Progression and Sequence Questions in English

(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
The $n^{th}$ term is given by $a_n = a + (n - 1)d$.
Given $a_{12} = -13$,we have $a + 11d = -13$ --- $(1)$.
The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Given $S_4 = 24$,we have $\frac{4}{2}[2a + 3d] = 24$,which simplifies to $2a + 3d = 12$ --- $(2)$.
Multiply equation $(1)$ by $2$ to get $2a + 22d = -26$ --- $(3)$.
Subtract equation $(2)$ from equation $(3)$:
$(2a + 22d) - (2a + 3d) = -26 - 12$
$19d = -38$,so $d = -2$.
Substitute $d = -2$ into equation $(1)$:
$a + 11(-2) = -13$
$a - 22 = -13$,so $a = 9$.
Now,calculate the sum of the first $10$ terms $(S_{10})$:
$S_{10} = \frac{10}{2}[2(9) + (10 - 1)(-2)]$
$S_{10} = 5[18 + 9(-2)]$
$S_{10} = 5[18 - 18] = 5(0) = 0$.

(A) Let $n$ be the number of terms,$a$ be the first term,$l$ be the last term,and $d$ be the common difference.
Given: $S_n = 525$,$a = 3$,$l = 39$.
The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$.
Substituting the values: $525 = \frac{n}{2}(3 + 39)$.
$525 = \frac{n}{2}(42) \Rightarrow 525 = 21n$.
$n = \frac{525}{21} = 25$.
Now,use the formula for the $n^{th}$ term: $l = a + (n - 1)d$.
$39 = 3 + (25 - 1)d$.
$39 - 3 = 24d$.
$36 = 24d$.
$d = \frac{36}{24} = \frac{3}{2}$.

(B) Let the first term be $a = 100$ 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 six terms is $S_6 = \frac{6}{2}[2(100) + 5d] = 3(200 + 5d) = 600 + 15d$.
The sum of the next six terms is the sum of the first $12$ terms minus the sum of the first $6$ terms: $S_{12} - S_6$.
$S_{12} = \frac{12}{2}[2(100) + 11d] = 6(200 + 11d) = 1200 + 66d$.
Sum of the next six terms $= (1200 + 66d) - (600 + 15d) = 600 + 51d$.
According to the problem,$S_6 = 5 \times (\text{sum of next six terms})$.
$600 + 15d = 5(600 + 51d)$.
$600 + 15d = 3000 + 255d$.
$15d - 255d = 3000 - 600$.
$-240d = 2400$.
$d = \frac{2400}{-240} = -10$.

(C) Given: First term $a = -14$ and fifth term $a_5 = 2$.
Let $d$ be the common difference of the $A.P.$
Using the formula $a_n = a + (n-1)d$,we have $a_5 = a + 4d = 2$.
Substituting $a = -14$: $-14 + 4d = 2 \Rightarrow 4d = 16 \Rightarrow d = 4$.
Let $n$ be the number of terms such that the sum $S_n = 40$.
Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:
$40 = \frac{n}{2}[2(-14) + (n-1)4]$
$80 = n[-28 + 4n - 4]$
$80 = n[4n - 32]$
$80 = 4n^2 - 32n$
Dividing by $4$: $n^2 - 8n - 20 = 0$.
Factoring the quadratic equation: $(n - 10)(n + 2) = 0$.
Thus,$n = 10$ or $n = -2$.
Since the number of terms $n$ cannot be negative,$n = 10$.

(A) The given series is an arithmetic progression where the first term $a = 51$,the common difference $d = -1$,and the last term $a_n = 21$.
Using the formula for the $n^{th}$ term of an arithmetic progression: $a_n = a + (n - 1)d$.
Substituting the values: $21 = 51 + (n - 1)(-1)$.
$21 = 51 - n + 1$.
$21 = 52 - n$.
$n = 52 - 21 = 31$.
Now,using the sum formula $S_n = \frac{n}{2}(a + a_n)$:
$S_{31} = \frac{31}{2}(51 + 21)$.
$S_{31} = \frac{31}{2}(72)$.
$S_{31} = 31 \times 36 = 1116$.

(B) Given the sum of $p$ terms is $S_p = 3p^2 + 4p$.
To find the $n^{th}$ term $(a_n)$,we use the relation $a_n = S_n - S_{n-1}$.
First,substitute $p = n$ to get $S_n = 3n^2 + 4n$.
Next,substitute $p = n-1$ to get $S_{n-1} = 3(n-1)^2 + 4(n-1)$.
Expanding $S_{n-1}$: $S_{n-1} = 3(n^2 - 2n + 1) + 4n - 4 = 3n^2 - 6n + 3 + 4n - 4 = 3n^2 - 2n - 1$.
Now,calculate $a_n = S_n - S_{n-1} = (3n^2 + 4n) - (3n^2 - 2n - 1)$.
$a_n = 3n^2 + 4n - 3n^2 + 2n + 1 = 6n + 1$.

(B) Given the $A.P.$ is $1, 4, 7, \ldots$ where the first term $a = 1$ and the common difference $d = 4 - 1 = 3$.
Let the sum of $n$ terms be $S_n = 715$.
The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values: $715 = \frac{n}{2}[2(1) + (n - 1)3]$.
$1430 = n[2 + 3n - 3]$.
$1430 = n[3n - 1]$.
$3n^2 - n - 1430 = 0$.
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{1 \pm \sqrt{(-1)^2 - 4(3)(-1430)}}{2(3)}$.
$n = \frac{1 \pm \sqrt{1 + 17160}}{6} = \frac{1 \pm \sqrt{17161}}{6} = \frac{1 \pm 131}{6}$.
Since $n$ must be a positive integer,$n = \frac{1 + 131}{6} = \frac{132}{6} = 22$.

(C) The even natural numbers divisible by $5$ are multiples of $10$,which are $10, 20, 30, 40, \ldots$
This sequence forms an Arithmetic Progression $(A.P.)$ where the first term $a = 10$ and the common difference $d = 10$.
The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 100$,we have:
$S_{100} = \frac{100}{2}[2 \times 10 + (100 - 1) \times 10]$
$S_{100} = 50[20 + 99 \times 10]$
$S_{100} = 50[20 + 990]$
$S_{100} = 50[1010]$
$S_{100} = 50500$.

(D) The first integer after $50$ that is divisible by $7$ is $56$,and the last integer before $500$ that is divisible by $7$ is $497.$
The sequence of integers between $50$ and $500$ divisible by $7$ is $56, 63, 70, \ldots, 497.$
This forms an Arithmetic Progression $(A.P.)$ where the first term $a = 56$ and the common difference $d = 7.$
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d.$
Substituting the values: $497 = 56 + (n - 1) \times 7.$
$497 - 56 = (n - 1) \times 7 \implies 441 = (n - 1) \times 7.$
$n - 1 = 441 / 7 = 63 \implies n = 64.$
The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(a + a_n).$
$S_{64} = \frac{64}{2}(56 + 497) = 32 \times 553 = 17696.$

(C) The smallest and the largest three-digit numbers divisible by $7$ are $105$ and $994$,respectively.
This forms an arithmetic progression: $105, 112, 119, \ldots, 994$.
Here,the first term $a = 105$,the common difference $d = 7$,and the last term $a_n = 994$.
Using the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.
$994 = 105 + (n - 1) \times 7$
$994 - 105 = 7(n - 1)$
$889 = 7(n - 1)$
$n - 1 = 127 \Rightarrow n = 128$.
Now,the sum of $n$ terms is given by $S_n = \frac{n}{2}(a + a_n)$.
$S_{128} = \frac{128}{2}(105 + 994)$
$S_{128} = 64 \times 1099 = 70336$.

(A) The four-digit odd numbers divisible by $9$ form an arithmetic progression ($A$.$P$.).
The smallest four-digit odd number divisible by $9$ is $1017$ (since $1008$ is even and $1017$ is odd).
The largest four-digit odd number divisible by $9$ is $9999$.
Since the numbers must be odd and divisible by $9$,the common difference $d$ between consecutive terms is $18$ (because $9 \times 2 = 18$).
Using the formula for the $n$-th term of an $A$.$P$.,$a_n = a + (n - 1)d$:
$9999 = 1017 + (n - 1) \times 18$
$8982 = (n - 1) \times 18$
$n - 1 = 8982 / 18 = 499$
$n = 500$
The sum $S_n$ of an $A$.$P$. is given by $S_n = \frac{n}{2}(a + l)$:
$S_{500} = \frac{500}{2}(1017 + 9999)$
$S_{500} = 250 \times 11016 = 2754000$.

(A) The given sequence is a geometric progression where the first term $a = \frac{1}{3}$ and the common ratio $r = \frac{1/9}{1/3} = \frac{1}{3}$.
Let the $n^{th}$ term be $a_n = \frac{1}{19683}$.
The formula for the $n^{th}$ term of a geometric sequence is $a_n = a \cdot r^{n-1}$.
Substituting the values,we get $\frac{1}{19683} = \frac{1}{3} \cdot (\frac{1}{3})^{n-1}$.
This simplifies to $\frac{1}{19683} = (\frac{1}{3})^n$.
Since $3^9 = 19683$,we have $(\frac{1}{3})^9 = \frac{1}{19683}$.
Comparing the powers,we get $n = 9$.

(A) The given geometric series is $5+25+125+\cdots$.
The first term $a = 5$ and the common ratio $r = \frac{25}{5} = 5$.
The formula for the $n^{th}$ term of a geometric progression is $a_n = a \cdot r^{n-1}$.
For the $10^{th}$ term,$n = 10$.
Substituting the values,we get $a_{10} = 5 \cdot 5^{10-1}$.
$a_{10} = 5 \cdot 5^9 = 5^{1+9} = 5^{10}$.

(B) The given sequence is a Geometric Progression $(G.P.)$ where the first term $a = 1$ and the common ratio $r = \frac{-1}{1} = -1$.
The formula for the $n^{th}$ term of a $G.P.$ is given by $a_n = a \cdot r^{n-1}$.
To find the $20^{th}$ term $(n = 20)$:
$a_{20} = 1 \cdot (-1)^{20-1}$
$a_{20} = 1 \cdot (-1)^{19}$
Since $19$ is an odd number,$(-1)^{19} = -1$.
Therefore,$a_{20} = 1 \cdot (-1) = -1$.

(C) The given series is a geometric progression $(GP)$.
Here,the first term $a = \frac{1}{4}$.
The common ratio $r = \frac{-1/2}{1/4} = -2$.
The formula for the $n^{th}$ term of a $GP$ is $a_n = a \cdot r^{n-1}$.
To find the $5^{th}$ term $(n = 5)$:
$a_5 = \frac{1}{4} \times (-2)^{5-1}$
$a_5 = \frac{1}{4} \times (-2)^4$
$a_5 = \frac{1}{4} \times 16$
$a_5 = 4$.

(B) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
The $n^{th}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.
Given that the $5^{th}$ term is $2$, we have:
$a_5 = a r^{5-1} = a r^4 = 2$ $...(1)$
We need to find the product of the first $9$ terms:
$P = a \times (a r) \times (a r^2) \times \dots \times (a r^8)$
$P = a^9 \times r^{(1 + 2 + 3 + \dots + 8)}$
The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$.
For $n=8$, the sum is $\frac{8 \times 9}{2} = 36$.
Therefore, $P = a^9 \times r^{36} = (a r^4)^9$.
Substituting the value from equation $(1)$:
$P = (2)^9 = 512$.

(C) The given progression is a geometric progression where the first term $a = 18$ and the common ratio $r = \frac{-12}{18} = \frac{-2}{3}$.
Let the $n^{th}$ term be $a_n = \frac{512}{729}$.
The formula for the $n^{th}$ term of a geometric progression is $a_n = a r^{n-1}$.
Substituting the values,we get $\frac{512}{729} = 18 \left( \frac{-2}{3} \right)^{n-1}$.
Dividing both sides by $18$,we have $\left( \frac{-2}{3} \right)^{n-1} = \frac{512}{729 \times 18} = \frac{256}{729 \times 9} = \frac{2^8}{3^6 \times 3^2} = \frac{2^8}{3^8} = \left( \frac{2}{3} \right)^8$.
Since the base is negative and the exponent is even,$\left( \frac{-2}{3} \right)^8 = \left( \frac{2}{3} \right)^8$.
Thus,$\left( \frac{-2}{3} \right)^{n-1} = \left( \frac{-2}{3} \right)^8$.
Comparing the exponents,$n - 1 = 8$,which gives $n = 9$.
Therefore,$\frac{512}{729}$ is the $9^{th}$ term of the progression.

(C) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
The $n^{th}$ term of a $G.P.$ is given by $a_n = ar^{n-1}$.
Given that the $3^{rd}$ term is the square of the first term:
$a_3 = (a_1)^2$
$ar^2 = a^2$
Since $a \neq 0$,we divide by $a$ to get $r^2 = a$ $....(1)$
Given that the $2^{nd}$ term is $8$:
$a_2 = ar = 8$ $....(2)$
Substitute $a = r^2$ from equation $(1)$ into equation $(2)$:
$(r^2)r = 8$
$r^3 = 8$
$r = 2$
Now,find $a$ using $a = r^2$:
$a = (2)^2 = 4$
We need to find the $6^{th}$ term $(a_6)$:
$a_6 = ar^5$
$a_6 = 4 \times (2)^5$
$a_6 = 4 \times 32 = 128$
Therefore,the $6^{th}$ term is $128$.

(C) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
The $n^{th}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.
Given that the $4^{th}$ term is $24$,we have:
$a r^{4-1} = 24 \Rightarrow a r^3 = 24$ $...(1)$
Given that the $8^{th}$ term is $384$,we have:
$a r^{8-1} = 384 \Rightarrow a r^7 = 384$ $...(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{a r^7}{a r^3} = \frac{384}{24}$
$r^4 = 16$
$r^4 = 2^4 \Rightarrow r = 2$ (Taking the positive real root for common ratio).
Substituting $r = 2$ into equation $(1)$:
$a(2)^3 = 24$
$a(8) = 24$
$a = \frac{24}{8} = 3$
Thus,the first term is $3$ and the common ratio is $2$.

(B) Let $r$ be the common ratio of the $G.P.$
The first term is $a = 1$.
The $n^{th}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.
Thus,the third term is $a_3 = a r^{3-1} = 1 \cdot r^2 = r^2$.
The fifth term is $a_5 = a r^{5-1} = 1 \cdot r^4 = r^4$.
Given that the sum of the third and fifth terms is $90$,we have:
$r^2 + r^4 = 90$
Rearranging the equation,we get:
$r^4 + r^2 - 90 = 0$
Let $x = r^2$. Then the equation becomes:
$x^2 + x - 90 = 0$
Factoring the quadratic equation:
$(x + 10)(x - 9) = 0$
This gives $x = -10$ or $x = 9$.
Since $x = r^2$,$r^2$ cannot be negative for real numbers. Therefore,$r^2 = 9$.
Taking the square root,we get $r = \pm 3$.

(A) If three numbers $a, b, c$ are in $G.P.,$ then $b^2 = ac.$
Here,$a = -\frac{2}{7},$ $b = x,$ and $c = -\frac{7}{2}.$
Substituting these values into the formula:
$x^2 = \left(-\frac{2}{7}\right) \times \left(-\frac{7}{2}\right)$
$x^2 = 1$
$x = \pm 1$
Therefore,the values of $x$ are $1$ and $-1.$

(C) The number of ancestors in each generation forms a Geometric Progression $(G.P.)$: $2, 4, 8, \ldots$ up to $10$ terms.
Here,the first term $a = 2$ and the common ratio $r = 2$.
The sum of the first $n$ terms of a $G.P.$ is given by the formula: $S_n = a(r^n - 1) / (r - 1)$.
For $n = 10$ generations,the total number of ancestors is:
$S_{10} = 2(2^{10} - 1) / (2 - 1)$
$S_{10} = 2(1024 - 1) / 1$
$S_{10} = 2 \times 1023 = 2046$.
Therefore,the total number of ancestors up to the $10^{th}$ generation is $2046$.

(D) Given: First term $a = 7$,last term $l = a_n = 448$,and sum $S_n = 889$.
Let $r$ be the common ratio.
The formula for the sum of a $G.P.$ is $S_n = \frac{a - lr}{1 - r}$.
Substituting the given values: $889 = \frac{7 - 448r}{1 - r}$.
Multiplying both sides by $(1 - r)$: $889(1 - r) = 7 - 448r$.
$889 - 889r = 7 - 448r$.
Rearranging the terms: $889 - 7 = 889r - 448r$.
$882 = 441r$.
$r = \frac{882}{441} = 2$.
Thus,the common ratio is $2$.

(C) Let the first term be $a$ and the common ratio be $r$.
The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Given,$\frac{S_3}{S_6} = \frac{125}{152}$.
Substituting the formula: $\frac{a(r^3 - 1) / (r - 1)}{a(r^6 - 1) / (r - 1)} = \frac{125}{152}$.
This simplifies to $\frac{r^3 - 1}{r^6 - 1} = \frac{125}{152}$.
Using the identity $r^6 - 1 = (r^3 - 1)(r^3 + 1)$,we get $\frac{r^3 - 1}{(r^3 - 1)(r^3 + 1)} = \frac{125}{152}$.
$\frac{1}{r^3 + 1} = \frac{125}{152}$.
$125(r^3 + 1) = 152$.
$125r^3 + 125 = 152$.
$125r^3 = 152 - 125 = 27$.
$r^3 = \frac{27}{125} = (\frac{3}{5})^3$.
Therefore,$r = 3/5$.

(A) The given expression is $\sum_{j=1}^{11} (2 + 3^j)$.
We can split the summation into two parts:
$\sum_{j=1}^{11} 2 + \sum_{j=1}^{11} 3^j$.
The first part is the sum of $11$ terms of $2$,which is $11 \times 2 = 22$.
The second part is a geometric progression with first term $a = 3$,common ratio $r = 3$,and number of terms $n = 11$.
The sum of a geometric progression is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values,we get $\frac{3(3^{11} - 1)}{3 - 1} = \frac{3}{2}(3^{11} - 1)$.
Adding both parts,the total sum is $22 + \frac{3}{2}(3^{11} - 1)$.

(B) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
Given: $a_1 + a_2 = 36 \Rightarrow a + ar = 36$
$\Rightarrow a(1 + r) = 36$ $...(1)$
Also,$a_1 \cdot a_3 = 9 \cdot a_2 \Rightarrow a \cdot ar^2 = 9 \cdot ar$
$\Rightarrow ar^2 = 9$ $...(2)$
From $(1)$,$a = \frac{36}{1+r}$. Substituting this into $(2)$:
$\frac{36}{1+r} \cdot r^2 = 9 \Rightarrow 4r^2 = 1 + r \Rightarrow 4r^2 - r - 1 = 0$
Solving for $r$: $r = \frac{1 \pm \sqrt{1 - 4(4)(-1)}}{8} = \frac{1 \pm \sqrt{17}}{8}$.
Wait,re-evaluating the condition: $a_1 a_3 = 9 a_2 \Rightarrow a(ar^2) = 9(ar) \Rightarrow a^2 r^2 = 9ar$. Since $a, r \neq 0$,$ar = 9$.
Substitute $ar = 9$ into $a + ar = 36$: $a + 9 = 36 \Rightarrow a = 27$.
Then $27r = 9 \Rightarrow r = \frac{1}{3}$.
The sum of the first $n$ terms is $S_n = \frac{a(1-r^n)}{1-r}$.
$S_8 = \frac{27(1 - (1/3)^8)}{1 - 1/3} = \frac{27(1 - 1/6561)}{2/3} = \frac{81}{2} \cdot \frac{6560}{6561} = \frac{3280}{81}$.

(B) The formula for the sum to infinity of a $G.P.$ is given by $S_{\infty} = \frac{a}{1-r}$,where $a$ is the first term and $r$ is the common ratio.
Given,$r = -\frac{4}{5}$ and $S_{\infty} = \frac{80}{9}$.
Substituting the values into the formula:
$\frac{80}{9} = \frac{a}{1 - (-\frac{4}{5})}$
$\frac{80}{9} = \frac{a}{1 + \frac{4}{5}}$
$\frac{80}{9} = \frac{a}{\frac{9}{5}}$
$a = \frac{80}{9} \times \frac{9}{5}$
$a = 16$.
Thus,the first term is $16$.

(C) The given series is $S = \left( \frac{3}{4} + \frac{3}{4^3} + \frac{3}{4^5} + \dots \right) - \left( \frac{5}{4^2} + \frac{5}{4^4} + \frac{5}{4^6} + \dots \right)$.
This consists of two infinite geometric series.
For the first series,the first term $a_1 = \frac{3}{4}$ and the common ratio $r_1 = \frac{1}{4^2} = \frac{1}{16}$.
The sum $S_1 = \frac{a_1}{1 - r_1} = \frac{3/4}{1 - 1/16} = \frac{3/4}{15/16} = \frac{3}{4} \times \frac{16}{15} = \frac{4}{5}$.
For the second series,the first term $a_2 = \frac{5}{4^2} = \frac{5}{16}$ and the common ratio $r_2 = \frac{1}{4^2} = \frac{1}{16}$.
The sum $S_2 = \frac{a_2}{1 - r_2} = \frac{5/16}{1 - 1/16} = \frac{5/16}{15/16} = \frac{5}{16} \times \frac{16}{15} = \frac{1}{3}$.
Therefore,the total sum $S = S_1 - S_2 = \frac{4}{5} - \frac{1}{3} = \frac{12 - 5}{15} = \frac{7}{15}$.

(B) The given expression is a product of powers with the same base $32$.
Using the property $a^m \cdot a^n = a^{m+n}$, we can write the product as:
$(32)^{1 + 1/6 + 1/36 + \ldots + \infty} = (32)^x$
where $x = 1 + 1/6 + 1/36 + \ldots + \infty$.
This is an infinite geometric series with the first term $a = 1$ and common ratio $r = 1/6$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
Substituting the values, $x = \frac{1}{1 - 1/6} = \frac{1}{5/6} = 6/5$.
Therefore, the product is $(32)^{6/5}$.
Since $32 = 2^5$, we have $(2^5)^{6/5} = 2^{5 \cdot (6/5)} = 2^6$.
$2^6 = 64$.

(B) The given sequence is $6, 4, 3, \ldots$ which is a $H.P.$
The sequence of reciprocals of its terms is $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \ldots$ which is an $A.P.$
Here,the first term $a = \frac{1}{6}$ and the common difference $d = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
For the $9^{th}$ term of the $A.P.$:
$a_9 = \frac{1}{6} + (9-1) \times \frac{1}{12}$
$a_9 = \frac{1}{6} + 8 \times \frac{1}{12} = \frac{1}{6} + \frac{8}{12} = \frac{1}{6} + \frac{2}{3} = \frac{1+4}{6} = \frac{5}{6}$.
Since the $H.P.$ is the reciprocal of the $A.P.$,the $9^{th}$ term of the $H.P.$ is the reciprocal of $a_9$.
Therefore,the $9^{th}$ term of the $H.P.$ is $\frac{6}{5}$.

(A) The first term of the $H.P.$ is $6$ and the second term is $3$.
Therefore,the first and second terms of the corresponding $A.P.$ are $\frac{1}{6}$ and $\frac{1}{3}$.
For this $A.P.$,the first term $a = \frac{1}{6}$ and the common difference $d = \frac{1}{3} - \frac{1}{6} = \frac{2-1}{6} = \frac{1}{6}$.
The $n^{th}$ term of the $A.P.$ is given by $a_n = a + (n-1)d$.
$a_n = \frac{1}{6} + (n-1)\frac{1}{6} = \frac{1 + n - 1}{6} = \frac{n}{6}$.
Since the $n^{th}$ term of an $H.P.$ is the reciprocal of the $n^{th}$ term of the corresponding $A.P.$,the $n^{th}$ term of the $H.P.$ is $\frac{6}{n}$.

(B) Since $x, y, z$ are in $G.P.$,we have $y^2 = xz$.
Taking the logarithm on both sides,we get $2 \log y = \log x + \log z$.
Adding $2$ to both sides,we get $2 + 2 \log y = 2 + \log x + \log z$.
This can be rewritten as $2(1 + \log y) = (1 + \log x) + (1 + \log z)$.
This implies that $(1 + \log x), (1 + \log y), (1 + \log z)$ are in $A.P.$
Since the reciprocals of terms in $A.P.$ are in $H.P.$,it follows that $\frac{1}{1 + \log x}, \frac{1}{1 + \log y}, \frac{1}{1 + \log z}$ are in $H.P.$

(C) The given series is $\frac{2}{5} + \frac{3}{5^{2}} + \frac{2}{5^{3}} + \frac{3}{5^{4}} + \dots \infty$.
We can split this into two separate infinite geometric series:
$S = \left( \frac{2}{5} + \frac{2}{5^{3}} + \frac{2}{5^{5}} + \dots \right) + \left( \frac{3}{5^{2}} + \frac{3}{5^{4}} + \frac{3}{5^{6}} + \dots \right)$.
For the first series,the first term $a_1 = \frac{2}{5}$ and the common ratio $r = \frac{1}{5^{2}} = \frac{1}{25}$.
The sum is $S_1 = \frac{a_1}{1 - r} = \frac{2/5}{1 - 1/25} = \frac{2/5}{24/25} = \frac{2}{5} \times \frac{25}{24} = \frac{5}{12}$.
For the second series,the first term $a_2 = \frac{3}{5^{2}} = \frac{3}{25}$ and the common ratio $r = \frac{1}{25}$.
The sum is $S_2 = \frac{a_2}{1 - r} = \frac{3/25}{1 - 1/25} = \frac{3/25}{24/25} = \frac{3}{24} = \frac{1}{8}$.
Total sum $S = S_1 + S_2 = \frac{5}{12} + \frac{1}{8} = \frac{10 + 3}{24} = \frac{13}{24}$.

(C) Given that the first term $a = 729$ and the $7^{th}$ term $a_{7} = 64$.
Let $r$ be the common ratio of the $G.P.$
The formula for the $n^{th}$ term is $a_{n} = a r^{n-1}$.
For $n = 7$,$a_{7} = a r^{6} = 64$.
Substituting $a = 729$,we get $729 r^{6} = 64$.
$r^{6} = \frac{64}{729} = \left(\frac{2}{3}\right)^{6}$.
Thus,$r = \frac{2}{3}$.
The sum of the first $n$ terms of a $G.P.$ is given by $S_{n} = \frac{a(1 - r^{n})}{1 - r}$.
For $n = 7$,$S_{7} = \frac{729(1 - (2/3)^{7})}{1 - 2/3}$.
$S_{7} = \frac{729(1 - 128/2187)}{1/3} = 3 \times 729 \times \left(\frac{2187 - 128}{2187}\right)$.
$S_{7} = 2187 \times \frac{2059}{2187} = 2059$.

(A) Let $r$ be the common ratio of the $G.P.$
The $n^{th}$ term is given by $l = ar^{n-1} \dots (1)$
The first $n$ terms of the $G.P.$ are $a, ar, ar^2, \dots, ar^{n-1}$.
The product $P$ is given by $P = a \times (ar) \times (ar^2) \times \dots \times (ar^{n-1})$.
$P = a^n \times r^{1+2+3+\dots+(n-1)}$.
Using the sum of the first $(n-1)$ natural numbers,$\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}$.
$P = a^n \times r^{\frac{n(n-1)}{2}} = (a^2 \times r^{n-1})^{n/2}$.
Since $l = ar^{n-1}$,we can write $a^2 r^{n-1} = a(ar^{n-1}) = al$.
Therefore,$P = (al)^{n/2}$.

(D) Let the $n$ numbers be $x_1, x_2, \dots, x_n$. Given that their average is $a$,so $\frac{\sum x_i}{n} = a$,which implies $\sum x_i = na$.
The numbers are increased by $2, 4, 8, \dots, 2^n$. This sequence is a Geometric Progression ($G$.$P$.) where the first term $A = 2$ and the common ratio $r = 2$.
The sum of these increases is $S_n = \frac{A(r^n - 1)}{r - 1} = \frac{2(2^n - 1)}{2 - 1} = 2(2^n - 1)$.
The sum of the new numbers is $\sum x_i + S_n = na + 2(2^n - 1)$.
The new average is $\frac{na + 2(2^n - 1)}{n} = a + \frac{2(2^n - 1)}{n}$.

(C) The positive multiples of $3$ less than $50$ are $3, 6, 9, \dots, 48$.
This is an arithmetic progression where the first term $a = 3$,the common difference $d = 3$,and the last term $l = 48$.
To find the number of terms $n$,we use the formula $l = a + (n - 1)d$:
$48 = 3 + (n - 1)3$
$45 = (n - 1)3$
$15 = n - 1$
$n = 16$.
The sum $S_n$ of an arithmetic progression is given by $S_n = \frac{n}{2}(a + l)$:
$S_{16} = \frac{16}{2}(3 + 48)$
$S_{16} = 8 \times 51 = 408$.

(B) The given expression is $\sum_{k=1}^{n} \left(1-\frac{k}{n+1}\right)$.
This can be expanded as $\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\cdots+\left(1-\frac{n}{n+1}\right)$.
There are $n$ terms in this series,so we can write it as $n - \left(\frac{1}{n+1} + \frac{2}{n+1} + \cdots + \frac{n}{n+1}\right)$.
Factoring out the denominator,we get $n - \frac{1}{n+1} (1+2+3+\cdots+n)$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we substitute this into the expression:
$n - \frac{1}{n+1} \cdot \frac{n(n+1)}{2}$.
Canceling $(n+1)$ from the numerator and denominator,we get $n - \frac{n}{2}$.
Thus,the sum is $\frac{n}{2}$.

(B) clock strikes the number of times corresponding to the hour (e.g.,$1$ time at $1$ o'clock,$2$ times at $2$ o'clock,...,$12$ times at $12$ o'clock).
The total number of strikes in $12$ hours is the sum of the first $12$ natural numbers:
Sum $= 1 + 2 + 3 + \dots + 12 = \frac{n(n+1)}{2} = \frac{12 \times 13}{2} = 78$.
Since a day consists of $24$ hours,the clock completes two $12$-hour cycles.
Total strikes in a day $= 2 \times 78 = 156$.

(D) Given that $a, 1, b$ are in arithmetic progression $(AP)$.
Therefore,$1 = \frac{a+b}{2}$,which implies $a+b = 2$ $......(1)$.
Given that $1, a, b$ are in geometric progression $(GP)$.
Therefore,$a^2 = 1 \times b$,which implies $b = a^2$ $......(2)$.
Substituting the value of $b$ from equation $(2)$ into equation $(1)$:
$a + a^2 = 2$
$a^2 + a - 2 = 0$
Factoring the quadratic equation:
$(a+2)(a-1) = 0$
So,$a = -2$ or $a = 1$.
If $a = 1$,then $b = (1)^2 = 1$. However,the problem states $a \neq b$,so we reject this case.
If $a = -2$,then $b = (-2)^2 = 4$.
Thus,the values are $a = -2$ and $b = 4$.

(A) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$.
To find the sum $11^{2} + 12^{2} + \cdots + 21^{2}$,we can express it as the difference between the sum of squares up to $21$ and the sum of squares up to $10$:
Sum $= \sum_{k=1}^{21} k^{2} - \sum_{k=1}^{10} k^{2}$.
For $n = 21$: $\frac{21(21+1)(2 \times 21 + 1)}{6} = \frac{21 \times 22 \times 43}{6} = 7 \times 11 \times 43 = 3311$.
For $n = 10$: $\frac{10(10+1)(2 \times 10 + 1)}{6} = \frac{10 \times 11 \times 21}{6} = 5 \times 11 \times 7 = 385$.
Therefore,the required sum is $3311 - 385 = 2926$.

(C) Given that $\log 2, \log (2^{x}-1)$ and $\log (2^{x}+3)$ are in an Arithmetic Progression $(A.P.)$.
For three terms $a, b, c$ to be in $A.P.$,the condition is $2b = a + c$.
Therefore,$2 \log (2^{x}-1) = \log 2 + \log (2^{x}+3)$.
Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^{n})$,we get:
$\log (2^{x}-1)^{2} = \log [2(2^{x}+3)]$.
Removing the logarithms from both sides:
$(2^{x}-1)^{2} = 2(2^{x}+3)$.
Let $2^{x} = y$. Then the equation becomes:
$(y-1)^{2} = 2(y+3)$.
$y^{2} - 2y + 1 = 2y + 6$.
$y^{2} - 4y - 5 = 0$.
Factoring the quadratic equation:
$(y-5)(y+1) = 0$.
So,$y = 5$ or $y = -1$.
Since $2^{x} = y$ and $2^{x}$ must be positive for all real $x$,we discard $y = -1$.
Thus,$2^{x} = 5$.
Taking the logarithm on both sides with base $2$:
$x = \log _{2} 5$.

(B) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$.
For $n = 10$,we substitute the value into the formula:
$1^{2}+2^{2}+3^{2}+\cdots+10^{2} = \frac{10(10+1)(2 \times 10+1)}{6}$.
$= \frac{10 \times 11 \times 21}{6}$.
$= \frac{2310}{6} = 385$.

(A) The given series is $S = 1 + 0.6 + 0.06 + 0.006 + 0.0006 + \dots$
We can write this as $S = 1 + (0.6 + 0.06 + 0.006 + 0.0006 + \dots)$.
The part inside the bracket is an infinite Geometric Progression ($G.P.$) where the first term $a = 0.6$ and the common ratio $r = \frac{0.06}{0.6} = 0.1 = \frac{1}{10}$.
The sum of an infinite $G.P.$ is given by the formula $S_{\infty} = \frac{a}{1 - r}$.
Substituting the values, we get $S_{\infty} = \frac{0.6}{1 - 0.1} = \frac{0.6}{0.9} = \frac{6}{9} = \frac{2}{3}$.
Therefore, the total sum $S = 1 + \frac{2}{3} = 1 \frac{2}{3}$.

(C) The given sequence is $0, 3, 8, 15, 24, 35, \dots$
Observe the differences between consecutive terms:
$3 - 0 = 3$
$8 - 3 = 5$
$15 - 8 = 7$
$24 - 15 = 9$
$35 - 24 = 11$
The differences are consecutive odd numbers: $3, 5, 7, 9, 11, \dots$
To find the $9^{th}$ term,we continue the pattern:
$7^{th}$ term $= 35 + 13 = 48$
$8^{th}$ term $= 48 + 15 = 63$
$9^{th}$ term $= 63 + 17 = 80$
Alternatively,the $n^{th}$ term of the sequence is given by $a_n = n^2 - 1$.
For $n = 9$,$a_9 = 9^2 - 1 = 81 - 1 = 80$.

(B) The general term of the series is given by $\frac{1}{n(n+1)}$.
Using partial fractions,we can write $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
Applying this to each term:
$\frac{1}{1 \times 2} = \frac{1}{1} - \frac{1}{2}$
$\frac{1}{2 \times 3} = \frac{1}{2} - \frac{1}{3}$
$\frac{1}{3 \times 4} = \frac{1}{3} - \frac{1}{4}$
...
$\frac{1}{99 \times 100} = \frac{1}{99} - \frac{1}{100}$
Summing these terms:
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \cdots + (\frac{1}{99} - \frac{1}{100})$
All intermediate terms cancel out,leaving only the first and the last term:
$= 1 - \frac{1}{100} = \frac{100-1}{100} = \frac{99}{100}$.

(C) To find the sum of all digits of numbers from $1$ to $100$,we consider numbers from $00$ to $99$ (since $00$ to $09$ are the same as $0$ to $9$ and $100$ contributes $1+0+0=1$).
There are $100$ numbers,each having $2$ digits,making a total of $200$ digit positions.
Each digit ($0$ through $9$) appears an equal number of times in the units and tens places.
Number of times each digit appears $= \frac{200}{10} = 20$ times.
Sum of digits from $0$ to $9 = 0+1+2+3+4+5+6+7+8+9 = 45$.
Total sum of digits for numbers $00$ to $99 = 20 \times 45 = 900$.
Finally,adding the sum of digits of $100$ $(1+0+0 = 1)$:
Total sum $= 900 + 1 = 901$.