Progression and Sequence Questions in English

(A) The given series is an arithmetic progression with the first term $a = 20$ and common difference $d = 19\frac{1}{3} - 20 = -\frac{2}{3}$.
The $n^{th}$ term of the series is given by $a_n = a + (n - 1)d = 20 + (n - 1)\left( -\frac{2}{3} \right)$.
For the sum to be maximum,we consider terms until they become non-negative,i.e.,$a_n \ge 0$.
$20 - \frac{2}{3}(n - 1) \ge 0$
$20 \ge \frac{2}{3}(n - 1)$
$30 \ge n - 1$
$n \le 31$.
Thus,the sum of the first $31$ terms is the maximum sum.
The sum $S_n = \frac{n}{2}[2a + (n - 1)d]$.
$S_{31} = \frac{31}{2}[2(20) + (31 - 1)(-\frac{2}{3})]$
$S_{31} = \frac{31}{2}[40 + 30(-\frac{2}{3})]$
$S_{31} = \frac{31}{2}[40 - 20] = \frac{31}{2} \times 20 = 310$.

(A) The numbers between $100$ and $1000$ that are divisible by $9$ form an Arithmetic Progression $(A.P.)$.
The first number greater than $100$ divisible by $9$ is $108$ $(a = 108)$.
The last number less than $1000$ divisible by $9$ is $999$ $(l = 999)$.
The common difference $d = 9$.
Using the formula for the $n^{th}$ term of an $A.P.$,$l = a + (n - 1)d$:
$999 = 108 + (n - 1)9$
$891 = (n - 1)9$
$n - 1 = 99$
$n = 100$.
The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$:
$S_{100} = \frac{100}{2}(108 + 999)$
$S_{100} = 50 \times 1107$
$S_{100} = 55350$.

(C) Given that the ratio of the sum of $m$ and $n$ terms is $\frac{S_m}{S_n} = \frac{m^2}{n^2}$.
We know that $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Therefore,$\frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2}$.
$\Rightarrow \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}$.
Cross-multiplying,we get $n[2a + (m - 1)d] = m[2a + (n - 1)d]$.
$2an + n(m - 1)d = 2am + m(n - 1)d$.
$2an + mnd - nd = 2am + mnd - md$.
$2an - 2am = nd - md$.
$2a(n - m) = d(n - m)$.
Thus,$d = 2a$.
The $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
The ratio of $m^{th}$ and $n^{th}$ term is $\frac{T_m}{T_n} = \frac{a + (m - 1)d}{a + (n - 1)d}$.
Substituting $d = 2a$,we get $\frac{a + (m - 1)2a}{a + (n - 1)2a} = \frac{a(1 + 2m - 2)}{a(1 + 2n - 2)} = \frac{2m - 1}{2n - 1}$.

(C) The given series is $S = \sum_{r=1}^n \log \left( \frac{a^r}{b^{r-1}} \right) = \log a + \log \left( \frac{a^2}{b} \right) + \log \left( \frac{a^3}{b^2} \right) + \dots + \log \left( \frac{a^n}{b^{n-1}} \right)$.
This is an arithmetic progression $(A.P.)$ where the first term $a_1 = \log a$ and the $n$-th term $a_n = \log \left( \frac{a^n}{b^{n-1}} \right)$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} (a_1 + a_n)$.
Substituting the values,we get $S_n = \frac{n}{2} \left[ \log a + \log \left( \frac{a^n}{b^{n-1}} \right) \right]$.
Using the property $\log x + \log y = \log(xy)$,we have $S_n = \frac{n}{2} \log \left( a \cdot \frac{a^n}{b^{n-1}} \right) = \frac{n}{2} \log \left( \frac{a^{n+1}}{b^{n-1}} \right)$.

(A) The given equation is $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$.
This is an arithmetic progression $(A.P.)$ where the first term $a = (x + 1)$ and the common difference $d = 3$.
Let $n$ be the number of terms. The last term $l = (x + 28)$.
The formula for the $n^{th}$ term is $l = a + (n - 1)d$.
Substituting the values: $(x + 28) = (x + 1) + (n - 1)3$.
$27 = (n - 1)3 \Rightarrow n - 1 = 9 \Rightarrow n = 10$.
The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$.
Substituting the values: $155 = \frac{10}{2}[(x + 1) + (x + 28)]$.
$155 = 5(2x + 29)$.
$31 = 2x + 29$.
$2x = 2 \Rightarrow x = 1$.

(C) The two-digit numbers that leave a remainder of $1$ when divided by $4$ are of the form $4n + 1$.
The smallest two-digit number of this form is $13$ (since $4 \times 3 + 1 = 13$) and the largest is $97$ (since $4 \times 24 + 1 = 97$).
These numbers form an Arithmetic Progression $(AP)$: $13, 17, 21, \dots, 97$.
Here,the first term $a = 13$,the last term $l = 97$,and the common difference $d = 4$.
Using the formula for the $n^{th}$ term of an $AP$: $l = a + (n - 1)d$.
$97 = 13 + (n - 1)4$
$84 = (n - 1)4$
$n - 1 = 21$
$n = 22$.
The sum of these $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.
$S_{22} = \frac{22}{2}(13 + 97) = 11(110) = 1210$.

(C) The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}\{2a + (n - 1)d\}$.
We need to find the value of $(S_{2n} - S_n)$.
$S_{2n} - S_n = \frac{2n}{2}\{2a + (2n - 1)d\} - \frac{n}{2}\{2a + (n - 1)d\}$
$= n\{2a + 2nd - d\} - \frac{n}{2}\{2a + nd - d\}$
$= \frac{n}{2}\{4a + 4nd - 2d - 2a - nd + d\}$
$= \frac{n}{2}\{2a + 3nd - d\} = \frac{n}{2}\{2a + (3n - 1)d\}$
Now,consider $S_{3n} = \frac{3n}{2}\{2a + (3n - 1)d\}$.
Therefore,$\frac{1}{3}S_{3n} = \frac{1}{3} \cdot \frac{3n}{2}\{2a + (3n - 1)d\} = \frac{n}{2}\{2a + (3n - 1)d\}$.
Thus,$(S_{2n} - S_n) = \frac{1}{3}S_{3n}$.

(D) Given equation: $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$
Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we can write $\log_{3^{1/n}} x = n \log_3 x$.
Substituting this into the equation:
$2 \log_3 x + 4 \log_3 x + 6 \log_3 x + \dots + 16 \log_3 x = 36$
Factor out $\log_3 x$:
$(\log_3 x) (2 + 4 + 6 + \dots + 16) = 36$
The sum of the arithmetic progression $2 + 4 + 6 + \dots + 16$ is given by $\frac{n}{2}(a + l)$,where $n=8$,$a=2$,and $l=16$:
Sum $= \frac{8}{2}(2 + 16) = 4 \times 18 = 72$.
So,$(\log_3 x) \times 72 = 36$
$\log_3 x = \frac{36}{72} = \frac{1}{2}$
$x = 3^{1/2} = \sqrt{3}$.

(A) The sum of the first $k$ terms of an arithmetic progression is given by ${S_k} = \frac{k}{2} \{2a + (k - 1)d\}$.
We are given the ratio $\frac{S_{kn}}{S_n} = \frac{\frac{kn}{2} \{2a + (kn - 1)d\}}{\frac{n}{2} \{2a + (n - 1)d\}}$.
Simplifying the expression,we get:
$\frac{S_{kn}}{S_n} = k \left\{ \frac{2a + (kn - 1)d}{2a + (n - 1)d} \right\} = k \left\{ \frac{(2a - d) + knd}{(2a - d) + nd} \right\}$.
For this expression to be independent of $n$,the term containing $n$ in the numerator must be proportional to the term containing $n$ in the denominator,or the constant terms must vanish such that $n$ cancels out.
If $2a - d = 0$,then the expression becomes:
$\frac{S_{kn}}{S_n} = k \left\{ \frac{knd}{nd} \right\} = k^2$.
Since $k^2$ is independent of $n$,the condition is $2a - d = 0$.

(A) The $n^{th}$ term of the series is given by $T_n = \frac{n}{x} + y$.
To find the sum of $r$ terms,we use the summation formula $S_r = \sum_{n=1}^{r} T_n$.
$S_r = \sum_{n=1}^{r} \left( \frac{n}{x} + y \right) = \frac{1}{x} \sum_{n=1}^{r} n + \sum_{n=1}^{r} y$.
Using the formula for the sum of the first $r$ natural numbers,$\sum_{n=1}^{r} n = \frac{r(r + 1)}{2}$.
Thus,$S_r = \frac{1}{x} \left( \frac{r(r + 1)}{2} \right) + ry$.
$S_r = \left\{ \frac{r(r + 1)}{2x} \right\} + ry$.

(D) Let $S$ be the sum of all integers from $1$ to $100$.
$S = \frac{100}{2}(1 + 100) = 50 \times 101 = 5050$.
Let $S_1$ be the sum of integers divisible by $3$ up to $100$: $3, 6, 9, \dots, 99$.
$S_1 = 3(1 + 2 + 3 + \dots + 33) = 3 \times \frac{33 \times 34}{2} = 3 \times 33 \times 17 = 1683$.
Let $S_2$ be the sum of integers divisible by $5$ up to $100$: $5, 10, 15, \dots, 100$.
$S_2 = 5(1 + 2 + 3 + \dots + 20) = 5 \times \frac{20 \times 21}{2} = 5 \times 10 \times 21 = 1050$.
Let $S_3$ be the sum of integers divisible by both $3$ and $5$ (i.e.,divisible by $15$) up to $100$: $15, 30, 45, 60, 75, 90$.
$S_3 = 15(1 + 2 + 3 + 4 + 5 + 6) = 15 \times \frac{6 \times 7}{2} = 15 \times 21 = 315$.
Using the Principle of Inclusion-Exclusion,the sum of integers divisible by $3$ or $5$ is $S_1 + S_2 - S_3 = 1683 + 1050 - 315 = 2418$.
The sum of integers not divisible by $3$ or $5$ is $S - (S_1 + S_2 - S_3) = 5050 - 2418 = 2632$.

(C) Let the first three terms of the arithmetic progression be $(a - d)$,$a$,and $(a + d)$.
According to the problem,the sum of the first and third term is $12$:
$(a - d) + (a + d) = 12$
$2a = 12$
$a = 6$
Now,the product of the first and second term is $24$:
$(a - d) \times a = 24$
Substitute $a = 6$ into the equation:
$(6 - d) \times 6 = 24$
$6 - d = 4$
$d = 2$
The first term is $(a - d) = 6 - 2 = 4$.

(C) For the first arithmetic progression $(2, 5, 8, \dots)$: First term $a_1 = 2$, common difference $d_1 = 3$. The sum of the first $2n$ terms is $S_{2n} = \frac{2n}{2} [2(2) + (2n - 1)3] = n [4 + 6n - 3] = n(6n + 1)$.
For the second arithmetic progression $(57, 59, 61, \dots)$: First term $a_2 = 57$, common difference $d_2 = 2$. The sum of the first $n$ terms is $S_n = \frac{n}{2} [2(57) + (n - 1)2] = \frac{n}{2} [114 + 2n - 2] = \frac{n}{2} [112 + 2n] = n(56 + n)$.
Given that $S_{2n} = S_n$, we have $n(6n + 1) = n(56 + n)$.
Since $n \neq 0$, we divide by $n$: $6n + 1 = 56 + n$.
$5n = 55$, which gives $n = 11$.

(D) The numbers divisible by $3$ between $250$ and $1000$ form an arithmetic progression: $252, 255, \dots, 999$.
Here,the first term $a = 252$,the last term $l = 999$,and the common difference $d = 3$.
Using the formula for the $n^{th}$ term: $l = a + (n - 1)d$.
$999 = 252 + (n - 1)3$.
$747 = (n - 1)3$.
$n - 1 = 249$.
$n = 250$.
The sum $S_n$ of an arithmetic progression is given by $S_n = \frac{n}{2}(a + l)$.
$S_{250} = \frac{250}{2}(252 + 999)$.
$S_{250} = 125 \times 1251$.
$S_{250} = 156375$.

(B) The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
Given the $7^{th}$ term,$a_7 = a + 6d = 40$.
The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 13$,$S_{13} = \frac{13}{2}[2a + (13 - 1)d] = \frac{13}{2}[2a + 12d]$.
Factoring out $2$,we get $S_{13} = \frac{13}{2} \times 2(a + 6d) = 13(a + 6d)$.
Substituting the value $a + 6d = 40$,we get $S_{13} = 13 \times 40 = 520$.

(D) Given that ${a_1}, {a_2}, \dots, {a_{n+1}}$ are in $A.P.$ with common difference $d = {a_{k+1}} - {a_k}$.
Let $S = \sum_{k=1}^{n} \frac{1}{{{a_k}{a_{k+1}}}}$.
Since ${a_{k+1}} - {a_k} = d$,we can write $\frac{1}{{{a_k}{a_{k+1}}}} = \frac{1}{d} \left( \frac{{a_{k+1}} - {a_k}}{{{a_k}{a_{k+1}}}} \right) = \frac{1}{d} \left( \frac{1}{{{a_k}}} - \frac{1}{{{a_{k+1}}}} \right)$.
Substituting this into the sum:
$S = \frac{1}{d} \left[ \left( \frac{1}{{{a_1}}} - \frac{1}{{{a_2}}} \right) + \left( \frac{1}{{{a_2}}} - \frac{1}{{{a_3}}} \right) + \dots + \left( \frac{1}{{{a_n}}} - \frac{1}{{{a_{n+1}}}} \right) \right]$.
This is a telescoping series,so all intermediate terms cancel out:
$S = \frac{1}{d} \left( \frac{1}{{{a_1}}} - \frac{1}{{{a_{n+1}}}} \right) = \frac{1}{d} \left( \frac{{{a_{n+1}} - {a_1}}}{{{a_1}{a_{n+1}}}} \right)$.
Since ${a_{n+1}} = {a_1} + nd$,we have ${a_{n+1}} - {a_1} = nd$.
Therefore,$S = \frac{1}{d} \left( \frac{nd}{{{a_1}{a_{n+1}}}} \right) = \frac{n}{{{a_1}{a_{n+1}}}}$.
Thus,the correct option is $D$.

(B) The sum of the first $n$ terms is given by $S_n = 5n^2 + 2n$.
To find the second term $(T_2)$,we use the relation $T_n = S_n - S_{n-1}$.
For $n = 2$,$T_2 = S_2 - S_1$.
First,calculate $S_2$: $S_2 = 5(2)^2 + 2(2) = 5(4) + 4 = 20 + 4 = 24$.
Next,calculate $S_1$: $S_1 = 5(1)^2 + 2(1) = 5 + 2 = 7$.
Therefore,$T_2 = 24 - 7 = 17$.

(A) Given that $a_1, a_2, a_3, \dots, a_{2n}$ form an $A.P.$ with common difference $d$.
Therefore,$a_2 - a_1 = a_4 - a_3 = \dots = a_{2n} - a_{2n - 1} = d$.
We need to evaluate the sum $S = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2n - 1}^2 - a_{2n}^2$.
Using the difference of squares formula $x^2 - y^2 = (x - y)(x + y)$,we get:
$S = (a_1 - a_2)(a_1 + a_2) + (a_3 - a_4)(a_3 + a_4) + \dots + (a_{2n - 1} - a_{2n})(a_{2n - 1} + a_{2n})$.
Since $a_k - a_{k+1} = -d$ for odd $k$,we have:
$S = -d(a_1 + a_2 + a_3 + a_4 + \dots + a_{2n - 1} + a_{2n})$.
The sum of an $A.P.$ with $2n$ terms is $\frac{2n}{2}(a_1 + a_{2n}) = n(a_1 + a_{2n})$.
Thus,$S = -d \cdot n(a_1 + a_{2n})$.
From the formula for the $n$-th term,$a_{2n} = a_1 + (2n - 1)d$,so $d = \frac{a_{2n} - a_1}{2n - 1}$.
Substituting $d$ into the expression for $S$:
$S = -\left( \frac{a_{2n} - a_1}{2n - 1} \right) \cdot n(a_1 + a_{2n}) = \frac{n(a_1 - a_{2n})(a_1 + a_{2n})}{2n - 1} = \frac{n}{2n - 1}(a_1^2 - a_{2n}^2)$.

(B) The sum of $n$ terms of an $A.P.$ is given by $S_n = 3n^2 + 5n$.
We know that the $m^{th}$ term $T_m$ is given by $T_m = S_m - S_{m-1}$.
Given $T_m = 164$,we have:
$164 = (3m^2 + 5m) - [3(m-1)^2 + 5(m-1)]$
$164 = (3m^2 + 5m) - [3(m^2 - 2m + 1) + 5m - 5]$
$164 = 3m^2 + 5m - [3m^2 - 6m + 3 + 5m - 5]$
$164 = 3m^2 + 5m - [3m^2 - m - 2]$
$164 = 3m^2 + 5m - 3m^2 + m + 2$
$164 = 6m + 2$
$6m = 162$
$m = 27$.

(D) The sum of the first $n$ terms of an $A.P.$ is given by the formula: ${S_n} = \frac{n}{2} \{ 2a + (n - 1)d \}$,where $a$ is the first term and $d$ is the common difference.
Given the expression: ${S_n} = nP + \frac{1}{2}n(n - 1)Q$.
We can rewrite this as: ${S_n} = \frac{n}{2} \{ 2P + (n - 1)Q \}$.
Comparing this with the standard formula,we identify that the common difference $d = Q$.
Alternatively,we can find the common difference using the terms:
${S_1} = P + 0 = P$
${S_2} = 2P + \frac{1}{2}(2)(1)Q = 2P + Q$
${T_1} = {S_1} = P$
${T_2} = {S_2} - {S_1} = (2P + Q) - P = P + Q$
Common difference $d = {T_2} - {T_1} = (P + Q) - P = Q$.

(B) Given that ${S_{2n}} = 3{S_n}$.
Using the formula for the sum of $n$ terms of an $A.P.$,${S_n} = \frac{n}{2}[2a + (n - 1)d]$.
Substituting this into the given equation:
$\frac{2n}{2}[2a + (2n - 1)d] = 3 \times \frac{n}{2}[2a + (n - 1)d]$
$2[2a + (2n - 1)d] = 3[2a + (n - 1)d]$
$4a + 4nd - 2d = 6a + 3nd - 3d$
$nd + d = 2a$
$2a = (n + 1)d$.
Now,we need to find the ratio $\frac{{{S_{3n}}}}{{{S_n}}}$:
$\frac{{{S_{3n}}}}{{{S_n}}} = \frac{\frac{3n}{2}[2a + (3n - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = 3 \times \frac{2a + (3n - 1)d}{2a + (n - 1)d}$.
Substituting $2a = (n + 1)d$ into the expression:
$= 3 \times \frac{(n + 1)d + (3n - 1)d}{(n + 1)d + (n - 1)d} = 3 \times \frac{(n + 1 + 3n - 1)d}{(n + 1 + n - 1)d} = 3 \times \frac{4nd}{2nd} = 3 \times 2 = 6$.

(D) Given that the $A.P.$ consists of consecutive integers,the common difference $d = 1$.
The first term $a = p^2 + 1$.
The number of terms $n = 2p + 1$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values:
$S_{2p+1} = \frac{2p + 1}{2} [2(p^2 + 1) + (2p + 1 - 1)(1)]$
$S_{2p+1} = \frac{2p + 1}{2} [2p^2 + 2 + 2p]$
$S_{2p+1} = (2p + 1)(p^2 + p + 1)$
We know that $(p + 1)^3 = p^3 + 3p^2 + 3p + 1$.
Expanding $p^3 + (p + 1)^3 = p^3 + p^3 + 3p^2 + 3p + 1 = 2p^3 + 3p^2 + 3p + 1$.
Alternatively,multiplying $(2p + 1)(p^2 + p + 1) = 2p^3 + 2p^2 + 2p + p^2 + p + 1 = 2p^3 + 3p^2 + 3p + 1$.
Thus,the sum is $p^3 + (p + 1)^3$.

(B) Let the first term be $a = 11$ and the common difference be $d$.
The sum of the first four terms is given by: $a + (a + d) + (a + 2d) + (a + 3d) = 56$.
Substituting $a = 11$: $11 + (11 + d) + (11 + 2d) + (11 + 3d) = 56$.
$44 + 6d = 56 \Rightarrow 6d = 12 \Rightarrow d = 2$.
The last four terms of an $A.P.$ with $n$ terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n$.
These are $(11 + (n-4)2), (11 + (n-3)2), (11 + (n-2)2), (11 + (n-1)2)$.
Sum $= 44 + 2(4n - 10) = 112$.
$44 + 8n - 20 = 112$.
$8n + 24 = 112$.
$8n = 88 \Rightarrow n = 11$.

(D) The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Here,the first term $a = 3$ and the common difference $d = 7 - 3 = 4$.
Given the sum $S_n = 406$,we substitute these values into the formula:
$406 = \frac{n}{2}[2(3) + (n - 1)4]$
$406 = \frac{n}{2}[6 + 4n - 4]$
$406 = \frac{n}{2}[4n + 2]$
$406 = n(2n + 1)$
$2n^2 + n - 406 = 0$
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{-1 \pm \sqrt{1^2 - 4(2)(-406)}}{2(2)}$
$n = \frac{-1 \pm \sqrt{1 + 3248}}{4}$
$n = \frac{-1 \pm \sqrt{3249}}{4}$
$n = \frac{-1 \pm 57}{4}$
Since the number of terms $n$ must be positive,we take the positive value:
$n = \frac{56}{4} = 14$.
Thus,the number of terms is $14$.

(B) Given that the number of terms $n = 15$,the first term $a = 5$,and the sum of terms $S_{15} = 390$.
Using the sum formula for an arithmetic progression: $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $390 = \frac{15}{2} [2(5) + (15 - 1)d]$.
$390 = \frac{15}{2} [10 + 14d]$.
$390 = 15(5 + 7d)$.
$26 = 5 + 7d$.
$7d = 21$,so $d = 3$.
The middle term of an arithmetic progression with $15$ terms is the $\frac{15+1}{2} = 8^{th}$ term.
The $n^{th}$ term formula is $a_n = a + (n - 1)d$.
For $n = 8$: $a_8 = 5 + (8 - 1)(3) = 5 + 7(3) = 5 + 21 = 26$.

(A) Let the first term be $a$ and the common difference be $d$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
According to the problem,$S_{10} = 4 \times S_5$.
Substituting the formula:
$\frac{10}{2}[2a + (10 - 1)d] = 4 \times \frac{5}{2}[2a + (5 - 1)d]$
$5[2a + 9d] = 4 \times 2.5[2a + 4d]$
$5[2a + 9d] = 10[2a + 4d]$
Dividing both sides by $5$:
$2a + 9d = 2[2a + 4d]$
$2a + 9d = 4a + 8d$
Rearranging the terms to find the ratio $a:d$:
$9d - 8d = 4a - 2a$
$d = 2a$
$\frac{a}{d} = \frac{1}{2}$
Therefore,the ratio of the first term to the common difference is $1:2$.

(B) Let the three numbers in $A.P.$ be $(a - d)$,$a$,and $(a + d)$.
According to the problem,the sum of the numbers is $18$:
$(a - d) + a + (a + d) = 18$
$3a = 18$
$a = 6$
The sum of their squares is $158$:
$(a - d)^2 + a^2 + (a + d)^2 = 158$
$(6 - d)^2 + 6^2 + (6 + d)^2 = 158$
$(36 - 12d + d^2) + 36 + (36 + 12d + d^2) = 158$
$108 + 2d^2 = 158$
$2d^2 = 50$
$d^2 = 25$
$d = \pm 5$
If $d = 5$,the numbers are $(6 - 5), 6, (6 + 5)$,which are $1, 6, 11$.
If $d = -5$,the numbers are $(6 - (-5)), 6, (6 + (-5))$,which are $11, 6, 1$.
In both cases,the greatest number is $11$.

(A) The numerator is an arithmetic progression with first term $a_1 = 3$ and common difference $d_1 = 2$. The sum of $n$ terms is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d_1] = \frac{n}{2}[6 + (n-1)2] = \frac{n}{2}[2n + 4] = n(n+2)$.
The denominator is an arithmetic progression with first term $a_2 = 5$,common difference $d_2 = 3$,and $10$ terms. The sum is $S_{10} = \frac{10}{2}[2(5) + (10-1)3] = 5[10 + 27] = 5 \times 37 = 185$.
Given the ratio is $7$,we have $\frac{n(n+2)}{185} = 7$.
$n^2 + 2n = 1295$.
$n^2 + 2n - 1295 = 0$.
Factoring the quadratic equation: $(n + 37)(n - 35) = 0$.
Since $n$ must be positive,$n = 35$.

(B) Let the sequence be $\frac{1}{3}, A_1, A_2, \frac{1}{24}$. This is an arithmetic progression $(A.P.)$ with $n=4$ terms.
The common difference $d$ is given by $d = \frac{b-a}{n+1}$,where $a = \frac{1}{3}$,$b = \frac{1}{24}$,and $n=2$ (number of means).
$d = \frac{\frac{1}{24} - \frac{1}{3}}{2+1} = \frac{\frac{1-8}{24}}{3} = \frac{-7/24}{3} = -\frac{7}{72}$.
Now,$A_1 = a + d = \frac{1}{3} - \frac{7}{72} = \frac{24-7}{72} = \frac{17}{72}$.
$A_2 = A_1 + d = \frac{17}{72} - \frac{7}{72} = \frac{10}{72} = \frac{5}{36}$.
Thus,the values are $\frac{17}{72}$ and $\frac{5}{36}$.

(C) Given that $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n} = \frac{a + b}{2}$.
Cross-multiplying,we get:
$2(a^{n + 1} + b^{n + 1}) = (a + b)(a^n + b^n)$
$2a^{n + 1} + 2b^{n + 1} = a^{n + 1} + ab^n + ba^n + b^{n + 1}$
Rearranging the terms:
$a^{n + 1} - ab^n - ba^n + b^{n + 1} = 0$
$a^n(a - b) - b^n(a - b) = 0$
$(a^n - b^n)(a - b) = 0$
Since $a \neq b$,we must have $a^n - b^n = 0$,which implies $a^n = b^n$.
Thus,$(a/b)^n = 1 = (a/b)^0$.
Therefore,$n = 0$.

(D) Let the two numbers be $a$ and $b$.
Given that one number is the reciprocal of the other,so $a = \frac{1}{b}$ or $ab = 1$.
The arithmetic mean of the two numbers is given as $\frac{a + b}{2} = \frac{13}{12}$.
This implies $a + b = \frac{13}{6}$.
Substituting $b = \frac{1}{a}$ into the equation,we get $a + \frac{1}{a} = \frac{13}{6}$.
Multiplying by $6a$,we get $6a^2 - 13a + 6 = 0$.
Factoring the quadratic equation: $6a^2 - 9a - 4a + 6 = 0 \Rightarrow 3a(2a - 3) - 2(2a - 3) = 0$.
$(3a - 2)(2a - 3) = 0$.
Thus,$a = \frac{2}{3}$ or $a = \frac{3}{2}$.
If $a = \frac{3}{2}$,then $b = \frac{2}{3}$. If $a = \frac{2}{3}$,then $b = \frac{3}{2}$.
Therefore,the numbers are $\frac{3}{2}$ and $\frac{2}{3}$.

(A) Let the two numbers be $a$ and $b$.
The arithmetic mean $A$ between $a$ and $b$ is given by $A = \frac{a+b}{2}$.
Let $A_1, A_2, \dots, A_n$ be the $n$ arithmetic means between $a$ and $b$. Then $a, A_1, A_2, \dots, A_n, b$ form an arithmetic progression $(A.P.)$ with $n+2$ terms.
The common difference $d$ is given by $d = \frac{b-a}{n+1}$.
The sum $S$ of $n$ arithmetic means is $S = A_1 + A_2 + \dots + A_n$.
Since $A_1, A_2, \dots, A_n$ are in $A.P.$,their sum is $S = \frac{n}{2}(A_1 + A_n)$.
Here,$A_1 = a + d = a + \frac{b-a}{n+1} = \frac{an + a + b - a}{n+1} = \frac{an + b}{n+1}$ and $A_n = b - d = b - \frac{b-a}{n+1} = \frac{bn + b - b + a}{n+1} = \frac{bn + a}{n+1}$.
Thus,$S = \frac{n}{2} \left( \frac{an + b + bn + a}{n+1} \right) = \frac{n}{2} \left( \frac{(a+b)(n+1)}{n+1} \right) = \frac{n(a+b)}{2}$.
Since $A = \frac{a+b}{2}$,we have $S = nA$.

(B) The first $n$ natural numbers are $1, 2, 3, \dots, n$.
The sum of the first $n$ natural numbers is given by the formula $S_n = \frac{n(n + 1)}{2}$.
The arithmetic mean is defined as the sum of the observations divided by the number of observations.
Therefore, $\text{Arithmetic Mean} = \frac{S_n}{n} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2}$.

(A) Let the $n$ arithmetic means between $a$ and $b$ be $A_1, A_2, \dots, A_n$.
These means form an arithmetic progression with $a$ as the first term and $b$ as the $(n+2)$-th term.
The sum of $n$ arithmetic means is given by the formula $S_n = \frac{n}{2}(A_1 + A_n)$.
Since $A_1 = a + d$ and $A_n = b - d$,where $d$ is the common difference,we have $A_1 + A_n = a + b$.
Therefore,the sum is $\frac{n}{2}(a + b)$.

(B) The resulting progression consists of $n + 2$ terms,where the first term $a = 2$ and the last term $l = 38$.
The sum of an arithmetic progression is given by the formula $S_m = \frac{m}{2}(a + l)$,where $m$ is the total number of terms.
Here,$m = n + 2$,$a = 2$,and $l = 38$.
Given that the sum is $200$,we have:
$200 = \frac{n + 2}{2}(2 + 38)$
$200 = \frac{n + 2}{2}(40)$
$200 = 20(n + 2)$
Dividing both sides by $20$:
$10 = n + 2$
$n = 10 - 2 = 8$.
Therefore,the value of $n$ is $8$.

(B) The mean of a series is calculated by dividing the sum of all terms by the total number of terms.
Given series: $a, a + nd, a + 2nd$.
Number of terms = $3$.
Sum of terms = $a + (a + nd) + (a + 2nd) = 3a + 3nd$.
Mean = $\frac{3a + 3nd}{3} = \frac{3(a + nd)}{3} = a + nd$.
Therefore,the correct option is $B$.

(C) Given $f(x + y, x - y) = xy$.
Let $u = x + y$ and $v = x - y$.
Solving for $x$ and $y$,we get $x = \frac{u + v}{2}$ and $y = \frac{u - v}{2}$.
Substituting these into the function definition,we have $f(u, v) = \left( \frac{u + v}{2} \right) \left( \frac{u - v}{2} \right) = \frac{u^2 - v^2}{4}$.
Thus,$f(x, y) = \frac{x^2 - y^2}{4}$ and $f(y, x) = \frac{y^2 - x^2}{4}$.
The arithmetic mean of $f(x, y)$ and $f(y, x)$ is $\frac{f(x, y) + f(y, x)}{2} = \frac{1}{2} \left( \frac{x^2 - y^2}{4} + \frac{y^2 - x^2}{4} \right) = \frac{1}{2} (0) = 0$.

(B) Given that $\log 2, \log (2^n - 1)$ and $\log (2^n + 3)$ are in $A.P.$
By the property of $A.P.$,if $a, b, c$ are in $A.P.$,then $2b = a + c$.
Therefore,$2 \log (2^n - 1) = \log 2 + \log (2^n + 3)$.
Using the logarithmic property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$,we get:
$\log (2^n - 1)^2 = \log [2(2^n + 3)]$.
Removing the logarithms from both sides:
$(2^n - 1)^2 = 2(2^n + 3)$.
Let $x = 2^n$. Then $(x - 1)^2 = 2(x + 3)$.
$x^2 - 2x + 1 = 2x + 6$.
$x^2 - 4x - 5 = 0$.
Factoring the quadratic equation: $(x - 5)(x + 1) = 0$.
So,$x = 5$ or $x = -1$.
Since $x = 2^n$ must be positive,we have $2^n = 5$.
Taking the logarithm on both sides,$n = \log_2 5$.

(B) Let the four terms of the $A.P.$ be $(a - 3d), (a - d), (a + d), (a + 3d)$.
According to the problem, the sum of the two extreme terms is $8$:
$(a - 3d) + (a + 3d) = 8$
$2a = 8 \Rightarrow a = 4$.
The product of the two middle terms is $15$:
$(a - d)(a + d) = 15$
$a^2 - d^2 = 15$.
Substituting $a = 4$ into the equation:
$4^2 - d^2 = 15$
$16 - d^2 = 15$
$d^2 = 1 \Rightarrow d = 1$ (taking positive value for the series).
The four terms are:
$a - 3d = 4 - 3(1) = 1$
$a - d = 4 - 1 = 3$
$a + d = 4 + 1 = 5$
$a + 3d = 4 + 3(1) = 7$
The terms are $1, 3, 5, 7$. The greatest number is $7$.

(C) Let the sides of the triangle be $(a - d)$,$a$,and $(a + d)$,where $d > 0$.
Since the triangle is right-angled,the hypotenuse is the longest side,which is $(a + d)$.
According to the Pythagorean theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides:
$(a + d)^2 = a^2 + (a - d)^2$
Expanding both sides:
$a^2 + d^2 + 2ad = a^2 + a^2 - 2ad + d^2$
Simplifying the equation:
$2ad = a^2 - 2ad$
$a^2 - 4ad = 0$
$a(a - 4d) = 0$
Since $a$ represents a side length,$a \neq 0$,so $a = 4d$.
Substituting $a = 4d$ into the sides $(a - d, a, a + d)$:
$(4d - d) : 4d : (4d + d) = 3d : 4d : 5d = 3:4:5$.

(A) Let the three numbers in $A.P.$ be $(a - d), a, (a + d)$.
According to the problem,the sum of these numbers is $33$:
$(a - d) + a + (a + d) = 33$
$3a = 33$
$a = 11$
The product of these numbers is $792$:
$(a - d) \cdot a \cdot (a + d) = 792$
$a(a^2 - d^2) = 792$
$11(11^2 - d^2) = 792$
$121 - d^2 = 72$
$d^2 = 121 - 72 = 49$
$d = 7$ (taking the positive value for the sequence).
The numbers are $(11 - 7), 11, (11 + 7)$,which are $4, 11, 18$.
Thus,the smallest number is $4$.

(C) Given that $a, b, c, d, e, f$ are in $A.P.$
Let the common difference be $K$.
Then,$b - a = c - b = d - c = e - d = f - e = K$.
We need to find the value of $e - c$.
Since $d - c = K$ and $e - d = K$,we have $e - d = d - c$.
Adding $d$ to both sides,we get $e = 2d - c$.
Therefore,$e - c = (2d - c) - c = 2d - 2c = 2(d - c)$.
Alternatively,using the property of $A.P.$,$e = c + 2K$ and $d = c + K$.
Thus,$e - c = (c + 2K) - c = 2K$.
Since $d - c = K$,$2K = 2(d - c)$.
Hence,$e - c = 2(d - c)$.

(B) Let the three numbers in the arithmetic sequence be $(a - d)$,$a$,and $(a + d)$.
According to the problem,the sum of these numbers is $15$:
$(a - d) + a + (a + d) = 15$
$3a = 15$
$a = 5$
The sum of their squares is $83$:
$(a - d)^2 + a^2 + (a + d)^2 = 83$
$(a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 83$
$3a^2 + 2d^2 = 83$
Substituting $a = 5$ into the equation:
$3(5^2) + 2d^2 = 83$
$3(25) + 2d^2 = 83$
$75 + 2d^2 = 83$
$2d^2 = 8$
$d^2 = 4$
$d = \pm 2$
If $d = 2$,the numbers are $(5 - 2), 5, (5 + 2)$,which are $3, 5, 7$.
If $d = -2$,the numbers are $(5 - (-2)), 5, (5 + (-2))$,which are $7, 5, 3$.
In both cases,the set of numbers is ${3, 5, 7}$.

(B) Let the four arithmetic means be $A_1, A_2, A_3,$ and $A_4$.
Then,the sequence $3, A_1, A_2, A_3, A_4, 23$ forms an Arithmetic Progression $(AP)$.
Here,the first term $a = 3$ and the sixth term $T_6 = 23$.
The formula for the $n^{th}$ term of an $AP$ is $T_n = a + (n-1)d$.
For $n = 6$,we have $23 = 3 + (6-1)d$.
$23 = 3 + 5d$
$20 = 5d$
$d = 4$.
Now,we calculate the means:
$A_1 = a + d = 3 + 4 = 7$
$A_2 = a + 2d = 3 + 8 = 11$
$A_3 = a + 3d = 3 + 12 = 15$
$A_4 = a + 4d = 3 + 16 = 19$
Thus,the four arithmetic means are $7, 11, 15, 19$.

(A) Let the three consecutive terms of an $A.P.$ be $(a - d)$,$a$,and $(a + d)$.
According to the given condition,the sum of these terms is $51$:
$(a - d) + a + (a + d) = 51$
$3a = 51$
$a = 17$
The product of the first term $(a - d)$ and the last term $(a + d)$ is $273$:
$(a - d)(a + d) = 273$
$a^2 - d^2 = 273$
Substituting $a = 17$:
$17^2 - d^2 = 273$
$289 - d^2 = 273$
$d^2 = 289 - 273 = 16$
$d = \pm 4$
If $d = 4$,the terms are $(17 - 4), 17, (17 + 4)$,which are $13, 17, 21$.
If $d = -4$,the terms are $(17 - (-4)), 17, (17 + (-4))$,which are $21, 17, 13$.
Both sets represent the same sequence. Thus,the numbers are $21, 17, 13$.

(B) Given that $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$
By the property of $A.P.$,the difference between consecutive terms is constant:
$\frac{1}{r + p} - \frac{1}{p + q} = \frac{1}{q + r} - \frac{1}{r + p}$
$\Rightarrow \frac{(p + q) - (r + p)}{(r + p)(p + q)} = \frac{(r + p) - (q + r)}{(q + r)(r + p)}$
$\Rightarrow \frac{q - r}{p + q} = \frac{p - q}{q + r}$
Cross-multiplying gives:
$(q - r)(q + r) = (p - q)(p + q)$
$q^2 - r^2 = p^2 - q^2$
$2q^2 = p^2 + r^2$
This condition implies that $p^2, q^2, r^2$ are in $A.P.$

(D) Let $d$ be the common difference of the $A.P.$
Then,$\log_y x = 1 + d \implies x = y^{1+d}$
$\log_z y = 1 + 2d \implies y = z^{1+2d}$
$-15 \log_x z = 1 + 3d \implies \log_x z = -\frac{1+3d}{15} \implies z = x^{-(1+3d)/15}$
Substituting these,we get $x = y^{1+d} = (z^{1+2d})^{1+d} = z^{(1+2d)(1+d)}$.
Also,$z = x^{-(1+3d)/15} \implies x = z^{-15/(1+3d)}$.
Equating the exponents of $z$: $(1+2d)(1+d) = -\frac{15}{1+3d}$.
$(1+d)(1+2d)(1+3d) = -15$.
Expanding the product: $(1 + 3d + 2d^2)(1 + 3d) = 1 + 3d + 3d + 9d^2 + 2d^2 + 6d^3 = 6d^3 + 11d^2 + 6d + 1 = -15$.
$6d^3 + 11d^2 + 6d + 16 = 0$.
Testing $d = -2$: $6(-8) + 11(4) + 6(-2) + 16 = -48 + 44 - 12 + 16 = 0$.
Thus,$d = -2$ is a root.
For $d = -2$: $\log_y x = 1 - 2 = -1 \implies x = y^{-1}$.
$\log_z y = 1 + 2(-2) = -3 \implies y = z^{-3}$.
Since $x = y^{-1}$ and $y = z^{-3}$,we have $x = (z^{-3})^{-1} = z^3$.
Therefore,all the given options are correct.

(B) Let the integer be $n$. The difference between the integer and its cube is given by $n^3 - n$.
We can factorize this expression as: $n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1)$.
This is the product of three consecutive integers: $(n - 1)$,$n$,and $(n + 1)$.
In any set of three consecutive integers,at least one must be a multiple of $2$ and exactly one must be a multiple of $3$.
Since $2$ and $3$ are coprime,their product $2 \times 3 = 6$ must divide the product of any three consecutive integers.
Therefore,$n^3 - n$ is always divisible by $6$.

(D) Given that $a, b, c$ are in $A.P.$,we have the property $2b = a + c$.
Now,substitute $2b = a + c$ into the given expression $(a + 2b - c)(2b + c - a)(c + a - b)$:
$1$. First term: $(a + 2b - c) = (a + (a + c) - c) = 2a$.
$2$. Second term: $(2b + c - a) = ((a + c) + c - a) = 2c$.
$3$. Third term: $(c + a - b) = (2b - b) = b$ (since $a + c = 2b$).
Multiplying these terms together:
$(2a) \times (2c) \times (b) = 4abc$.
Therefore,the correct option is $D$.

(D) Let the four numbers in arithmetic progression be $A_1, A_2, A_3, A_4$.
Given that $A_1 + A_4 = 8$ $(i)$ and $A_2 \times A_3 = 15$ $(ii)$.
In an arithmetic progression,the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first and last terms.
Therefore,$A_2 + A_3 = A_1 + A_4 = 8$ $(iii)$.
From $(ii)$ and $(iii)$,we have $A_2 + \frac{15}{A_2} = 8$,which simplifies to $A_2^2 - 8A_2 + 15 = 0$.
Solving the quadratic equation,we get $(A_2 - 3)(A_2 - 5) = 0$,so $A_2 = 3$ or $A_2 = 5$.
If $A_2 = 3$,then $A_3 = 5$. If $A_2 = 5$,then $A_3 = 3$.
Using the property $A_2 = \frac{A_1 + A_3}{2}$,we find $A_1 = 2A_2 - A_3$.
For $A_2 = 3$ and $A_3 = 5$,$A_1 = 2(3) - 5 = 1$. Then $A_4 = 8 - 1 = 7$.
The series is $1, 3, 5, 7$. The least number is $1$.