Progression and Sequence Questions in English

(C) To determine the type of sequence,we check the difference between consecutive terms.
Let the sequence be $a_1, a_2, a_3, \dots$ where $a_1 = \frac{5}{\sqrt{7}}$,$a_2 = \frac{6}{\sqrt{7}}$,and $a_3 = \sqrt{7}$.
The difference between the second and first term is $d_1 = a_2 - a_1 = \frac{6}{\sqrt{7}} - \frac{5}{\sqrt{7}} = \frac{1}{\sqrt{7}}$.
The difference between the third and second term is $d_2 = a_3 - a_2 = \sqrt{7} - \frac{6}{\sqrt{7}}$.
Since $\sqrt{7} = \frac{\sqrt{7} \times \sqrt{7}}{\sqrt{7}} = \frac{7}{\sqrt{7}}$,we have $d_2 = \frac{7}{\sqrt{7}} - \frac{6}{\sqrt{7}} = \frac{1}{\sqrt{7}}$.
Since $d_1 = d_2 = \frac{1}{\sqrt{7}}$,the common difference is constant.
Therefore,the sequence is an $A.P.$ (Arithmetic Progression).

(B) The given series is $\left( 3 - \frac{1}{n} \right) + \left( 3 - \frac{2}{n} \right) + \left( 3 - \frac{3}{n} \right) + \dots$,which is an Arithmetic Progression ($A$.$P$.).
The first term $a = \left( 3 - \frac{1}{n} \right)$.
The common difference $d = \left( 3 - \frac{2}{n} \right) - \left( 3 - \frac{1}{n} \right) = -\frac{2}{n} + \frac{1}{n} = -\frac{1}{n}$.
The $p^{th}$ term of an $A$.$P$. is given by the formula $T_p = a + (p - 1)d$.
Substituting the values:
$T_p = \left( 3 - \frac{1}{n} \right) + (p - 1)\left( -\frac{1}{n} \right)$
$T_p = 3 - \frac{1}{n} - \frac{p}{n} + \frac{1}{n}$
$T_p = 3 - \frac{p}{n}$.
Alternatively,by observation,the $k^{th}$ term is $\left( 3 - \frac{k}{n} \right)$. Therefore,the $p^{th}$ term is $\left( 3 - \frac{p}{n} \right)$.

(A) The given series is $2\sqrt{2} + \sqrt{2} + 0 + \dots$
This is an Arithmetic Progression $(A.P.)$ where the first term $a = 2\sqrt{2}$ and the common difference $d = \sqrt{2} - 2\sqrt{2} = -\sqrt{2}$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
For the $8^{th}$ term $(n = 8)$:
$a_8 = 2\sqrt{2} + (8 - 1)(-\sqrt{2})$
$a_8 = 2\sqrt{2} + 7(-\sqrt{2})$
$a_8 = 2\sqrt{2} - 7\sqrt{2}$
$a_8 = -5\sqrt{2}$.

(B) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
Given that the $9^{th}$ term is zero:
$a_9 = a + (9 - 1)d = 0$
$a + 8d = 0 \Rightarrow a = -8d$.
Now,we need to find the ratio of the $29^{th}$ term to the $19^{th}$ term:
$a_{29} = a + 28d = -8d + 28d = 20d$.
$a_{19} = a + 18d = -8d + 18d = 10d$.
Ratio $= \frac{a_{29}}{a_{19}} = \frac{20d}{10d} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(A) The sequence $f(n) = an + b;\;n \in N$ is an $A.P.$ (Arithmetic Progression).
By substituting $n = 1, 2, 3, 4, \dots$,we obtain the sequence:
$(a + b), (2a + b), (3a + b), \dots$
Here,the first term $A = (a + b)$ and the common difference $d = (2a + b) - (a + b) = a$.
Since the common difference is constant,the sequence is an $A.P.$
Alternatively,the $n^{th}$ term of an $A.P.$ is always a linear expression in $n$ of the form $an + b$ for all $n \in N$.

(A) The given sequence is an arithmetic progression where the first term $a = -8 + 18i$ and the common difference $d = (-6 + 15i) - (-8 + 18i) = 2 - 3i$.
The $n^{th}$ term of an arithmetic progression is given by $T_n = a + (n - 1)d$.
$T_n = (-8 + 18i) + (n - 1)(2 - 3i)$
$T_n = -8 + 18i + 2n - 2 - 3ni + 3i$
$T_n = (2n - 10) + i(21 - 3n)$
For the term to be purely imaginary,the real part must be zero.
$2n - 10 = 0$
$2n = 10$
$n = 5$
Thus,the $5^{th}$ term is purely imaginary.

(C) Given that the $n^{th}$ term is $T_n = 2n - 1$.
To find the sum of the first $n$ terms $(S_n)$,we can use the formula $S_n = \sum_{k=1}^{n} T_k$.
$S_n = \sum_{k=1}^{n} (2k - 1)$
$S_n = 2 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1$
Using the sum of the first $n$ natural numbers formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:
$S_n = 2 \left[ \frac{n(n+1)}{2} \right] - n$
$S_n = n(n + 1) - n$
$S_n = n^2 + n - n = n^2$.
Alternatively,the sequence is $1, 3, 5, \dots, (2n - 1)$,which is the sum of the first $n$ odd numbers,which is known to be $n^2$.

(A) The given series is $(1 \times 3) + (3 \times 5) + (5 \times 7) + (7 \times 9) + \dots$
Let the $n^{th}$ term be $T_n$.
The first factors of each term are $1, 3, 5, 7, \dots$,which form an Arithmetic Progression $(AP)$ with first term $a = 1$ and common difference $d = 2$.
The $n^{th}$ term of this $AP$ is $a_n = a + (n - 1)d = 1 + (n - 1)2 = 1 + 2n - 2 = 2n - 1$.
The second factors of each term are $3, 5, 7, 9, \dots$,which form an $AP$ with first term $a = 3$ and common difference $d = 2$.
The $n^{th}$ term of this $AP$ is $b_n = a + (n - 1)d = 3 + (n - 1)2 = 3 + 2n - 2 = 2n + 1$.
Therefore,the $n^{th}$ term of the series is $T_n = a_n \times b_n = (2n - 1)(2n + 1)$.

(B) The given series is $101, 99, 97, \dots, 47$.
This is an arithmetic progression $(AP)$ where the first term $a = 101$ and the common difference $d = 99 - 101 = -2$.
The last term $l$ (or $T_n$) is $47$.
The formula for the $n^{th}$ term of an $AP$ is $T_n = a + (n - 1)d$.
Substituting the values,we get $47 = 101 + (n - 1)(-2)$.
Subtracting $101$ from both sides,we get $47 - 101 = (n - 1)(-2)$,which simplifies to $-54 = (n - 1)(-2)$.
Dividing both sides by $-2$,we get $27 = n - 1$.
Therefore,$n = 27 + 1 = 28$.
Thus,the number of terms in the series is $28$.

(B) Let the first term be $a$ and the common difference be $d$.
Given that,the ${p^{th}}$ term is ${T_p} = a + (p - 1)d = q$ ..... $(i)$
And the ${q^{th}}$ term is ${T_q} = a + (q - 1)d = p$ ..... $(ii)$
Subtracting equation $(ii)$ from $(i)$:
$(a + (p - 1)d) - (a + (q - 1)d) = q - p$
$(p - 1 - q + 1)d = -(p - q)$
$(p - q)d = -(p - q)$
$d = -1$
Substituting $d = -1$ in equation $(i)$:
$a + (p - 1)(-1) = q$
$a - p + 1 = q$
$a = p + q - 1$
Now,the ${r^{th}}$ term is ${T_r} = a + (r - 1)d$
${T_r} = (p + q - 1) + (r - 1)(-1)$
${T_r} = p + q - 1 - r + 1$
${T_r} = p + q - r$
Thus,the ${r^{th}}$ term is $p + q - r$.

(A) Given the equation $\tan n\theta = \tan m\theta$.
The general solution for $\tan x = \tan y$ is $x = N\pi + y$,where $N$ is an integer.
Applying this to the given equation: $n\theta = N\pi + m\theta$.
Rearranging for $\theta$: $(n - m)\theta = N\pi$,which gives $\theta = \frac{N\pi}{n - m}$.
For different integer values of $N = 1, 2, 3, \dots$,the values of $\theta$ are $\frac{\pi}{n - m}, \frac{2\pi}{n - m}, \frac{3\pi}{n - m}, \dots$.
These values form an Arithmetic Progression $(A.P.)$ because the difference between consecutive terms is constant: $d = \frac{(N+1)\pi}{n - m} - \frac{N\pi}{n - m} = \frac{\pi}{n - m}$.

(A) The given series is $3.8 + 6.11 + 9.14 + 12.17 + .....$
The series consists of products of two terms.
The first factors are $3, 6, 9, 12, .....$,which form an Arithmetic Progression $(AP)$ with the first term $a = 3$ and common difference $d = 3$.
The $n^{th}$ term of this sequence is $a_n = a + (n - 1)d = 3 + (n - 1)3 = 3n$.
The second factors are $8, 11, 14, 17, .....$,which form an $AP$ with the first term $a = 8$ and common difference $d = 3$.
The $n^{th}$ term of this sequence is $b_n = 8 + (n - 1)3 = 8 + 3n - 3 = 3n + 5$.
Therefore,the $n^{th}$ term of the given series is the product of the $n^{th}$ terms of the two sequences:
$T_n = (3n) \times (3n + 5) = 3n(3n + 5)$.

(B) To find the sum of integers from $1$ to $100$ divisible by $2$ or $5$,we use the Principle of Inclusion-Exclusion: $S = S_2 + S_5 - S_{10}$,where $S_n$ is the sum of integers divisible by $n$.
$1$. Sum of integers divisible by $2$ $(2, 4, ..., 100)$: This is an arithmetic progression with $n = 50$,$a = 2$,$l = 100$. Sum $= \frac{50}{2}(2 + 100) = 25 \times 102 = 2550$.
$2$. Sum of integers divisible by $5$ $(5, 10, ..., 100)$: This is an arithmetic progression with $n = 20$,$a = 5$,$l = 100$. Sum $= \frac{20}{2}(5 + 100) = 10 \times 105 = 1050$.
$3$. Sum of integers divisible by both $2$ and $5$ (i.e.,divisible by $10$) $(10, 20, ..., 100)$: This is an arithmetic progression with $n = 10$,$a = 10$,$l = 100$. Sum $= \frac{10}{2}(10 + 100) = 5 \times 110 = 550$.
Total sum $= 2550 + 1050 - 550 = 3050$.

(C) The first series is $63, 65, 67, 69, \dots$ which is an Arithmetic Progression $(AP)$ with first term $a_1 = 63$ and common difference $d_1 = 2$.
The $m^{th}$ term of the first series is $T_m = a_1 + (m - 1)d_1 = 63 + (m - 1)2 = 63 + 2m - 2 = 2m + 61$.
The second series is $3, 10, 17, 24, \dots$ which is an $AP$ with first term $a_2 = 3$ and common difference $d_2 = 7$.
The $m^{th}$ term of the second series is $T'_m = a_2 + (m - 1)d_2 = 3 + (m - 1)7 = 3 + 7m - 7 = 7m - 4$.
Given that the $m^{th}$ terms are equal,we have $2m + 61 = 7m - 4$.
Rearranging the terms,$61 + 4 = 7m - 2m$.
$65 = 5m$.
$m = 13$.

(B) The given series is $\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + \dots$
This can be rewritten as $1\sqrt 2 + 2\sqrt 2 + 3\sqrt 2 + 4\sqrt 2 + \dots$
This is an arithmetic progression where the first term $a = \sqrt 2$ and the common difference $d = \sqrt 2$.
The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.
For $n = 24$ terms:
$S_{24} = \frac{24}{2} [2(\sqrt 2) + (24 - 1)\sqrt 2]$
$S_{24} = 12 [2\sqrt 2 + 23\sqrt 2]$
$S_{24} = 12 [25\sqrt 2]$
$S_{24} = 300\sqrt 2$.

(C) Given that the terms $2x, x + 8, 3x + 1$ are in $A.P.$
For any three terms $a, b, c$ to be in $A.P.$,the middle term must be the arithmetic mean of the other two,i.e.,$b = \frac{a + c}{2}$.
Applying this to the given terms:
$x + 8 = \frac{2x + (3x + 1)}{2}$
Multiply both sides by $2$:
$2(x + 8) = 5x + 1$
$2x + 16 = 5x + 1$
Rearranging the terms to solve for $x$:
$16 - 1 = 5x - 2x$
$15 = 3x$
$x = \frac{15}{3}$
$x = 5$

(D) Given that the sum of $n$ terms is $S_n = nA + n^2B$.
To find the terms of the $A.P.$,we use the relation $T_n = S_n - S_{n-1}$.
First,find $S_1$ and $S_2$:
$S_1 = A(1) + B(1)^2 = A + B$
$S_2 = A(2) + B(2)^2 = 2A + 4B$
Now,calculate the first two terms:
$T_1 = S_1 = A + B$
$T_2 = S_2 - S_1 = (2A + 4B) - (A + B) = A + 3B$
The common difference $d$ is given by $d = T_2 - T_1$:
$d = (A + 3B) - (A + B) = 2B$
Thus,the common difference is $2B$.

(B) The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$,where $a$ is the first term and $d$ is the common difference.
Given $T_9 = a + 8d = 35$ (Equation $1$).
Given $T_{19} = a + 18d = 75$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(a + 18d) - (a + 8d) = 75 - 35$
$10d = 40$
$d = 4$.
Substituting $d = 4$ into Equation $1$:
$a + 8(4) = 35$
$a + 32 = 35$
$a = 3$.
The $20^{th}$ term is $T_{20} = a + 19d$.
$T_{20} = 3 + 19(4) = 3 + 76 = 79$.

(A) The given series is $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$
We can rewrite the terms as:
$27 = \frac{27}{1}$
$9 = \frac{27}{3}$
$5\frac{2}{5} = \frac{27}{5}$
$3\frac{6}{7} = \frac{27}{7}$
Thus,the series is $\frac{27}{1} + \frac{27}{3} + \frac{27}{5} + \frac{27}{7} + \dots$
The numerator is constant $(27)$,and the denominators form an arithmetic progression: $1, 3, 5, 7, \dots$
The $n^{th}$ term of the denominator is $a + (n-1)d = 1 + (n-1)2 = 2n - 1$.
Therefore,the $n^{th}$ term of the series is $T_n = \frac{27}{2n - 1}$.
For the $9^{th}$ term $(n = 9)$:
$T_9 = \frac{27}{2(9) - 1} = \frac{27}{18 - 1} = \frac{27}{17} = 1\frac{10}{17}$.

(D) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $b = \frac{a + c}{2}$.
Substituting this into the expression $\frac{(a - c)^2}{(b^2 - ac)}$:
$\frac{(a - c)^2}{(\frac{a + c}{2})^2 - ac} = \frac{(a - c)^2}{\frac{a^2 + c^2 + 2ac}{4} - ac}$
$= \frac{4(a - c)^2}{a^2 + c^2 + 2ac - 4ac} = \frac{4(a - c)^2}{a^2 + c^2 - 2ac}$
$= \frac{4(a - c)^2}{(a - c)^2} = 4$.
Alternatively,let $a = 1, b = 2, c = 3$ (which are in $A.P.$).
Then $\frac{(1 - 3)^2}{(2^2 - 1 \times 3)} = \frac{(-2)^2}{4 - 3} = \frac{4}{1} = 4$.

(D) Given that $\log_3 2, \log_3(2^x - 5)$,and $\log_3(2^x - 7/2)$ are in $A.P.$
Since they are in $A.P.$,the middle term is the arithmetic mean of the other two: $2 \log_3(2^x - 5) = \log_3 2 + \log_3(2^x - 7/2)$.
Using the property $\log_b a + \log_b c = \log_b(ac)$,we get: $\log_3(2^x - 5)^2 = \log_3[2(2^x - 7/2)]$.
Equating the arguments: $(2^x - 5)^2 = 2(2^x) - 7$.
Let $y = 2^x$. Then $(y - 5)^2 = 2y - 7$.
$y^2 - 10y + 25 = 2y - 7$.
$y^2 - 12y + 32 = 0$.
Factoring the quadratic: $(y - 8)(y - 4) = 0$,so $y = 8$ or $y = 4$.
If $y = 8$,then $2^x = 8 \Rightarrow x = 3$.
If $y = 4$,then $2^x = 4 \Rightarrow x = 2$.
Check the domain: For $\log_3(2^x - 5)$ to be defined,$2^x - 5 > 0$,so $2^x > 5$.
If $x = 2$,$2^2 = 4$,which is not $> 5$. Thus,$x = 2$ is rejected.
If $x = 3$,$2^3 = 8$,which is $> 5$. Thus,$x = 3$ is the only solution.
Since $3$ is not listed in options $A, B,$ or $C$,the correct option is $D$.

(C) Let the first term of the arithmetic progression $(A.P.)$ be $A$ and the common difference be $D$.
The $p^{th}$ term is given by $a = A + (p - 1)D$ .....$(i)$
The $q^{th}$ term is given by $b = A + (q - 1)D$ .....$(ii)$
The $r^{th}$ term is given by $c = A + (r - 1)D$ .....$(iii)$
We need to evaluate the expression $E = a(q - r) + b(r - p) + c(p - q)$.
Substituting the values of $a, b,$ and $c$:
$E = [A + (p - 1)D](q - r) + [A + (q - 1)D](r - p) + [A + (r - 1)D](p - q)$
Expanding the terms:
$E = A(q - r + r - p + p - q) + D[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)]$
Since $(q - r + r - p + p - q) = 0$,the first part becomes $A(0) = 0$.
For the second part:
$(p - 1)(q - r) = pq - pr - q + r$
$(q - 1)(r - p) = qr - qp - r + p$
$(r - 1)(p - q) = rp - rq - p + q$
Summing these:
$(pq - qp) + (-pr + rp) + (-qr + rq) + (-q + q) + (-r + r) + (-p + p) = 0$
Thus,$E = A(0) + D(0) = 0$.

(A) Let the ${n^{th}}$ terms of the two arithmetic progressions be $a_n = 3n + 8$ and $b_n = 7n + 15$.
To find the ${12^{th}}$ term of the first $A.P.$,substitute $n = 12$ into $a_n$:
$a_{12} = 3(12) + 8 = 36 + 8 = 44$.
To find the ${12^{th}}$ term of the second $A.P.$,substitute $n = 12$ into $b_n$:
$b_{12} = 7(12) + 15 = 84 + 15 = 99$.
The ratio of their ${12^{th}}$ terms is $\frac{a_{12}}{b_{12}} = \frac{44}{99}$.
Simplifying the fraction by dividing both numerator and denominator by $11$,we get $\frac{44 \div 11}{99 \div 11} = \frac{4}{9}$.

(B) Given the recurrence relation $a_n = a_{n-1} - 1$ for $n > 2$ with $a_2 = 2$.
Step $1$: Calculate $a_3$:
$a_3 = a_2 - 1 = 2 - 1 = 1$.
Step $2$: Calculate $a_4$:
$a_4 = a_3 - 1 = 1 - 1 = 0$.
Step $3$: Calculate $a_5$:
$a_5 = a_4 - 1 = 0 - 1 = -1$.
Thus,the value of $a_5$ is $-1$.

(C) Let $D$ be the common difference of the $A.P.$
Since $a, b, c, d, e$ are in $A.P.$,we can write:
$b = a + D$
$c = a + 2D$
$d = a + 3D$
$e = a + 4D$
Substituting these values into the expression $a - 4b + 6c - 4d + e$:
$= a - 4(a + D) + 6(a + 2D) - 4(a + 3D) + (a + 4D)$
$= a - 4a - 4D + 6a + 12D - 4a - 12D + a + 4D$
$= (a - 4a + 6a - 4a + a) + (-4D + 12D - 12D + 4D)$
$= (0)a + (0)D$
$= 0$

(C) Let $a$ be the first term and $x$ be the common difference of the $A.P.$
Given that the sixth term $a_6 = a + 5x = 2$,so $a = 2 - 5x$.
Let the product $P = a_1 a_4 a_5 = a(a + 3x)(a + 4x)$.
Substituting $a = 2 - 5x$ into the product:
$P = (2 - 5x)(2 - 5x + 3x)(2 - 5x + 4x)$
$P = (2 - 5x)(2 - 2x)(2 - x)$
$P = (2 - 5x)(4 - 6x + 2x^2) = 8 - 12x + 4x^2 - 20x + 30x^2 - 10x^3$
$P = -10x^3 + 34x^2 - 32x + 8 = 2(-5x^3 + 17x^2 - 16x + 4)$.
To find the minimum,differentiate $P$ with respect to $x$:
$\frac{dP}{dx} = 2(-15x^2 + 34x - 16) = 0$.
Solving $-15x^2 + 34x - 16 = 0$ or $15x^2 - 34x + 16 = 0$:
Using the quadratic formula,$x = \frac{34 \pm \sqrt{34^2 - 4(15)(16)}}{2(15)} = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm \sqrt{196}}{30} = \frac{34 \pm 14}{30}$.
So,$x = \frac{48}{30} = \frac{8}{5}$ or $x = \frac{20}{30} = \frac{2}{3}$.
Checking the second derivative $\frac{d^2P}{dx^2} = 2(-30x + 34)$:
For $x = \frac{8}{5}$,$\frac{d^2P}{dx^2} = 2(-48 + 34) < 0$ (Local maximum).
For $x = \frac{2}{3}$,$\frac{d^2P}{dx^2} = 2(-20 + 34) > 0$ (Local minimum).
Thus,$P$ is least for $x = \frac{2}{3}$.

(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.
According to the problem,$p \cdot T_p = q \cdot T_q$.
Substituting the formula: $p\{a + (p - 1)d\} = q\{a + (q - 1)d\}$.
Expanding the terms: $ap + p(p - 1)d = aq + q(q - 1)d$.
Rearranging the terms: $a(p - q) + \{p(p - 1) - q(q - 1)\}d = 0$.
$a(p - q) + \{p^2 - p - q^2 + q\}d = 0$.
$a(p - q) + \{(p^2 - q^2) - (p - q)\}d = 0$.
$a(p - q) + \{(p - q)(p + q) - (p - q)\}d = 0$.
Dividing by $(p - q)$ (since $p \neq q$): $a + (p + q - 1)d = 0$.
Since $T_{p+q} = a + (p + q - 1)d$,we conclude that $T_{p+q} = 0$.

(A) Let the first terms of the two arithmetic series be $a_1$ and $a_2$,and their common differences be $d_1$ and $d_2$ respectively.
Given the ratio of the sum of $n$ terms: $\frac{S_{n_1}}{S_{n_2}} = \frac{2n + 3}{6n + 5}$.
The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
So,$\frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{2n + 3}{6n + 5}$.
Simplifying,we get $\frac{a_1 + \frac{n-1}{2}d_1}{a_2 + \frac{n-1}{2}d_2} = \frac{2n + 3}{6n + 5}$.
We want the ratio of the $13^{th}$ terms,which is $\frac{T_{13_1}}{T_{13_2}} = \frac{a_1 + 12d_1}{a_2 + 12d_2}$.
Comparing $\frac{n-1}{2} = 12$,we get $n - 1 = 24$,so $n = 25$.
Substituting $n = 25$ in the ratio: $\frac{T_{13_1}}{T_{13_2}} = \frac{2(25) + 3}{6(25) + 5} = \frac{50 + 3}{150 + 5} = \frac{53}{155}$.

(C) Let the first term of the $A.P.$ be $A$ and the common difference be $D$.
The $m^{th}$ term is given by $a_m = A + (m - 1)D$.
The $(m+k)^{th}$ term is $a_{m+k} = A + (m+k-1)D$.
The $(m-k)^{th}$ term is $a_{m-k} = A + (m-k-1)D$.
Adding these two terms:
$a_{m+k} + a_{m-k} = [A + (m+k-1)D] + [A + (m-k-1)D]$
$a_{m+k} + a_{m-k} = 2A + (m+k-1+m-k-1)D$
$a_{m+k} + a_{m-k} = 2A + (2m-2)D$
$a_{m+k} + a_{m-k} = 2[A + (m-1)D]$
$a_{m+k} + a_{m-k} = 2a_m$
Therefore,$a_m = \frac{a_{m+k} + a_{m-k}}{2}$.

(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $r^{th}$ term is given by ${T_r} = a + (r - 1)d$.
Given,${T_m} = a + (m - 1)d = \frac{1}{n}$ --- $(1)$
Given,${T_n} = a + (n - 1)d = \frac{1}{m}$ --- $(2)$
Subtracting equation $(2)$ from $(1)$:
$(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$
Since $m \neq n$,we get $d = \frac{1}{mn}$.
Substituting $d$ in equation $(1)$:
$a + (m - 1)\frac{1}{mn} = \frac{1}{n}$
$a + \frac{1}{n} - \frac{1}{mn} = \frac{1}{n}$
$a = \frac{1}{mn}$.
Now,the $mn^{th}$ term is ${T_{mn}} = a + (mn - 1)d$.
${T_{mn}} = \frac{1}{mn} + (mn - 1)\frac{1}{mn} = \frac{1}{mn} + 1 - \frac{1}{mn} = 1$.

(B) Since the given numbers are in $A.P.$, the middle term is the arithmetic mean of the other two terms.
$2 \log_9(3^{1-x} + 2) = \log_3(4 \cdot 3^x - 1) + 1$
Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$, we get:
$2 \cdot \frac{1}{2} \log_3(3^{1-x} + 2) = \log_3(4 \cdot 3^x - 1) + \log_3 3$
$\log_3(3^{1-x} + 2) = \log_3(3(4 \cdot 3^x - 1))$
$3^{1-x} + 2 = 12 \cdot 3^x - 3$
Let $y = 3^x$. Then $3^{1-x} = \frac{3}{y}$.
$\frac{3}{y} + 2 = 12y - 3$
$3 + 2y = 12y^2 - 3y$
$12y^2 - 5y - 3 = 0$
$(4y - 3)(3y + 1) = 0$
Since $y = 3^x > 0$, we have $y = \frac{3}{4}$.
$3^x = \frac{3}{4} \implies x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4$.

(D) Let the common difference of the $A.P.$ be $d_0$.
Then,$b = a + d_0$,$c = a + 2d_0$,$d = a + 3d_0$,and $e = a + 4d_0$.
Substitute these values into the expression $a + b + 4c - 4d + e$:
$= a + (a + d_0) + 4(a + 2d_0) - 4(a + 3d_0) + (a + 4d_0)$
$= a + a + d_0 + 4a + 8d_0 - 4a - 12d_0 + a + 4d_0$
$= (a + a + 4a - 4a + a) + (d_0 + 8d_0 - 12d_0 + 4d_0)$
$= 3a + (1 + 8 - 12 + 4)d_0$
$= 3a + 1d_0$
$= 3a + d_0$
Since the expression depends on the common difference $d_0$,it cannot be expressed solely in terms of $a$ unless $d_0$ is known. Therefore,the correct option is $D$.

(C) Let $S_n$ and $S'_n$ be the sums of $n$ terms of two $A.P.'s$ with first terms $a, a'$ and common differences $d, d'$ respectively.
The ratio of the sum of $n$ terms is given by:
$\frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2a' + (n - 1)d']} = \frac{7n + 1}{4n + 27}$
$\frac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2}d'} = \frac{7n + 1}{4n + 27}$
We want to find the ratio of the $11^{th}$ terms,which is $\frac{T_{11}}{T'_{11}} = \frac{a + 10d}{a' + 10d'}$.
Comparing $\frac{n - 1}{2}$ with $10$,we get $n - 1 = 20$,so $n = 21$.
Substituting $n = 21$ in the ratio:
$\frac{T_{11}}{T'_{11}} = \frac{7(21) + 1}{4(21) + 27} = \frac{147 + 1}{84 + 27} = \frac{148}{111} = \frac{4}{3}$.
Thus,the ratio is $4:3$.

(D) The given series is $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms.
This is an arithmetic progression where the first term $a = \frac{1}{2}$.
The common difference $d = \frac{1}{3} - \frac{1}{2} = \frac{2-3}{6} = -\frac{1}{6}$.
The number of terms $n = 9$.
The sum of an arithmetic progression is given by the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $S_9 = \frac{9}{2} [2(\frac{1}{2}) + (9 - 1)(-\frac{1}{6})]$.
$S_9 = \frac{9}{2} [1 + 8(-\frac{1}{6})] = \frac{9}{2} [1 - \frac{8}{6}] = \frac{9}{2} [1 - \frac{4}{3}]$.
$S_9 = \frac{9}{2} [-\frac{1}{3}] = -\frac{9}{6} = -\frac{3}{2}$.

(C) Let the number of sides of the polygon be $n$.
The sum of the interior angles of a polygon with $n$ sides is given by $(n - 2) \times 180^{\circ}$.
Since the angles are in $A.P.$ with first term $a = 120^{\circ}$ and common difference $d = 5^{\circ}$,the sum of the $n$ angles is $\frac{n}{2} [2a + (n - 1)d]$.
Equating the two expressions:
$\frac{n}{2} [2(120^{\circ}) + (n - 1)5^{\circ}] = (n - 2)180^{\circ}$
$n[240 + 5n - 5] = 360(n - 2)$
$5n^2 + 235n = 360n - 720$
$5n^2 - 125n + 720 = 0$
Dividing by $5$:
$n^2 - 25n + 144 = 0$
$(n - 9)(n - 16) = 0$
So,$n = 9$ or $n = 16$.
If $n = 16$,the largest angle is $T_{16} = a + 15d = 120^{\circ} + 15(5^{\circ}) = 120^{\circ} + 75^{\circ} = 195^{\circ}$.
Since an interior angle of a convex polygon cannot exceed $180^{\circ}$,$n = 16$ is rejected.
Therefore,the number of sides is $n = 9$.

(C) Let the first term be $a$ and the common difference be $d$.
Given that the $p^{th}$ term $T_p = a + (p - 1)d = \frac{1}{q} \dots (i)$
And the $q^{th}$ term $T_q = a + (q - 1)d = \frac{1}{p} \dots (ii)$
Subtracting equation $(ii)$ from $(i)$:
$(p - q)d = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq}$
Thus,$d = \frac{1}{pq}$.
Substituting $d$ in equation $(i)$:
$a + (p - 1)\frac{1}{pq} = \frac{1}{q} \implies a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q} \implies a = \frac{1}{pq}$.
The sum of $pq$ terms $S_{pq} = \frac{pq}{2} [2a + (pq - 1)d]$.
Substituting the values of $a$ and $d$:
$S_{pq} = \frac{pq}{2} [2(\frac{1}{pq}) + (pq - 1)(\frac{1}{pq})]$
$S_{pq} = \frac{pq}{2} [\frac{2 + pq - 1}{pq}] = \frac{pq}{2} [\frac{pq + 1}{pq}] = \frac{pq + 1}{2}$.

(D) The sequence of the first $n$ natural numbers is $1, 2, 3, 4, \dots, n$,which forms an Arithmetic Progression $(A.P.)$ with the first term $a = 1$ and common difference $d = 1$.
The formula for the sum of the first $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values $a = 1$ and $d = 1$ into the formula:
$S_n = \frac{n}{2}[2(1) + (n - 1)(1)]$
$S_n = \frac{n}{2}[2 + n - 1]$
$S_n = \frac{n(n + 1)}{2}$.

(A) Given: First term $a = 2$,common difference $d = 4$,and number of terms $n = 40$.
The formula for the sum of the first $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values into the formula:
$S_{40} = \frac{40}{2}[2(2) + (40 - 1)4]$
$S_{40} = 20[4 + 39 \times 4]$
$S_{40} = 20[4 + 156]$
$S_{40} = 20[160]$
$S_{40} = 3200$.
Therefore,the sum of the first $40$ terms is $3200$.

(D) The given series is $S = 1 - 2 + 3 - 4 + 5 - 6 + \dots + (-1)^{n-1}n$.
Case $I$: If $n$ is odd,let $n = 2m + 1$.
The sum is $(1 - 2) + (3 - 4) + \dots + ((2m - 1) - 2m) + (2m + 1)$.
This simplifies to $(-1) + (-1) + \dots + (-1) \text{ (m times)} + (2m + 1) = -m + 2m + 1 = m + 1$.
Since $n = 2m + 1$,we have $m = \frac{n - 1}{2}$,so the sum is $\frac{n - 1}{2} + 1 = \frac{n + 1}{2}$.
Case $II$: If $n$ is even,let $n = 2m$.
The sum is $(1 - 2) + (3 - 4) + \dots + ((2m - 1) - 2m)$.
This is a sum of $m$ terms,each equal to $-1$,so the sum is $-m$.
Since $n = 2m$,we have $m = \frac{n}{2}$,so the sum is $-\frac{n}{2}$.
Thus,the sum is $\frac{n + 1}{2}$ if $n$ is odd and $-\frac{n}{2}$ if $n$ is even. Option $(d)$ correctly represents this conditional behavior.

(C) Given,the first term $A = a$.
The second term is $A + d = b$,where $d$ is the common difference.
Thus,$d = b - a$.
The last term $l = 2a$.
The formula for the $n^{th}$ term is $l = A + (n - 1)d$.
Substituting the values: $2a = a + (n - 1)(b - a)$.
$a = (n - 1)(b - a) \implies n - 1 = \frac{a}{b - a}$.
$n = \frac{a}{b - a} + 1 = \frac{a + b - a}{b - a} = \frac{b}{b - a}$.
The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(A + l)$.
Substituting $n = \frac{b}{b - a}$,$A = a$,and $l = 2a$:
$S_n = \frac{b}{2(b - a)}(a + 2a) = \frac{b}{2(b - a)}(3a) = \frac{3ab}{2(b - a)}$.

(C) The sum of the first $n$ even numbers is given by the formula $S_{Even} = 2 + 4 + 6 + \dots + 2n = n(n + 1)$.
The sum of the first $n$ odd numbers is given by the formula $S_{Odd} = 1 + 3 + 5 + \dots + (2n - 1) = n^2$.
Therefore,the ratio of the sum of the first $n$ even numbers to the sum of the first $n$ odd numbers is $\frac{S_{Even}}{S_{Odd}} = \frac{n(n + 1)}{n^2} = \frac{n + 1}{n}$.
Thus,the ratio is $(n + 1):n$.

(A) Given that $a_1, a_2, ..., a_n$ are in $A.P.$ with common difference $d = a_{i+1} - a_i$ for all $i$.
We know that $a_n = a_1 + (n-1)d$,which implies $(n-1)d = a_n - a_1$.
Rationalizing each term of the sum $S = \sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}}$:
$S = \sum_{i=1}^{n-1} \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{a_{i+1} - a_i} = \sum_{i=1}^{n-1} \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{d}$
$S = \frac{1}{d} [(\sqrt{a_2} - \sqrt{a_1}) + (\sqrt{a_3} - \sqrt{a_2}) + ... + (\sqrt{a_n} - \sqrt{a_{n-1}})]$
This is a telescoping sum,so $S = \frac{1}{d} (\sqrt{a_n} - \sqrt{a_1})$.
Multiplying the numerator and denominator by $(\sqrt{a_n} + \sqrt{a_1})$:
$S = \frac{1}{d} \cdot \frac{(\sqrt{a_n} - \sqrt{a_1})(\sqrt{a_n} + \sqrt{a_1})}{\sqrt{a_n} + \sqrt{a_1}} = \frac{1}{d} \cdot \frac{a_n - a_1}{\sqrt{a_n} + \sqrt{a_1}}$
Since $a_n - a_1 = (n-1)d$,we get:
$S = \frac{1}{d} \cdot \frac{(n-1)d}{\sqrt{a_n} + \sqrt{a_1}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$.

(B) Given that $a_1, a_2, \dots, a_n$ are in $A.P.$ with common difference $d = a_2 - a_1 = a_3 - a_2 = \dots = a_n - a_{n-1}$.
The given series is $S = \sin d (\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n)$.
We can write $\sin d = \sin(a_{k+1} - a_k)$.
Thus,the general term is $\frac{\sin(a_{k+1} - a_k)}{\sin a_k \sin a_{k+1}} = \frac{\sin a_{k+1} \cos a_k - \cos a_{k+1} \sin a_k}{\sin a_k \sin a_{k+1}} = \cot a_k - \cot a_{k+1}$.
Substituting this into the sum:
$S = (\cot a_1 - \cot a_2) + (\cot a_2 - \cot a_3) + \dots + (\cot a_{n-1} - \cot a_n)$.
This is a telescoping series,so all intermediate terms cancel out:
$S = \cot a_1 - \cot a_n$.

(B) The given series is an Arithmetic Progression $(A.P.)$ where the first term $a = 2$ and the common difference $d = 5 - 2 = 3$.
Let the number of terms be $n$.
The sum of the first $n$ terms of an $A.P.$ is given by the formula: $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Given $S_n = 60100$, we have:
$60100 = \frac{n}{2} [2(2) + (n - 1)3]$
$120200 = n [4 + 3n - 3]$
$120200 = n(3n + 1)$
$3n^2 + n - 120200 = 0$
Solving the quadratic equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{-1 \pm \sqrt{1^2 - 4(3)(-120200)}}{2(3)}$
$n = \frac{-1 \pm \sqrt{1 + 1442400}}{6}$
$n = \frac{-1 \pm \sqrt{1442401}}{6}$
$n = \frac{-1 \pm 1201}{6}$
Since $n$ must be positive, $n = \frac{1200}{6} = 200$.
Thus, the number of terms is $200$.

(B) The natural numbers between $1$ and $100$ that are multiples of $3$ form an arithmetic progression: $3, 6, 9, \dots, 99$.
Here,the first term $a = 3$,the common difference $d = 3$,and the last term $l = 99$.
To find the number of terms $n$,we use the formula $l = a + (n - 1)d$:
$99 = 3 + (n - 1)3$
$96 = (n - 1)3$
$32 = n - 1$
$n = 33$.
The sum $S_n$ of an arithmetic progression is given by $S_n = \frac{n}{2}(a + l)$:
$S_{33} = \frac{33}{2}(3 + 99)$
$S_{33} = \frac{33}{2}(102)$
$S_{33} = 33 \times 51 = 1683$.

(C) The given series is $1, 3, 5, 7, \dots$ which is an Arithmetic Progression $(AP)$.
Here,the first term $a = 1$ and the common difference $d = 3 - 1 = 2$.
The sum of the first $n$ terms of an $AP$ is given by the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $S_n = \frac{n}{2} [2(1) + (n - 1)2]$.
$S_n = \frac{n}{2} [2 + 2n - 2]$.
$S_n = \frac{n}{2} [2n]$.
$S_n = n^2$.

(A) The given series is an Arithmetic Progression $(AP)$ where the first term $a = 54$ and the common difference $d = 51 - 54 = -3$.
Let the number of terms be $n$. The sum of the first $n$ terms of an $AP$ is given by the formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Given $S_n = 513$, we have:
$513 = \frac{n}{2} [2(54) + (n - 1)(-3)]$
$513 = \frac{n}{2} [108 - 3n + 3]$
$1026 = n(111 - 3n)$
$1026 = 111n - 3n^2$
$3n^2 - 111n + 1026 = 0$
Dividing the equation by $3$, we get:
$n^2 - 37n + 342 = 0$
Factoring the quadratic equation:
$(n - 18)(n - 19) = 0$
Thus, $n = 18$ or $n = 19$.
Checking $n = 19$: The $19^{th}$ term is $a_{19} = a + 18d = 54 + 18(-3) = 54 - 54 = 0$. Since the $19^{th}$ term is $0$, the sum of $18$ terms and $19$ terms is the same $(513)$. However, in standard mathematical problems of this type, we usually provide the smallest integer value or note both. Given the options, $18$ is the correct choice.

(A) Given that the sum of $n$ terms is $S_n = 2n^2 + 5n$.
The $n^{th}$ term $T_n$ is given by the formula $T_n = S_n - S_{n-1}$ for $n > 1$.
First,calculate $S_{n-1}$:
$S_{n-1} = 2(n-1)^2 + 5(n-1)$
$S_{n-1} = 2(n^2 - 2n + 1) + 5n - 5$
$S_{n-1} = 2n^2 - 4n + 2 + 5n - 5$
$S_{n-1} = 2n^2 + n - 3$
Now,find $T_n$:
$T_n = (2n^2 + 5n) - (2n^2 + n - 3)$
$T_n = 2n^2 + 5n - 2n^2 - n + 3$
$T_n = 4n + 3$
Alternatively,for $n=1$,$T_1 = S_1 = 2(1)^2 + 5(1) = 7$. Using the formula $4n+3$,$4(1)+3 = 7$,which matches.

(D) Given the $n^{th}$ term of the $A.P.$ is $T_n = 3n - 1$.
To find the first five terms,we substitute $n = 1, 2, 3, 4, 5$:
For $n = 1: T_1 = 3(1) - 1 = 2$
For $n = 2: T_2 = 3(2) - 1 = 5$
For $n = 3: T_3 = 3(3) - 1 = 8$
For $n = 4: T_4 = 3(4) - 1 = 11$
For $n = 5: T_5 = 3(5) - 1 = 14$
The first five terms are $2, 5, 8, 11, 14$.
The sum of the first five terms is $S_5 = 2 + 5 + 8 + 11 + 14 = 40$.
Alternatively,using the sum formula: $S_n = \sum_{k=1}^{n} (3k - 1) = 3 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 3 \frac{n(n+1)}{2} - n$.
For $n = 5: S_5 = \frac{3(5)(6)}{2} - 5 = 45 - 5 = 40$.

(C) Given that the first term $a = 10$,the last term $l = 50$,and the sum of terms $S = 300$.
The formula for the sum of an $A.P.$ when the first and last terms are known is $S = \frac{n}{2}(a + l)$.
Substituting the given values into the formula:
$300 = \frac{n}{2}(10 + 50)$
$300 = \frac{n}{2}(60)$
$300 = n \times 30$
$n = \frac{300}{30} = 10$.
Therefore,the number of terms is $10$.