ $...(1)$
The $7$th term is $a_7 = a + 6d$ and the $3$rd term is $a_3 = a + 2d$.
According to the problem,$a_7 = 2a_3 + 1$.
So,$a + 6d = 2(a + 2d) + 1$
$a + 6d = 2a + 4d + 1$
$2d - a = 1$ $...(2)$
Substitute $a = \frac{3}{2}d$ from equation $(1)$ into equation $(2)$:
$2d - \frac{3}{2}d = 1$
$\frac{4d - 3d}{2} = 1$
$\frac{d}{2} = 1 \Rightarrow d = 2$
Now,substitute $d = 2$ into equation $(1)$:
$a = \frac{3}{2} \times 2 = 3$
Thus,the first term is $3$ and the common difference is $2$.

(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d.$
Given that the $9^{th}$ term is $99$:
$a + 8d = 99$ $...(1)$
Given that the $99^{th}$ term is $9$:
$a + 98d = 9$ $...(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(a + 98d) - (a + 8d) = 9 - 99$
$90d = -90$
$d = -1$
Substituting $d = -1$ into equation $(1)$:
$a + 8(-1) = 99$
$a - 8 = 99$
$a = 107$
Now,find the $108^{th}$ term $(a_{108})$:
$a_{108} = a + (108 - 1)d$
$a_{108} = 107 + 107(-1)$
$a_{108} = 107 - 107 = 0$
Thus,the $108^{th}$ term is $0$.

(C) Let $A$ be the first term and $D$ be the common difference of the $A.P.$
The $p^{th}$ term is $a = A + (p-1)D$ $...(1)$
The $q^{th}$ term is $b = A + (q-1)D$ $...(2)$
The $r^{th}$ term is $c = A + (r-1)D$ $...(3)$
Now,substitute these values into the expression $a(q-r) + b(r-p) + c(p-q)$:
$= [A + (p-1)D](q-r) + [A + (q-1)D](r-p) + [A + (r-1)D](p-q)$
$= A(q-r + r-p + p-q) + D[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]$
$= A(0) + D[pq - pr - q + r + qr - qp - r + p + rp - rq - p + q]$
$= 0 + D[0] = 0$
Thus,the value of the expression is $0$.

(D) The distances covered by the body in the first,second,third,fourth,... seconds form an Arithmetic Progression $(A.P.)$: $16, 48, 80, 112, \dots$
Here,the first term $a = 16$ and the common difference $d = 48 - 16 = 32$.
The distance covered in the $n^{th}$ second is given by the $n^{th}$ term of the $A.P.$: $T_n = a + (n - 1)d$.
For the $11^{th}$ second,$n = 11$.
$T_{11} = 16 + (11 - 1) \times 32$
$T_{11} = 16 + 10 \times 32$
$T_{11} = 16 + 320 = 336\, m$.

(A) The distances covered by the ball in successive seconds are $36\, m, 32\, m, 28\, m, \dots$
This sequence forms an Arithmetic Progression $(A.P.)$ where the first term $a = 36$ and the common difference $d = 32 - 36 = -4$.
The distance covered during the $n^{th}$ second is given by the formula for the $n^{th}$ term of an $A.P.$:
$T_n = a + (n - 1)d$
For the $8^{th}$ second,$n = 8$:
$T_8 = 36 + (8 - 1)(-4)$
$T_8 = 36 + 7(-4)$
$T_8 = 36 - 28$
$T_8 = 8\, m$
Therefore,the ball will travel $8\, m$ during the $8^{th}$ second.

(B) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
We know that the $n^{th}$ term is given by $t_{n} = a + (n - 1)d$.
Given $t_{2} = 2$,we have $a + d = 2$ $...(1)$.
Given $t_{7} = 22$,we have $a + 6d = 22$ $...(2)$.
Subtracting equation $(1)$ from equation $(2)$:
$(a + 6d) - (a + d) = 22 - 2$
$5d = 20$,which gives $d = 4$.
Substituting $d = 4$ in equation $(1)$:
$a + 4 = 2$,so $a = -2$.
The sum of the first $n$ terms is $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 35$:
$S_{35} = \frac{35}{2}[2(-2) + (35 - 1)4]$
$S_{35} = \frac{35}{2}[-4 + 34 \times 4]$
$S_{35} = \frac{35}{2}[-4 + 136]$
$S_{35} = \frac{35}{2} \times 132$
$S_{35} = 35 \times 66 = 2310$.

(C) Let $a$ be the first term and $d$ be the common difference of an $A.P.$
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
Given $a_5 = 30$,we have $a + 4d = 30$ --- $(1)$
Given $a_{12} = 65$,we have $a + 11d = 65$ --- $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(a + 11d) - (a + 4d) = 65 - 30$
$7d = 35$
$d = 5$
Substituting $d = 5$ in equation $(1)$:
$a + 4(5) = 30$
$a + 20 = 30$
$a = 10$
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
For $n = 20$:
$S_{20} = \frac{20}{2}[2(10) + (20-1)5]$
$S_{20} = 10[20 + 19 \times 5]$
$S_{20} = 10[20 + 95]$
$S_{20} = 10 \times 115 = 1150$.