Arithmetic progression Questions in English

(C) To check if the sequence is an $A.P.$,we calculate the difference between consecutive terms.
First difference: $d_1 = \frac{6}{\sqrt{7}} - \frac{5}{\sqrt{7}} = \frac{1}{\sqrt{7}}$.
Second difference: $d_2 = \sqrt{7} - \frac{6}{\sqrt{7}} = \frac{(\sqrt{7} \times \sqrt{7}) - 6}{\sqrt{7}} = \frac{7 - 6}{\sqrt{7}} = \frac{1}{\sqrt{7}}$.
Since the common difference $d = \frac{1}{\sqrt{7}}$ is constant,the sequence is an $A.P.$

(B) The given series is $\left( 3 - \frac{1}{n} \right) + \left( 3 - \frac{2}{n} \right) + \left( 3 - \frac{3}{n} \right) + \dots$
This is an Arithmetic Progression ($A$.$P$.).
Here,the first term $a = \left( 3 - \frac{1}{n} \right)$.
The common difference $d = \left( 3 - \frac{2}{n} \right) - \left( 3 - \frac{1}{n} \right) = -\frac{1}{n}$.
The $p^{th}$ term of an $A$.$P$. is given by $T_p = a + (p - 1)d$.
Substituting the values,we get:
$T_p = \left( 3 - \frac{1}{n} \right) + (p - 1)\left( -\frac{1}{n} \right)$
$T_p = 3 - \frac{1}{n} - \frac{p}{n} + \frac{1}{n}$
$T_p = 3 - \frac{p}{n}$.
Alternatively,by inspection:
$1^{st}$ term: $3 - \frac{1}{n}$
$2^{nd}$ term: $3 - \frac{2}{n}$
$3^{rd}$ term: $3 - \frac{3}{n}$
Therefore,the $p^{th}$ term is $3 - \frac{p}{n}$.

(A) The given series $2\sqrt{2} + \sqrt{2} + 0 + \dots$ is an Arithmetic Progression $(A.P.)$.
Here,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) Given that the $9^{th}$ term $a_9 = a + (9 - 1)d = 0$.
$a + 8d = 0 \Rightarrow a = -8d$.
We need to find the ratio of the $29^{th}$ term to the $19^{th}$ term:
$\frac{a_{29}}{a_{19}} = \frac{a + 28d}{a + 18d}$.
Substituting $a = -8d$ into the expression:
$\frac{-8d + 28d}{-8d + 18d} = \frac{20d}{10d} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

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

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

(C) Given that the $n^{th}$ term is $T_n = 2n - 1$.
The first term $a = T_1 = 2(1) - 1 = 1$.
The last term $l = T_n = 2n - 1$.
The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$.
Substituting the values,we get $S_n = \frac{n}{2}(1 + 2n - 1) = \frac{n}{2}(2n) = n^2$.
Alternatively,$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (2k - 1) = 2\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 2 \cdot \frac{n(n+1)}{2} - n = n^2 + n - n = n^2$.

(B) The given series is an arithmetic progression: $101, 99, 97, \dots, 47$.
Here,the first term $a = 101$,the common difference $d = 99 - 101 = -2$,and the last term $l = 47$.
We use the formula for the $n^{th}$ term of an arithmetic progression: $T_n = a + (n - 1)d$.
Substituting the values,we get: $47 = 101 + (n - 1)(-2)$.
Subtracting $101$ from both sides: $47 - 101 = (n - 1)(-2)$.
$-54 = (n - 1)(-2)$.
Dividing by $-2$: $27 = n - 1$.
Therefore,$n = 28$.

(B) 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 $(i)$,we get $(p - q)d = q - p$,which implies $d = -1$.
Substituting $d = -1$ into equation $(i)$,we get $a + (p - 1)(-1) = q$,so $a = p + q - 1$.
Now,the $r^{th}$ term is given by $T_r = a + (r - 1)d$.
Substituting the values of $a$ and $d$,we get $T_r = (p + q - 1) + (r - 1)(-1) = p + q - 1 - r + 1 = p + q - r$.
Thus,the $r^{th}$ term is $p + q - r$.

(A) The given series is $3 \cdot 8 + 6 \cdot 11 + 9 \cdot 14 + 12 \cdot 17 + \dots$
The first factors of the terms are $3, 6, 9, 12, \dots$,which form an arithmetic progression with the first term $a = 3$ and common difference $d = 3$. The ${n^{th}}$ term is $a_n = 3 + (n - 1)3 = 3n$.
The second factors of the terms are $8, 11, 14, 17, \dots$,which form an arithmetic progression with the first term $a = 8$ and common difference $d = 3$. The ${n^{th}}$ term is $b_n = 8 + (n - 1)3 = 3n + 5$.
Therefore,the ${n^{th}}$ term of the given series is $T_n = a_n \cdot b_n = 3n(3n + 5)$.

(B) Let $S_2$ be the sum of integers divisible by $2$,$S_5$ be the sum of integers divisible by $5$,and $S_{10}$ be the sum of integers divisible by both $2$ and $5$ (i.e.,divisible by $10$).
$S_2 = 2 + 4 + \dots + 100 = \frac{50}{2}(2 + 100) = 25 \times 102 = 2550$.
$S_5 = 5 + 10 + \dots + 100 = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050$.
$S_{10} = 10 + 20 + \dots + 100 = \frac{10}{2}(10 + 100) = 5 \times 110 = 550$.
By the Principle of Inclusion-Exclusion,the required sum is $S_2 + S_5 - S_{10} = 2550 + 1050 - 550 = 3050$.

(C) For the first series $63 + 65 + 67 + 69 + \dots$,the first term $a_1 = 63$ and common difference $d_1 = 2$. The $m^{th}$ term is $T_m = a_1 + (m-1)d_1 = 63 + (m-1)2 = 63 + 2m - 2 = 2m + 61$.
For the second series $3 + 10 + 17 + 24 + \dots$,the first term $a_2 = 3$ and common difference $d_2 = 7$. The $m^{th}$ term 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:
$2m + 61 = 7m - 4$
$61 + 4 = 7m - 2m$
$65 = 5m$
$m = 13$.

(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 $n$-th term is $a_n = n\sqrt{2}$.
The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} k\sqrt{2} = \sqrt{2} \sum_{k=1}^{n} k$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.
For $n = 24$,the sum is $S_{24} = \sqrt{2} \times \frac{24 \times 25}{2}$.
$S_{24} = \sqrt{2} \times 12 \times 25 = 300\sqrt{2}$.

(C) Given that $2x, x + 8, 3x + 1$ are in $A.P.$
For three terms $a, b, c$ to be in $A.P.$,the middle term must satisfy $b = \frac{a + c}{2}$.
Therefore,$x + 8 = \frac{(2x) + (3x + 1)}{2}$.
$x + 8 = \frac{5x + 1}{2}$.
Multiplying both sides by $2$,we get $2(x + 8) = 5x + 1$.
$2x + 16 = 5x + 1$.
$16 - 1 = 5x - 2x$.
$15 = 3x$.
$x = \frac{15}{3} = 5$.

(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 $T_n = S_n - S_{n-1}$.
For $n=1$,$T_1 = S_1 = A(1) + (1)^2B = A + B$.
For $n=2$,$S_2 = A(2) + (2)^2B = 2A + 4B$.
Thus,$T_2 = S_2 - S_1 = (2A + 4B) - (A + B) = A + 3B$.
The common difference $d$ is given by $d = T_2 - T_1$.
$d = (A + 3B) - (A + B) = 2B$.
Therefore,the common difference is $2B$.

(B) The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$.
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 $T_{20} = a + 19d$.
$T_{20} = 3 + 19(4) = 3 + 76 = 79$.

(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 $\frac{(a - c)^2}{(b^2 - ac)}$:
$\frac{(a - c)^2}{(\frac{a + c}{2})^2 - ac} = \frac{(a - c)^2}{\frac{a^2 + c^2 + 2ac}{4} - ac}$
$= \frac{(a - c)^2}{\frac{a^2 + c^2 + 2ac - 4ac}{4}} = \frac{4(a - c)^2}{a^2 + c^2 - 2ac}$
$= \frac{4(a - c)^2}{(a - c)^2} = 4$.
Alternatively,let $a = 1, b = 2, c = 3$. These are in $A.P.$
The expression becomes $\frac{(1 - 3)^2}{(2^2 - 1 \times 3)} = \frac{(-2)^2}{4 - 3} = \frac{4}{1} = 4$.

(D) Given that $\log_3 2, \log_3(2^x - 5)$ and $\log_3(2^x - \frac{7}{2})$ are in $A.P.$
By the property of $A.P.$,$2b = a + c$,so:
$2 \log_3(2^x - 5) = \log_3 2 + \log_3(2^x - \frac{7}{2})$
Using the property $\log a + \log b = \log(ab)$:
$\log_3(2^x - 5)^2 = \log_3(2(2^x - \frac{7}{2}))$
$(2^x - 5)^2 = 2^{x+1} - 7$
Let $2^x = y$. Then $(y - 5)^2 = 2y - 7$
$y^2 - 10y + 25 = 2y - 7$
$y^2 - 12y + 32 = 0$
$(y - 8)(y - 4) = 0$
So,$y = 8$ or $y = 4$.
If $2^x = 8$,then $x = 3$.
If $2^x = 4$,then $x = 2$.
Check the domain of the logarithmic terms:
For $x = 2$,$\log_3(2^2 - 5) = \log_3(-1)$,which is undefined.
For $x = 3$,$\log_3(2^3 - 5) = \log_3(3) = 1$ and $\log_3(2^3 - 3.5) = \log_3(4.5)$,which are defined.
Thus,$x = 3$ is the only solution.

(C) Let the first term be $A$ and the common difference be $D$ for the arithmetic progression.
The $p^{th}$ term is $A + (p - 1)D = a$ $(i)$
The $q^{th}$ term is $A + (q - 1)D = b$ $(ii)$
The $r^{th}$ term is $A + (r - 1)D = c$ $(iii)$
Subtracting $(ii)$ from $(i)$,$(i)$ from $(iii)$,and $(iii)$ from $(ii)$ gives:
$a - b = (p - q)D$
$b - c = (q - r)D$
$c - a = (r - p)D$
Now,consider the expression $E = a(q - r) + b(r - p) + c(p - q)$.
Substituting the values of $(q - r)$,$(r - p)$,and $(p - q)$ from the differences above:
$E = a\left(\frac{b - c}{D}\right) + b\left(\frac{c - a}{D}\right) + c\left(\frac{a - b}{D}\right)$
$E = \frac{1}{D} [a(b - c) + b(c - a) + c(a - b)]$
$E = \frac{1}{D} [ab - ac + bc - ab + ca - bc]$
$E = \frac{1}{D} [0] = 0$.

(A) Let the $n^{th}$ terms of the two $A.P.$'s be $a_n = 3n + 8$ and $b_n = 7n + 15$.
For the $12^{th}$ term,substitute $n = 12$ into both expressions.
$a_{12} = 3(12) + 8 = 36 + 8 = 44$.
$b_{12} = 7(12) + 15 = 84 + 15 = 99$.
The ratio of their $12^{th}$ terms is $\frac{a_{12}}{b_{12}} = \frac{44}{99}$.
Dividing both numerator and denominator by $11$,we get $\frac{44 \div 11}{99 \div 11} = \frac{4}{9}$.

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

(A) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is $T_n = a + (n - 1)d$.
Given: $p \cdot T_p = q \cdot T_q$
$p\{a + (p - 1)d\} = q\{a + (q - 1)d\}$
$ap + p(p - 1)d = aq + q(q - 1)d$
$a(p - q) + d\{p^2 - p - q^2 + q\} = 0$
$a(p - q) + d\{(p^2 - q^2) - (p - q)\} = 0$
$a(p - q) + d\{(p - q)(p + q) - (p - q)\} = 0$
Since $p \neq q$,we can divide by $(p - q)$:
$a + d(p + q - 1) = 0$
This expression represents the $(p + q)^{th}$ term,$T_{p+q} = a + (p + q - 1)d$.
Therefore,$T_{p+q} = 0$.

(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 sums of $n$ terms: $\frac{S_{n_1}}{S_{n_2}} = \frac{2n + 3}{6n + 5}$.
Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$,we have:
$\frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{2n + 3}{6n + 5}$
$\frac{a_1 + \frac{n - 1}{2}d_1}{a_2 + \frac{n - 1}{2}d_2} = \frac{2n + 3}{6n + 5}$.
To find the ratio of the $13^{th}$ terms,we need the expression to be in the form $\frac{a_1 + 12d_1}{a_2 + 12d_2}$.
Setting $\frac{n - 1}{2} = 12$,we get $n - 1 = 24$,so $n = 25$.
Substituting $n = 25$ into 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 be $a$ and the common difference be $d$.
Given $T_m = a + (m - 1)d = \frac{1}{n}$ and $T_n = a + (n - 1)d = \frac{1}{m}$.
Subtracting the two equations: $(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$.
Thus,$d = \frac{1}{mn}$.
Substituting $d$ into the first equation: $a + (m - 1)\frac{1}{mn} = \frac{1}{n} \implies a + \frac{1}{n} - \frac{1}{mn} = \frac{1}{n} \implies a = \frac{1}{mn}$.
Now,$T_{mn} = a + (mn - 1)d = \frac{1}{mn} + (mn - 1)\frac{1}{mn} = \frac{1 + mn - 1}{mn} = \frac{mn}{mn} = 1$.

(D) Let the common difference of the $A.P.$ be $k$.
Then,$b = a + k$,$c = a + 2k$,$d = a + 3k$,and $e = a + 4k$.
Substitute these values into the expression $a + b + 4c - 4d + e$:
$= a + (a + k) + 4(a + 2k) - 4(a + 3k) + (a + 4k)$
$= a + a + k + 4a + 8k - 4a - 12k + a + 4k$
$= (a + a + 4a - 4a + a) + (k + 8k - 12k + 4k)$
$= 3a + k$
Since the result depends on the common difference $k$,it is not possible to express the value solely in terms of $a$.

(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.
Given $\frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2a' + (n - 1)d']} = \frac{7n + 1}{4n + 27}$.
This simplifies to $\frac{a + \frac{n-1}{2}d}{a' + \frac{n-1}{2}d'} = \frac{7n + 1}{4n + 27}$.
To find the ratio of the $11^{th}$ terms,we need the expression $\frac{a + 10d}{a' + 10d'}$.
Comparing $\frac{n-1}{2} = 10$,we get $n-1 = 20$,so $n = 21$.
Substituting $n = 21$ in the ratio:
$\frac{T_{11}}{T'_{11}} = \frac{7(21) + 1}{4(21) + 27} = \frac{147 + 1}{84 + 27} = \frac{148}{111}$.
Dividing both by $37$,we get $\frac{148 \div 37}{111 \div 37} = \frac{4}{3}$.
Thus,the ratio is $4:3$.

(D) The given series is $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}, \dots$
Here,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 $S_n = \frac{n}{2} [2a + (n-1)d]$.
Substituting the values:
$S_9 = \frac{9}{2} [2(\frac{1}{2}) + (9-1)(-\frac{1}{6})]$
$S_9 = \frac{9}{2} [1 + 8(-\frac{1}{6})]$
$S_9 = \frac{9}{2} [1 - \frac{4}{3}]$
$S_9 = \frac{9}{2} [-\frac{1}{3}] = -\frac{3}{2}$.

(C) Let the number of sides of the polygon be $n$.
The sum of the interior angles of a polygon with $n$ sides is $(n - 2) \times 180^o$.
Since the angles are in $A.P.$ with first term $a = 120^o$ and common difference $d = 5^o$,the sum of the angles is given by $\frac{n}{2}[2a + (n - 1)d]$.
Equating the two expressions:
$\frac{n}{2}[2(120) + (n - 1)5] = (n - 2)180$
$n[240 + 5n - 5] = 360(n - 2)$
$5n^2 + 235n = 360n - 720$
$5n^2 - 125n + 720 = 0$
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 $T_{16} = a + 15d = 120^o + 15(5^o) = 120^o + 75^o = 195^o$.
Since an interior angle of a convex polygon must be less than $180^o$,$n = 16$ 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 $T_p = a + (p - 1)d = \frac{1}{q}$ $(i)$
And the $q^{th}$ term $T_q = a + (q - 1)d = \frac{1}{p}$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$(p - q)d = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq}$
Thus,$d = \frac{1}{pq}$.
Substituting $d$ in $(i)$:
$a + (p - 1)\frac{1}{pq} = \frac{1}{q} \implies a = \frac{1}{q} - \frac{p - 1}{pq} = \frac{p - p + 1}{pq} = \frac{1}{pq}$.
The sum of $pq$ terms $S_{pq} = \frac{pq}{2} [2a + (pq - 1)d]$.
$S_{pq} = \frac{pq}{2} [\frac{2}{pq} + (pq - 1)\frac{1}{pq}] = \frac{pq}{2} [\frac{2 + pq - 1}{pq}] = \frac{pq + 1}{2}$.

(D) The sequence of 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 sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
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 $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values:
$S_{40} = \frac{40}{2}[2(2) + (40 - 1)4]$
$S_{40} = 20[4 + 39 \times 4]$
$S_{40} = 20[4 + 156]$
$S_{40} = 20 \times 160 = 3200$.

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

(C) The sum of the first $n$ even numbers is given by the arithmetic progression $2, 4, 6, \dots, 2n$.
Sum $S_{E} = \frac{n}{2}(2 + 2n) = n(n + 1)$.
The sum of the first $n$ odd numbers is given by the arithmetic progression $1, 3, 5, \dots, (2n - 1)$.
Sum $S_{O} = \frac{n}{2}(1 + 2n - 1) = \frac{n}{2}(2n) = n^2$.
The ratio of the sums is $\frac{S_{E}}{S_{O}} = \frac{n(n + 1)}{n^2} = \frac{n + 1}{n}$.
Thus,the ratio is $(n + 1):n$.

(A) Given that $a_1, a_2, a_3, ......., a_n$ are in $A.P.$ with common difference $d = a_{i+1} - a_i$.
We know that $a_n = a_1 + (n - 1)d$,which implies $d = \frac{a_n - a_1}{n - 1}$.
Rationalizing each term of the sum:
$\frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} = \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{a_{i+1} - a_i} = \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{d}$.
Summing these terms from $i = 1$ to $n-1$:
$\sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} = \frac{1}{d} ((\sqrt{a_2} - \sqrt{a_1}) + (\sqrt{a_3} - \sqrt{a_2}) + ....... + (\sqrt{a_n} - \sqrt{a_{n-1}}))$.
This is a telescoping sum,which simplifies to:
$\frac{1}{d} (\sqrt{a_n} - \sqrt{a_1})$.
Substituting $d = \frac{a_n - a_1}{n - 1}$:
$= \frac{n - 1}{a_n - a_1} (\sqrt{a_n} - \sqrt{a_1}) = \frac{n - 1}{(\sqrt{a_n} - \sqrt{a_1})(\sqrt{a_n} + \sqrt{a_1})} (\sqrt{a_n} - \sqrt{a_1}) = \frac{n - 1}{\sqrt{a_n} + \sqrt{a_1}}$.

(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 $n$ 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 $100$ which 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 $(AP)$ with first term $a = 1$ and 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 $a = 1$ and $d = 2$:
$S_n = \frac{n}{2} \{2(1) + (n - 1)2\}$
$S_n = \frac{n}{2} \{2 + 2n - 2\}$
$S_n = \frac{n}{2} \{2n\}$
$S_n = n^2$.

(A) The given series is an Arithmetic Progression $(AP)$ with first term $a = 54$ and common difference $d = 51 - 54 = -3$.
Let the number of terms be $n$. The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2} \{2a + (n - 1)d\}$.
Given $S_n = 513$,we have:
$513 = \frac{n}{2} \{2(54) + (n - 1)(-3)\}$
$1026 = n(108 - 3n + 3)$
$1026 = n(111 - 3n)$
$1026 = 111n - 3n^2$
$3n^2 - 111n + 1026 = 0$
Dividing by $3$:
$n^2 - 37n + 342 = 0$
Factoring the quadratic equation:
$(n - 18)(n - 19) = 0$
Thus,$n = 18$ or $n = 19$.
Checking $n = 19$: The $19^{th}$ term is $a_{19} = 54 + (19 - 1)(-3) = 54 - 54 = 0$.
Since the sum of $18$ terms and $19$ terms are both $513$,and $18$ is the first value,$n = 18$ is the standard solution.

(A) Given that $S_n = 2n^2 + 5n$.
The $n^{th}$ term $T_n$ is given by the formula $T_n = S_n - S_{n-1}$ for $n > 1$.
$T_n = (2n^2 + 5n) - [2(n-1)^2 + 5(n-1)]$
$T_n = 2n^2 + 5n - [2(n^2 - 2n + 1) + 5n - 5]$
$T_n = 2n^2 + 5n - [2n^2 - 4n + 2 + 5n - 5]$
$T_n = 2n^2 + 5n - [2n^2 + n - 3]$
$T_n = 2n^2 + 5n - 2n^2 - n + 3$
$T_n = 4n + 3$.
Thus,the $n^{th}$ term is $4n + 3$.

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

(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 $A.P.$ 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 $10$.

(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 all positive terms of the series,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.
$S_{31} = \frac{31}{2} [2a + (31 - 1)d] = \frac{31}{2} [2(20) + 30(-\frac{2}{3})]$
$S_{31} = \frac{31}{2} [40 - 20] = \frac{31}{2} \times 20 = 310$.

(A) The numbers between $100$ and $1000$ divisible by $9$ form an arithmetic progression $(A.P.)$.
The first term $a = 108$ and the last term $l = 999$.
The common difference $d = 9$.
To find the number of terms $n$,we use the formula $l = a + (n - 1)d$:
$999 = 108 + (n - 1)9$
$891 = (n - 1)9$
$n - 1 = 99$
$n = 100$.
The sum $S_n$ is given by $S_n = \frac{n}{2}(a + l)$:
$S_{100} = \frac{100}{2}(108 + 999)$
$S_{100} = 50 \times 1107 = 55350$.

(C) Given that $\frac{S_m}{S_n} = \frac{m^2}{n^2}$.
We know that $S_m = \frac{m}{2}[2a + (m-1)d]$.
So,$\frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{m^2}{n^2}$.
$\Rightarrow \frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n}$.
Cross-multiplying,we get $n[2a + (m-1)d] = m[2a + (n-1)d]$.
$2an + n(m-1)d = 2am + m(n-1)d$.
$2an - 2am = m(n-1)d - n(m-1)d$.
$2a(n-m) = d[mn - m - mn + n] = d(n-m)$.
Thus,$d = 2a$.
The ratio of the $m^{th}$ 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 $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} = \log a + \log \left( {\frac{{{a^2}}}{b}} \right) + \log \left( {\frac{{{a^3}}}{{{b^2}}}} \right) + \dots + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)$.
This is an arithmetic progression $(A.P.)$ where the first term $A = \log a$ and the last term $L = \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}(A + L)$.
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 where the first term $a = x + 1$,the common difference $d = 3$,and the last term $l = x + 28$.
Let $n$ be the number of terms. 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 arithmetic progression is $S_n = \frac{n}{2}(a + l)$.
Substituting the values: $\frac{10}{2}[(x + 1) + (x + 28)] = 155$.
$5(2x + 29) = 155$.
$2x + 29 = 31$.
$2x = 2 \Rightarrow x = 1$.

(C) The two-digit numbers that leave a remainder of $1$ when divided by $4$ are of the form $4k + 1$.
The smallest two-digit number of this form is $13$ $(4 \times 3 + 1)$ and the largest is $97$ $(4 \times 24 + 1)$.
This forms an arithmetic progression $(AP)$ with first term $a = 13$,last term $l = 97$,and common difference $d = 4$.
Using the formula for the $n$-th term: $l = a + (n - 1)d$
$97 = 13 + (n - 1)4$
$84 = (n - 1)4$
$n - 1 = 21$
$n = 22$
The sum of the $AP$ 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 evaluate $S_{2n} - S_n$:
$S_{2n} - S_n = \frac{2n}{2}\{2a + (2n - 1)d\} - \frac{n}{2}\{2a + (n - 1)d\}$
$= n\{2a + 2nd - d\} - \frac{n}{2}\{2a + nd - d\}$
$= \frac{n}{2}\{4a + 4nd - 2d - 2a - nd + d\}$
$= \frac{n}{2}\{2a + 3nd - d\} = \frac{n}{2}\{2a + (3n - 1)d\}$
Now,$S_{3n} = \frac{3n}{2}\{2a + (3n - 1)d\}$.
Therefore,$S_{2n} - S_n = \frac{1}{3} \times \frac{3n}{2}\{2a + (3n - 1)d\} = \frac{1}{3}S_{3n}$.

(A) The sum of the first $k$ terms of an arithmetic progression is given by $S_k = \frac{k}{2} \{2a + (k - 1)d\}$.
Therefore,the ratio $\frac{S_{kn}}{S_n}$ is given by:
$\frac{S_{kn}}{S_n} = \frac{\frac{kn}{2} \{2a + (kn - 1)d\}}{\frac{n}{2} \{2a + (n - 1)d\}}$
Simplifying the expression:
$= k \left\{ \frac{2a + knd - d}{2a + nd - d} \right\} = k \left\{ \frac{(2a - d) + knd}{(2a - d) + nd} \right\}$
For this expression to be independent of $n$,the term $(2a - d)$ must be equal to $0$.
If $2a - d = 0$,then the expression becomes:
$= k \left\{ \frac{knd}{nd} \right\} = k^2$
Since $k^2$ is independent of $n$,the condition is $2a - d = 0$.