Arithmetic progression Questions in English

(A) Given the recursive formula:
$a_{1} = 1$
$a_{n} = a_{n-1} + 2$ for $n \ge 2$
Calculating the terms:
$a_{1} = 1$
$a_{2} = a_{1} + 2 = 1 + 2 = 3$
$a_{3} = a_{2} + 2 = 3 + 2 = 5$
$a_{4} = a_{3} + 2 = 5 + 2 = 7$
$a_{5} = a_{4} + 2 = 7 + 2 = 9$
The first five terms are $1, 3, 5, 7, 9$.
The corresponding series is $1 + 3 + 5 + 7 + 9 + \ldots$

(A) Substituting $n = 1, 2, 3, 4, 5$ in the formula $a_{n} = \frac{2n - 3}{6}$,we get:
$a_{1} = \frac{2(1) - 3}{6} = \frac{-1}{6}$
$a_{2} = \frac{2(2) - 3}{6} = \frac{1}{6}$
$a_{3} = \frac{2(3) - 3}{6} = \frac{3}{6} = \frac{1}{2}$
$a_{4} = \frac{2(4) - 3}{6} = \frac{5}{6}$
$a_{5} = \frac{2(5) - 3}{6} = \frac{7}{6}$
Thus,the first five terms are $-\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}$.

(A) Given the $n^{\text{th}}$ term formula: $a_{n} = 4n - 3$.
To find the $17^{\text{th}}$ term,substitute $n = 17$:
$a_{17} = 4(17) - 3 = 68 - 3 = 65$.
To find the $24^{\text{th}}$ term,substitute $n = 24$:
$a_{24} = 4(24) - 3 = 96 - 3 = 93$.
Thus,the $17^{\text{th}}$ and $24^{\text{th}}$ terms are $65$ and $93$ respectively.

(A) Let the first term be $a$ and the common difference be $d$.
We are given:
$a_m = a + (m - 1)d = n$ --- $(1)$
$a_n = a + (n - 1)d = m$ --- $(2)$
Subtracting $(2)$ from $(1)$:
$(a + (m - 1)d) - (a + (n - 1)d) = n - m$
$(m - 1 - n + 1)d = n - m$
$(m - n)d = -(m - n)$
Since $m \neq n$,we can divide by $(m - n)$:
$d = -1$
Substituting $d = -1$ into $(1)$:
$a + (m - 1)(-1) = n$
$a - m + 1 = n$
$a = n + m - 1$
Now,the $p^{\text{th}}$ term is given by:
$a_p = a + (p - 1)d$
$a_p = (n + m - 1) + (p - 1)(-1)$
$a_p = n + m - 1 - p + 1$
$a_p = n + m - p$

(B) Given the sum of $n$ terms of an $A.P.$ is $S_n = nP + \frac{1}{2}n(n-1)Q$.
For $n=1$,$S_1 = a_1 = P(1) + \frac{1}{2}(1)(0)Q = P$.
For $n=2$,$S_2 = a_1 + a_2 = P(2) + \frac{1}{2}(2)(1)Q = 2P + Q$.
Since $a_2 = S_2 - S_1$,we have $a_2 = (2P + Q) - P = P + Q$.
The common difference $d$ is given by $d = a_2 - a_1$.
Substituting the values,$d = (P + Q) - P = Q$.

(A) Let $a_1, a_2$ and $d_1, d_2$ be the first terms and common differences of the first and second arithmetic progressions,respectively.
Given the ratio of the sum of $n$ terms:
$\frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{3n + 8}{7n + 15}$
$\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n + 8}{7n + 15}$ --- $(1)$
We need the ratio of the $12^{\text{th}}$ terms,which is $\frac{a_1 + 11d_1}{a_2 + 11d_2}$.
To match the form in equation $(1)$,we multiply the numerator and denominator by $2$:
$\frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{a_1 + 11d_1}{a_2 + 11d_2}$
Comparing $(n-1) = 22$,we get $n = 23$.
Substituting $n = 23$ in equation $(1)$:
$\frac{a_1 + 11d_1}{a_2 + 11d_2} = \frac{3(23) + 8}{7(23) + 15} = \frac{69 + 8}{161 + 15} = \frac{77}{176}$
Dividing both by $11$,we get $\frac{7}{16}$.
Thus,the required ratio is $7 : 16$.

(A) The income follows an arithmetic progression $(A.P.)$ where the first term $a = 3,00,000$,the common difference $d = 10,000$,and the number of years $n = 20$.
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:
$S_{20} = \frac{20}{2} [2(3,00,000) + (20 - 1)(10,000)]$
$S_{20} = 10 [6,00,000 + 19(10,000)]$
$S_{20} = 10 [6,00,000 + 1,90,000]$
$S_{20} = 10 [7,90,000] = 79,00,000$.
Thus,the total amount received in $20$ years is $Rs. \,79,00,000$.

(A) Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ be six numbers between $3$ and $24$ such that $3, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, 24$ are in $A.P.$
Here,the first term $a = 3$,the last term $l = 24$,and the total number of terms $n = 6 + 2 = 8$.
The formula for the $n^{th}$ term is $a_{n} = a + (n - 1)d$.
Substituting the values: $24 = 3 + (8 - 1)d$.
$24 - 3 = 7d \implies 21 = 7d \implies d = 3$.
The numbers are:
$A_{1} = a + d = 3 + 3 = 6$
$A_{2} = a + 2d = 3 + 6 = 9$
$A_{3} = a + 3d = 3 + 9 = 12$
$A_{4} = a + 4d = 3 + 12 = 15$
$A_{5} = a + 5d = 3 + 15 = 18$
$A_{6} = a + 6d = 3 + 18 = 21$
Thus,the six numbers are $6, 9, 12, 15, 18, 21$.

(A) The odd integers from $1$ to $2001$ are $1, 3, 5, \dots, 2001$.
This sequence forms an Arithmetic Progression $(A.P.)$.
Here,the first term $a = 1$ and the common difference $d = 2$.
Let $n$ be the number of terms. The $n^{th}$ term is given by $a + (n - 1)d = 2001$.
$1 + (n - 1)(2) = 2001$
$2(n - 1) = 2000$
$n - 1 = 1000$
$n = 1001$
The sum of $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[a + l]$,where $l$ is the last term.
$S_{1001} = \frac{1001}{2}[1 + 2001]$
$S_{1001} = \frac{1001}{2} \times 2002$
$S_{1001} = 1001 \times 1001 = 1002001$.
Thus,the sum of odd integers from $1$ to $2001$ is $1002001$.

(A) The natural numbers lying between $100$ and $1000$ which are multiples of $5$ are $105, 110, \dots, 995.$
This forms an arithmetic progression where the first term $a = 105,$ common difference $d = 5,$ and the last term $l = 995.$
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d.$
$995 = 105 + (n - 1)5.$
$890 = (n - 1)5.$
$n - 1 = 178 \Rightarrow n = 179.$
Now,the sum $S_n = \frac{n}{2}(a + l).$
$S_{179} = \frac{179}{2}(105 + 995).$
$S_{179} = \frac{179}{2}(1100).$
$S_{179} = 179 \times 550 = 98450.$
Thus,the sum is $98450.$

(A) Let the first term be $a = 2$ 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 five terms is $S_5 = \frac{5}{2}[2(2) + (5-1)d] = \frac{5}{2}[4 + 4d] = 5(2 + d) = 10 + 5d$.
The sum of the next five terms is the sum of the first ten terms minus the sum of the first five terms: $S_{10} - S_5$.
$S_{10} = \frac{10}{2}[2(2) + (10-1)d] = 5[4 + 9d] = 20 + 45d$.
Sum of the next five terms $= (20 + 45d) - (10 + 5d) = 10 + 40d$.
According to the given condition,$S_5 = \frac{1}{4}(\text{sum of next five terms})$.
$10 + 5d = \frac{1}{4}(10 + 40d)$.
$40 + 20d = 10 + 40d$.
$30 = 20d \Rightarrow d = 1.5$.
Wait,re-evaluating the sum of terms: $S_5 = 2 + (2+d) + (2+2d) + (2+3d) + (2+4d) = 10 + 10d$.
Sum of next five terms $= (2+5d) + (2+6d) + (2+7d) + (2+8d) + (2+9d) = 10 + 35d$.
Given $10 + 10d = \frac{1}{4}(10 + 35d)$.
$40 + 40d = 10 + 35d$.
$5d = -30 \Rightarrow d = -6$.
The $20^{th}$ term is $a_{20} = a + (20-1)d = 2 + 19(-6) = 2 - 114 = -112$.
Thus,the $20^{th}$ term is $-112$.

(C) Let the sum of $n$ terms of the given $A.P.$ be $S_n = -25$.
The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.
Here,the first term $a = -6$ and the common difference $d = -\frac{11}{2} - (-6) = -\frac{11}{2} + 6 = \frac{1}{2}$.
Substituting the values into the formula:
$-25 = \frac{n}{2}[2(-6) + (n-1)(\frac{1}{2})]$
$-50 = n[-12 + \frac{n}{2} - \frac{1}{2}]$
$-50 = n[\frac{n-25}{2}]$
$-100 = n^2 - 25n$
$n^2 - 25n + 100 = 0$
Factoring the quadratic equation:
$n^2 - 20n - 5n + 100 = 0$
$n(n-20) - 5(n-20) = 0$
$(n-20)(n-5) = 0$
Thus,$n = 20$ or $n = 5$.
Both values are positive integers,so both are valid.

The general term of an $A.P.$ is $a_n = a + (n-1)d$.
According to the given information:
$a_p = a + (p-1)d = \frac{1}{q}$ $(1)$
$a_q = a + (q-1)d = \frac{1}{p}$ $(2)$
Subtracting $(2)$ from $(1)$:
$(p-1)d - (q-1)d = \frac{1}{q} - \frac{1}{p}$
$(p-q)d = \frac{p-q}{pq}$
Since $p \neq q$,we have $d = \frac{1}{pq}$.
Substituting $d$ into $(1)$:
$a + (p-1)\frac{1}{pq} = \frac{1}{q}$
$a = \frac{1}{q} - \frac{p-1}{pq} = \frac{p - (p-1)}{pq} = \frac{1}{pq}$.
The sum of the first $pq$ terms is $S_{pq} = \frac{pq}{2}[2a + (pq-1)d]$.
$S_{pq} = \frac{pq}{2}[2(\frac{1}{pq}) + (pq-1)(\frac{1}{pq})]$
$S_{pq} = \frac{pq}{2}[\frac{2 + pq - 1}{pq}]$
$S_{pq} = \frac{pq}{2}[\frac{pq+1}{pq}] = \frac{1}{2}(pq+1)$.

(A) Let the sum of $n$ terms of the given $A.P.$ be $116$.
$S_{n} = \frac{n}{2}[2a + (n - 1)d]$
Here,$a = 25$ and $d = 22 - 25 = -3$.
$\therefore 116 = \frac{n}{2}[2 \times 25 + (n - 1)(-3)]$
$\Rightarrow 232 = n(50 - 3n + 3)$
$\Rightarrow 232 = 53n - 3n^{2}$
$\Rightarrow 3n^{2} - 53n + 232 = 0$
$\Rightarrow 3n^{2} - 24n - 29n + 232 = 0$
$\Rightarrow 3n(n - 8) - 29(n - 8) = 0$
$\Rightarrow (n - 8)(3n - 29) = 0$
Since $n$ must be a positive integer,$n = 8$.
$\therefore a_{8} = a + (8 - 1)d = 25 + 7(-3) = 25 - 21 = 4$.
Thus,the last term of the $A.P.$ is $4$.

(A) The $k^{\text{th}}$ term of the $A.P.$ is given by $a_k = 5k+1$.
To find the first term $a_1$,substitute $k=1$:
$a_1 = 5(1)+1 = 6$.
To find the second term $a_2$,substitute $k=2$:
$a_2 = 5(2)+1 = 11$.
The common difference $d$ is $a_2 - a_1 = 11 - 6 = 5$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting $a=6$ and $d=5$:
$S_n = \frac{n}{2}[2(6) + (n-1)5]$
$S_n = \frac{n}{2}[12 + 5n - 5]$
$S_n = \frac{n}{2}[5n + 7]$.

(D) The sum of $n$ terms of an $A.P.$ is given by $S_n = pn + qn^2$.
The first term $a_1$ is $S_1 = p(1) + q(1)^2 = p + q$.
The sum of the first two terms $S_2$ is $p(2) + q(2)^2 = 2p + 4q$.
The second term $a_2$ is $S_2 - S_1 = (2p + 4q) - (p + q) = p + 3q$.
The common difference $d$ is $a_2 - a_1 = (p + 3q) - (p + q) = 2q$.
Alternatively,comparing $S_n = \frac{d}{2}n^2 + (a - \frac{d}{2})n$ with $S_n = qn^2 + pn$,we get $\frac{d}{2} = q$,which implies $d = 2q$.

(A) Let $a_1, a_2$ and $d_1, d_2$ be the first terms and common differences of the first and second arithmetic progressions,respectively.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
According to the given condition:
$\frac{S_{n,1}}{S_{n,2}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{5n+4}{9n+6}$
$\Rightarrow \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n+4}{9n+6}$
We need to find the ratio of the $18^{th}$ terms,which is $\frac{a_1 + 17d_1}{a_2 + 17d_2}$.
To get the coefficient $17$ for $d_1$ and $d_2$,we set $\frac{n-1}{2} = 17$,which gives $n-1 = 34$,so $n = 35$.
Substituting $n = 35$ in the ratio:
$\frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{5(35)+4}{9(35)+6}$
$\Rightarrow \frac{2(a_1 + 17d_1)}{2(a_2 + 17d_2)} = \frac{175+4}{315+6}$
$\Rightarrow \frac{a_1 + 17d_1}{a_2 + 17d_2} = \frac{179}{321}$
Thus,the ratio of their $18^{th}$ terms is $179 : 321$.

(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given that $S_p = S_q$,we have:
$\frac{p}{2}[2a + (p-1)d] = \frac{q}{2}[2a + (q-1)d]$
Multiplying both sides by $2$:
$p[2a + (p-1)d] = q[2a + (q-1)d]$
$2ap + p(p-1)d = 2aq + q(q-1)d$
Rearranging the terms:
$2a(p-q) + [p(p-1) - q(q-1)]d = 0$
$2a(p-q) + [p^2 - p - q^2 + q]d = 0$
$2a(p-q) + [(p^2 - q^2) - (p-q)]d = 0$
$2a(p-q) + [(p-q)(p+q) - (p-q)]d = 0$
Dividing by $(p-q)$ (assuming $p \neq q$):
$2a + (p+q-1)d = 0$
Now,the sum of the first $(p+q)$ terms is:
$S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$
Substituting $2a + (p+q-1)d = 0$:
$S_{p+q} = \frac{p+q}{2} \times 0 = 0$
Thus,the sum of the first $(p+q)$ terms is $0$.

Let $A$ be the first term and $D$ be the common difference of the $A.P.$
According to the given information:
$S_{p} = \frac{p}{2}[2A + (p-1)D] = a \Rightarrow \frac{a}{p} = A + \frac{(p-1)D}{2} \dots (1)$
$S_{q} = \frac{q}{2}[2A + (q-1)D] = b \Rightarrow \frac{b}{q} = A + \frac{(q-1)D}{2} \dots (2)$
$S_{r} = \frac{r}{2}[2A + (r-1)D] = c \Rightarrow \frac{c}{r} = A + \frac{(r-1)D}{2} \dots (3)$
Now,consider the expression $\frac{a}{p}(q-r) + \frac{b}{q}(r-p) + \frac{c}{r}(p-q)$.
Substituting the values from $(1), (2),$ and $(3)$:
$= [A + \frac{(p-1)D}{2}](q-r) + [A + \frac{(q-1)D}{2}](r-p) + [A + \frac{(r-1)D}{2}](p-q)$
$= A(q-r+r-p+p-q) + \frac{D}{2}[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]$
$= A(0) + \frac{D}{2}[pq - pr - q + r + qr - qp - r + p + rp - rq - p + q]$
$= 0 + \frac{D}{2}[0] = 0$.
Thus,the result is proved.

(N/A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
Given that the ratio of the sum of $m$ terms to the sum of $n$ terms is $\frac{m^2}{n^2}$.
$\frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{m^2}{n^2}$
$\frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n}$
We need to find the ratio of the $m^{th}$ term to the $n^{th}$ term,which is $\frac{a + (m-1)d}{a + (n-1)d}$.
To get this form,we divide the numerator and denominator of the left side of the equation by $2$:
$\frac{a + \frac{(m-1)}{2}d}{a + \frac{(n-1)}{2}d} = \frac{m}{n}$
Comparing the terms,we substitute $m-1 = 2(m'-1)$ where $m'$ is the term index. Alternatively,replace $m$ with $(2m-1)$ and $n$ with $(2n-1)$ in the ratio $\frac{2a+(m-1)d}{2a+(n-1)d} = \frac{m}{n}$:
$\frac{2a + (2m-1-1)d}{2a + (2n-1-1)d} = \frac{2m-1}{2n-1}$
$\frac{2a + (2m-2)d}{2a + (2n-2)d} = \frac{2m-1}{2n-1}$
$\frac{2[a + (m-1)d]}{2[a + (n-1)d]} = \frac{2m-1}{2n-1}$
$\frac{a + (m-1)d}{a + (n-1)d} = \frac{2m-1}{2n-1}$
Thus,the ratio of the $m^{th}$ term to the $n^{th}$ term is $(2m-1):(2n-1)$.

(C) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
Given the sum of $n$ terms $S_{n} = 3n^{2} + 5n$.
The $n^{\text{th}}$ term $a_{n}$ is given by $S_{n} - S_{n-1}$.
$a_{n} = (3n^{2} + 5n) - [3(n-1)^{2} + 5(n-1)]$
$a_{n} = 3n^{2} + 5n - [3(n^{2} - 2n + 1) + 5n - 5]$
$a_{n} = 3n^{2} + 5n - [3n^{2} - 6n + 3 + 5n - 5]$
$a_{n} = 3n^{2} + 5n - 3n^{2} + n + 2 = 6n + 2$.
Given the $m^{\text{th}}$ term $a_{m} = 164$.
$6m + 2 = 164$
$6m = 162$
$m = \frac{162}{6} = 27$.
Thus,the value of $m$ is $27$.

(A) Let $A_{1}, A_{2}, A_{3}, A_{4},$ and $A_{5}$ be five numbers between $8$ and $26$ such that $8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26$ is an $A.P.$
Here,the first term $a = 8$ and the last term $l = 26$. The total number of terms $n = 5 + 2 = 7$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_{n} = a + (n - 1)d$.
Substituting the values: $26 = 8 + (7 - 1)d$.
$26 = 8 + 6d$ $\Rightarrow 6d = 18$ $\Rightarrow d = 3$.
The five numbers are:
$A_{1} = a + d = 8 + 3 = 11$
$A_{2} = a + 2d = 8 + 6 = 14$
$A_{3} = a + 3d = 8 + 9 = 17$
$A_{4} = a + 4d = 8 + 12 = 20$
$A_{5} = a + 5d = 8 + 15 = 23$
Thus,the required five numbers are $11, 14, 17, 20, 23$.

(B) The $A.M.$ of $a$ and $b$ is given by $\frac{a+b}{2}$.
According to the given condition:
$\frac{a+b}{2} = \frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$
Cross-multiplying,we get:
$(a+b)(a^{n-1}+b^{n-1}) = 2(a^{n}+b^{n})$
Expanding the left side:
$a^{n} + ab^{n-1} + ba^{n-1} + b^{n} = 2a^{n} + 2b^{n}$
Rearranging the terms:
$ab^{n-1} + a^{n-1}b = a^{n} + b^{n}$
Grouping terms:
$ab^{n-1} - b^{n} = a^{n} - a^{n-1}b$
$b^{n-1}(a-b) = a^{n-1}(a-b)$
Assuming $a \neq b$,we divide by $(a-b)$:
$b^{n-1} = a^{n-1}$
$\left(\frac{a}{b}\right)^{n-1} = 1 = \left(\frac{a}{b}\right)^{0}$
Equating the exponents:
$n-1 = 0$
$n = 1$

(B) Let $A_{1}, A_{2}, \ldots, A_{m}$ be $m$ numbers such that $1, A_{1}, A_{2}, \ldots, A_{m}, 31$ is an $A.P.$
Here,$a=1$,$b=31$,and the total number of terms is $n=m+2$.
The common difference $d$ is given by $b = a + (n-1)d$,so $31 = 1 + (m+2-1)d$.
$30 = (m+1)d \Rightarrow d = \frac{30}{m+1}$ (Equation $1$).
The $k^{\text{th}}$ inserted number is $A_{k} = a + kd = 1 + kd$.
Given the ratio of the $7^{\text{th}}$ and $(m-1)^{\text{th}}$ inserted numbers is $5:9$:
$\frac{1+7d}{1+(m-1)d} = \frac{5}{9}$.
Substituting $d = \frac{30}{m+1}$:
$\frac{1+7(\frac{30}{m+1})}{1+(m-1)(\frac{30}{m+1})} = \frac{5}{9}$.
$\frac{m+1+210}{m+1+30m-30} = \frac{5}{9}$.
$\frac{m+211}{31m-29} = \frac{5}{9}$.
$9(m+211) = 5(31m-29)$.
$9m + 1899 = 155m - 145$.
$146m = 2044$.
$m = \frac{2044}{146} = 14$.
Thus,the value of $m$ is $14$.

(A) The first installment of the loan is $a = 100$.
The increase in the installment every month is the common difference $d = 5$.
The sequence of installments forms an Arithmetic Progression $(A.P.)$: $100, 105, 110, \dots$
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
For the $30^{th}$ installment,$n = 30$:
$a_{30} = 100 + (30 - 1) \times 5$
$a_{30} = 100 + 29 \times 5$
$a_{30} = 100 + 145$
$a_{30} = 245$
Thus,the amount to be paid in the $30^{th}$ installment is $Rs. 245$.

(A) The angles of the polygon form an $A.P.$ with the first term $a = 120^{\circ}$ and common difference $d = 5^{\circ}$.
The sum of all interior angles of a polygon with $n$ sides is given by $S_{n} = 180^{\circ}(n-2)$.
Using the sum formula for an $A.P.$,$S_{n} = \frac{n}{2}[2a + (n-1)d]$.
Equating the two,we get $\frac{n}{2}[2(120^{\circ}) + (n-1)5^{\circ}] = 180^{\circ}(n-2)$.
Multiplying by $2$,$n[240 + 5n - 5] = 360(n-2)$.
$n(5n + 235) = 360n - 720$.
$5n^{2} + 235n = 360n - 720$.
$5n^{2} - 125n + 720 = 0$.
Dividing by $5$,$n^{2} - 25n + 144 = 0$.
Factoring the quadratic equation,$(n-9)(n-16) = 0$.
Thus,$n = 9$ or $n = 16$.
However,for $n = 16$,the largest angle is $a + (n-1)d = 120^{\circ} + 15(5^{\circ}) = 120^{\circ} + 75^{\circ} = 195^{\circ}$. Since an interior angle of a convex polygon must be less than $180^{\circ}$,$n = 16$ is rejected.
Therefore,the number of sides is $n = 9$.

Let $a$ and $d$ be the first term and the common difference of the $A.P.$ respectively.
It is known that the $k^{th}$ term of an $A.P.$ is given by $a_{k} = a + (k - 1)d$.
Therefore,$a_{m+n} = a + (m + n - 1)d$ and $a_{m-n} = a + (m - n - 1)d$.
Also,the $m^{th}$ term is $a_{m} = a + (m - 1)d$.
Now,the sum of the $(m+n)^{th}$ and $(m-n)^{th}$ terms is:
$a_{m+n} + a_{m-n} = [a + (m + n - 1)d] + [a + (m - n - 1)d]$
$= 2a + (m + n - 1 + m - n - 1)d$
$= 2a + (2m - 2)d$
$= 2a + 2(m - 1)d$
$= 2[a + (m - 1)d]$
$= 2a_{m}$
Thus,the sum of the $(m+n)^{th}$ and $(m-n)^{th}$ terms of an $A.P.$ is equal to twice the $m^{th}$ term.

(A) Let the three numbers in $A.P.$ be $(a-d), a,$ and $(a+d).$
According to the given information:
$(a-d) + a + (a+d) = 24$
$3a = 24$
$a = 8$
Also,the product is given as:
$(a-d) \times a \times (a+d) = 440$
$(8-d) \times 8 \times (8+d) = 440$
$(8-d)(8+d) = \frac{440}{8}$
$64 - d^2 = 55$
$d^2 = 64 - 55 = 9$
$d = \pm 3$
If $d = 3,$ the numbers are $(8-3), 8, (8+3),$ which are $5, 8, 11.$
If $d = -3,$ the numbers are $(8-(-3)), 8, (8+(-3)),$ which are $11, 8, 5.$
Thus,the numbers are $5, 8, 11.$

Let $a$ be the first term and $d$ be the common difference of the $A.P.$
$S_{1} = \frac{n}{2}[2a + (n - 1)d]$ $(1)$
$S_{2} = \frac{2n}{2}[2a + (2n - 1)d] = n[2a + (2n - 1)d]$ $(2)$
$S_{3} = \frac{3n}{2}[2a + (3n - 1)d]$ $(3)$
Now,calculate $S_{2} - S_{1}$:
$S_{2} - S_{1} = n[2a + (2n - 1)d] - \frac{n}{2}[2a + (n - 1)d]$
$= \frac{n}{2} [2(2a + 2nd - d) - (2a + nd - d)]$
$= \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d]$
$= \frac{n}{2} [2a + 3nd - d] = \frac{n}{2} [2a + (3n - 1)d]$
Therefore,$3(S_{2} - S_{1}) = 3 \times \frac{n}{2} [2a + (3n - 1)d] = \frac{3n}{2} [2a + (3n - 1)d] = S_{3}$.
Hence,the result is proved.

(A) The numbers between $200$ and $400$ that are divisible by $7$ form an arithmetic progression: $203, 210, 217, \dots, 399.$
Here,the first term $a = 203$ and the last term $l = 399.$
The common difference $d = 7.$
Using the formula for the $n^{th}$ term of an $A.P.$,$a_n = a + (n - 1)d:$
$399 = 203 + (n - 1)7$
$196 = (n - 1)7$
$n - 1 = 28$
$n = 29.$
The sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l):$
$S_{29} = \frac{29}{2}(203 + 399)$
$S_{29} = \frac{29}{2}(602)$
$S_{29} = 29 \times 301 = 8729.$
Thus,the required sum is $8729.$

(A) Let $S_2$ be the sum of integers divisible by $2$ and $S_5$ be the sum of integers divisible by $5.$
The integers divisible by $2$ are $2, 4, 6, \ldots, 100.$ This is an $A.P.$ with $a=2, d=2, n=50.$
$S_2 = \frac{50}{2}(2 + 100) = 25 \times 102 = 2550.$
The integers divisible by $5$ are $5, 10, 15, \ldots, 100.$ This is an $A.P.$ with $a=5, d=5, n=20.$
$S_5 = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050.$
The integers divisible by both $2$ and $5$ (i.e.,divisible by $10$) are $10, 20, \ldots, 100.$ This is an $A.P.$ with $a=10, d=10, n=10.$
$S_{10} = \frac{10}{2}(10 + 100) = 5 \times 110 = 550.$
By the Principle of Inclusion-Exclusion,the required sum is $S_2 + S_5 - S_{10}.$
Required sum $= 2550 + 1050 - 550 = 3050.$

(A) The two-digit numbers,which when divided by $4$ yield $1$ as a remainder,are $13, 17, \ldots, 97$.
This series forms an $A.P.$ with the first term $a = 13$ and common difference $d = 4$.
Let $n$ be the number of terms in this $A.P.$
The $n^{th}$ term formula is $a_{n} = a + (n - 1)d$.
Substituting the values,we get $97 = 13 + (n - 1)(4)$.
$84 = (n - 1)(4) \implies n - 1 = 21 \implies n = 22$.
The sum of $n$ terms of an $A.P.$ is $S_{n} = \frac{n}{2}[a + a_{n}]$.
$S_{22} = \frac{22}{2}[13 + 97] = 11 \times 110 = 1210$.
Thus,the required sum is $1210$.

(A) Let the $A.P.$ be $a, a+d, a+2d, \ldots, a+(n-1)d$.
The sum of the first four terms is $a + (a+d) + (a+2d) + (a+3d) = 4a + 6d = 56$.
Given $a = 11$,we have $4(11) + 6d = 56$ $\Rightarrow 44 + 6d = 56$ $\Rightarrow 6d = 12$ $\Rightarrow d = 2$.
The sum of the last four terms is $[a+(n-4)d] + [a+(n-3)d] + [a+(n-2)d] + [a+(n-1)d] = 4a + (4n - 10)d = 112$.
Substituting $a = 11$ and $d = 2$:
$4(11) + (4n - 10)(2) = 112$
$44 + 8n - 20 = 112$
$8n + 24 = 112$
$8n = 88$
$n = 11$.
Thus,the number of terms is $11$.

(N/A) Let $A$ be the first term and $D$ be the common difference of the $A.P.$
The $n^{\text{th}}$ term of an $A.P.$ is given by $a_n = A + (n-1)D$.
Therefore,
$a_p = A + (p-1)D = a$ $(1)$
$a_q = A + (q-1)D = b$ $(2)$
$a_r = A + (r-1)D = c$ $(3)$
Subtracting $(2)$ from $(1)$:
$(p-q)D = a-b \Rightarrow D = \frac{a-b}{p-q}$ $(4)$
Subtracting $(3)$ from $(2)$:
$(q-r)D = b-c \Rightarrow D = \frac{b-c}{q-r}$ $(5)$
Equating $(4)$ and $(5)$:
$\frac{a-b}{p-q} = \frac{b-c}{q-r}$
Cross-multiplying:
$(a-b)(q-r) = (b-c)(p-q)$
$aq - ar - bq + br = bp - bq - cp + cq$
Rearranging the terms to one side:
$aq - ar - bq + br - bp + bq + cp - cq = 0$
$a(q-r) + b(r-p) + c(p-q) = 0$
Thus,the result is proved.

(A) Given that $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in $A.P.$
$\therefore b\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)$
$\Rightarrow \frac{b(a+c)}{ac}-\frac{a(b+c)}{bc}=\frac{c(a+b)}{ab}-\frac{b(a+c)}{ac}$
$\Rightarrow \frac{b^2a+b^2c-a^2b-a^2c}{abc}=\frac{c^2a+c^2b-b^2a-b^2c}{abc}$
$\Rightarrow b^2a-a^2b+b^2c-a^2c=c^2a-b^2a+c^2b-b^2c$
$\Rightarrow ab(b-a)+c(b^2-a^2)=a(c^2-b^2)+bc(c-b)$
$\Rightarrow ab(b-a)+c(b-a)(b+a)=a(c-b)(c+b)+bc(c-b)$
$\Rightarrow (b-a)(ab+cb+ca)=(c-b)(ac+ab+bc)$
Assuming $ab+bc+ca \neq 0$,we get $b-a=c-b$.
Thus,$2b=a+c$,which implies $a, b, c$ are in $A.P.$

(A) The farmer pays $Rs. 6000$ in cash.
Therefore,the unpaid amount is $Rs. 12000 - Rs. 6000 = Rs. 6000$.
The farmer pays the balance in $12$ annual instalments of $Rs. 500$ each,plus $12\%$ interest on the remaining unpaid amount.
The unpaid amounts at the end of each year are $6000, 5500, 5000, \dots, 500$.
The total interest paid is $12\%$ of $(6000 + 5500 + 5000 + \dots + 500)$.
This is an Arithmetic Progression $(A.P.)$ where the first term $a = 500$,the last term $l = 6000$,and the number of terms $n = 12$.
The sum of the $A.P.$ is $S_n = \frac{n}{2}(a + l) = \frac{12}{2}(500 + 6000) = 6(6500) = 39000$.
Total interest $= 12\% \text{ of } 39000 = \frac{12}{100} \times 39000 = 12 \times 390 = Rs. 4680$.
Total cost of the tractor $= \text{Cash paid} + \text{Balance} + \text{Total interest} = 6000 + 6000 + 4680 = Rs. 16680$.

(A) Shamshad Ali buys a scooter for $Rs. 22000$ and pays $Rs. 4000$ in cash.
$\therefore$ Unpaid amount $= Rs. 22000 - Rs. 4000 = Rs. 18000$.
He pays $Rs. 1000$ annually plus $10\%$ interest on the remaining balance.
The interest paid each year is:
$10\%$ of $18000, 10\%$ of $17000, 10\%$ of $16000, \dots, 10\%$ of $1000$.
Total interest $= 10\% \times (18000 + 17000 + 16000 + \dots + 1000)$.
This is an $A.P.$ where $a = 1000$,$l = 18000$,and $d = 1000$.
Number of terms $n = 18$.
Sum $= \frac{n}{2}(a + l) = \frac{18}{2}(1000 + 18000) = 9 \times 19000 = 171000$.
Total interest $= 10\% \text{ of } 171000 = Rs. 17100$.
Total cost of the scooter $= \text{Principal} + \text{Total Interest} = 22000 + 17100 = Rs. 39100$.

(A) The principal amount $P = Rs. 10000$ and the annual simple interest $I = \frac{5}{100} \times 10000 = Rs. 500$.
The amount in the $n^{\text{th}}$ year is given by $A_n = P + (n-1)I$.
For the $15^{\text{th}}$ year,$n = 15$:
$A_{15} = 10000 + (15-1) \times 500 = 10000 + 14 \times 500 = 10000 + 7000 = Rs. 17000$.
The total amount after $20$ years is the amount at the end of the $20^{\text{th}}$ year:
$A_{20} = P + 20 \times I = 10000 + 20 \times 500 = 10000 + 10000 = Rs. 20000$.

(A) Let $x$ be the number of days in which $150$ workers would finish the work.
The total work is $150x$.
According to the given information,the work is completed in $(x+8)$ days with a decreasing number of workers:
$150x = 150 + 146 + 142 + \dots$ up to $(x+8)$ terms.
The series $150 + 146 + 142 + \dots$ is an $A.P.$ with first term $a = 150$,common difference $d = -4$,and number of terms $n = x+8$.
The sum of $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[2a + (n-1)d]$.
$150x = \frac{x+8}{2}[2(150) + (x+8-1)(-4)]$
$150x = \frac{x+8}{2}[300 + (x+7)(-4)]$
$150x = (x+8)[150 + (x+7)(-2)]$
$150x = (x+8)(150 - 2x - 14)$
$150x = (x+8)(136 - 2x)$
$75x = (x+8)(68 - x)$
$75x = 68x - x^2 + 544 - 8x$
$x^2 + 75x - 60x - 544 = 0$
$x^2 + 15x - 544 = 0$
$(x + 32)(x - 17) = 0$
Since $x > 0$,we have $x = 17$.
The total number of days taken is $x + 8 = 17 + 8 = 25$ days.

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

(C) The given series is an Arithmetic Progression $(AP)$ with first term $a = 20$ and 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 $n^{\text{th}}$ term $T_n = a + (n - 1)d = 20 + (60)(-\frac{2}{5}) = 20 - 24 = -4$.
If $n = 40$,the $n^{\text{th}}$ term $T_n = 20 + (39)(-\frac{2}{5}) = 20 - 15.6 = 4.4$ (which is not negative).
Since the $n^{\text{th}}$ term must be negative,we take $n = 61$,and the $n^{\text{th}}$ term is $-4$.

(D) Let the first term be $a = 3$ and the common difference be $d$.
Given that the sum of the first $25$ terms is equal to the sum of the next $15$ terms.
Let $S_n$ denote the sum of the first $n$ terms.
The sum of the first $25$ terms is $S_{25}$.
The sum of the next $15$ terms is $S_{40} - S_{25}$.
According to the problem,$S_{25} = S_{40} - S_{25}$,which implies $2S_{25} = S_{40}$.
Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$:
$2 \times \frac{25}{2}[2(3) + (25-1)d] = \frac{40}{2}[2(3) + (40-1)d]$
$25[6 + 24d] = 20[6 + 39d]$
Divide both sides by $5$:
$5[6 + 24d] = 4[6 + 39d]$
$30 + 120d = 24 + 156d$
$30 - 24 = 156d - 120d$
$6 = 36d$
$d = \frac{6}{36} = \frac{1}{6}$.

(C) The formula for the $n^{th}$ term of an $A.P.$ is $a_{n} = a_{1} + (n-1)d$.
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 factors of $299$ are $1, 13, 23, 299$.
For $n-1$ to be in the range $[14, 49]$,the only possible value is $n-1 = 23$,which gives $n = 24$.
Then $d = 13$.
We need to find $(S_{n-4}, a_{n-4})$. Since $n = 24$,$n-4 = 20$.
$a_{20} = a_{1} + 19d = 1 + 19(13) = 1 + 247 = 248$.
$S_{20} = \frac{20}{2}(a_{1} + a_{20}) = 10(1 + 248) = 10(249) = 2490$.
Thus,the ordered pair is $(2490, 248)$.

(A) Let the terms be $a_1 = 3^{2 \sin 2 \alpha - 1}$,$a_2 = 14$,and $a_3 = 3^{4 - 2 \sin 2 \alpha}$.
Since they are in $A.P.$,$2a_2 = a_1 + a_3$.
$2(14) = 3^{2 \sin 2 \alpha - 1} + 3^{4 - 2 \sin 2 \alpha} = 28$.
Let $x = 3^{2 \sin 2 \alpha}$. Then the equation becomes $\frac{x}{3} + \frac{81}{x} = 28$.
$x^2 - 84x + 243 = 0$.
$(x - 81)(x - 3) = 0$,so $x = 81$ or $x = 3$.
If $x = 3$,$3^{2 \sin 2 \alpha} = 3^1 \implies 2 \sin 2 \alpha = 1 \implies \sin 2 \alpha = 0.5$.
Then $a_1 = 3^{1-1} = 1$ and $a_2 = 14$. The common difference $d = 14 - 1 = 13$.
The $6^{th}$ term $T_6 = a_1 + 5d = 1 + 5(13) = 1 + 65 = 66$.
If $x = 81$,$3^{2 \sin 2 \alpha} = 3^4 \implies 2 \sin 2 \alpha = 4 \implies \sin 2 \alpha = 2$,which is impossible.

(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 $A.P.$,we have $a_{40} = a + 39d = -159$ and $a_{100} = a + 99d = -399$.
Subtracting these equations: $(a + 99d) - (a + 39d) = -399 - (-159)$ $\Rightarrow 60d = -240$ $\Rightarrow d = -4$.
Substituting $d = -4$ into $a + 39d = -159$: $a + 39(-4) = -159$ $\Rightarrow a - 156 = -159$ $\Rightarrow a = -3$.
Now,$a_{70} = a + 69d = -3 + 69(-4) = -3 - 276 = -279$.
Given $b_{100} = a_{70}$,we have $b_{100} = -279$.
Using the formula for the $n^{th}$ term of the second $A.P.$: $b_{100} = b_{1} + 99(d + 2) = -279$.
Substituting $d = -4$: $b_{1} + 99(-4 + 2) = -279 \Rightarrow b_{1} + 99(-2) = -279$.
$b_{1} - 198 = -279 \Rightarrow b_{1} = -279 + 198 = -81$.

(D) Let the first term be $a$ and the common difference be $d$.
$S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$
$S_{4n} = \frac{4n}{2}[2a + (4n-1)d] = 2n[2a + (4n-1)d]$
Given $S_{2} - S_{1} = 1000$,where $S_{1} = S_{2n}$ and $S_{2} = S_{4n}$:
$2n[2a + (4n-1)d] - n[2a + (2n-1)d] = 1000$
$n[4a + 2(4n-1)d - 2a - (2n-1)d] = 1000$
$n[2a + (8n - 2 - 2n + 1)d] = 1000$
$n[2a + (6n - 1)d] = 1000$
$2a + (6n - 1)d = \frac{1000}{n}$
Now,the sum of the first $6n$ terms is $S_{6n} = \frac{6n}{2}[2a + (6n-1)d]$
$S_{6n} = 3n \times \frac{1000}{n} = 3000$

(C) The sum of the first $n$ terms of an $A.P.$ is given by $S_{n} = \frac{n}{2}(2a_{1} + (n-1)d)$.
Given $\frac{S_{10}}{S_{p}} = \frac{100}{p^{2}}$,we have $\frac{\frac{10}{2}(2a_{1} + 9d)}{\frac{p}{2}(2a_{1} + (p-1)d)} = \frac{100}{p^{2}}$.
Simplifying,$\frac{5(2a_{1} + 9d)}{\frac{p}{2}(2a_{1} + (p-1)d)} = \frac{100}{p^{2}} \implies \frac{2a_{1} + 9d}{2a_{1} + (p-1)d} = \frac{10}{p}$.
Cross-multiplying gives $p(2a_{1} + 9d) = 10(2a_{1} + (p-1)d)$.
$2a_{1}p + 9dp = 20a_{1} + 10dp - 10d$.
$10d - 20a_{1} = 10dp - 9dp - 2a_{1}p = dp - 2a_{1}p = p(d - 2a_{1})$.
Alternatively,$2a_{1}(p - 10) = d(10p - 9p - 10) = d(p - 10)$.
Since $p \neq 10$,we divide by $(p - 10)$ to get $2a_{1} = d$,or $\frac{a_{1}}{d} = \frac{1}{2}$.
We need to find $\frac{a_{11}}{a_{10}} = \frac{a_{1} + 10d}{a_{1} + 9d}$.
Substituting $d = 2a_{1}$,we get $\frac{a_{1} + 10(2a_{1})}{a_{1} + 9(2a_{1})} = \frac{a_{1} + 20a_{1}}{a_{1} + 18a_{1}} = \frac{21a_{1}}{19a_{1}} = \frac{21}{19}$.

(B) Given $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}} = \frac{4}{9}$.
Since $a_{n+1} = a_{n} + d$,we have $\frac{1}{a_{n} a_{n+1}} = \frac{1}{d} \left( \frac{1}{a_{n}} - \frac{1}{a_{n+1}} \right)$.
Thus,$\frac{1}{d} \sum_{n=1}^{20} \left( \frac{1}{a_{n}} - \frac{1}{a_{n+1}} \right) = \frac{1}{d} \left( \frac{1}{a_{1}} - \frac{1}{a_{21}} \right) = \frac{4}{9}$.
$\frac{1}{d} \left( \frac{a_{21} - a_{1}}{a_{1} a_{21}} \right) = \frac{1}{d} \left( \frac{20d}{a_{1} a_{21}} \right) = \frac{20}{a_{1} a_{21}} = \frac{4}{9} \implies a_{1} a_{21} = 45$.
Sum of $21$ terms $S_{21} = \frac{21}{2} (a_{1} + a_{21}) = 189 \implies a_{1} + a_{21} = 18$.
We have $a_{1} + a_{21} = 18$ and $a_{1} a_{21} = 45$. The roots of $x^{2} - 18x + 45 = 0$ are $a_{1}, a_{21}$.
$(x - 15)(x - 3) = 0 \implies \{a_{1}, a_{21}\} = \{3, 15\}$.
Case $1$: $a_{1} = 3, a_{21} = 15 \implies 3 + 20d = 15 \implies d = 0.6$.
Case $2$: $a_{1} = 15, a_{21} = 3 \implies 15 + 20d = 3 \implies d = -0.6$.
$a_{6} a_{16} = (a_{1} + 5d)(a_{1} + 15d)$.
For Case $1$: $(3 + 5(0.6))(3 + 15(0.6)) = (3 + 3)(3 + 9) = 6 \times 12 = 72$.
For Case $2$: $(15 + 5(-0.6))(15 + 15(-0.6)) = (15 - 3)(15 - 9) = 12 \times 6 = 72$.
Thus,$a_{6} a_{16} = 72$.

(A) The given series is $\log _{9^{1/2}} x + \log _{9^{1/3}} x + \log _{9^{1/4}} x + \dots$
Using the property $\log_{a^b} x = \frac{1}{b} \log_a x$,the terms become:
$2 \log_9 x + 3 \log_9 x + 4 \log_9 x + \dots$
This is an arithmetic series with $21$ terms where the first term $a = 2 \log_9 x$ and the common difference $d = \log_9 x$.
The sum of $n$ terms is $S_n = \frac{n}{2} [2a + (n-1)d]$.
For $n = 21$,$S_{21} = \frac{21}{2} [2(2 \log_9 x) + (21-1) \log_9 x] = 504$.
$S_{21} = \frac{21}{2} [4 \log_9 x + 20 \log_9 x] = \frac{21}{2} [24 \log_9 x] = 21 \times 12 \log_9 x = 252 \log_9 x$.
Given $252 \log_9 x = 504$,we get $\log_9 x = 2$.
Therefore,$x = 9^2 = 81$.

(D) The sum of the first $n$ terms of an arithmetic progression is given by $S_{n} = \frac{n}{2} \{2a + (n-1)d\}$.
Given $S_{10} = 530$,we have $\frac{10}{2} \{2a + 9d\} = 530 \Rightarrow 2a + 9d = 106 \quad \dots(1)$.
Given $S_{5} = 140$,we have $\frac{5}{2} \{2a + 4d\} = 140 \Rightarrow 2a + 4d = 56 \quad \dots(2)$.
Subtracting equation $(2)$ from $(1)$,we get $5d = 50$,which implies $d = 10$.
Substituting $d = 10$ into equation $(2)$,$2a + 4(10) = 56$ $\Rightarrow 2a = 16$ $\Rightarrow a = 8$.
Now,$S_{20} - S_{6} = \frac{20}{2} \{2a + 19d\} - \frac{6}{2} \{2a + 5d\}$.
$= 10(2(8) + 19(10)) - 3(2(8) + 5(10))$.
$= 10(16 + 190) - 3(16 + 50)$.
$= 10(206) - 3(66) = 2060 - 198 = 1862$.