Geometric progression Questions in English

(B) The number of ancestors in each generation forms a geometric progression: $2, 4, 8, \dots$
Here,the first term $a = 2$,the common ratio $r = 2$,and the number of generations $n = 10$.
Using the sum formula for a geometric progression: $S_{n} = \frac{a(r^{n} - 1)}{r - 1}$.
Substituting the values: $S_{10} = \frac{2(2^{10} - 1)}{2 - 1}$.
$S_{10} = 2(1024 - 1) = 2 \times 1023 = 2046$.
Thus,the total number of ancestors is $2046$.

(A) Let $G_{1}, G_{2}, G_{3}$ be three numbers between $1$ and $256$ such that $1, G_{1}, G_{2}, G_{3}, 256$ is a $G.P.$
Here,the first term $a = 1$ and the fifth term $a_{5} = ar^{4} = 256$.
Substituting $a = 1$,we get $r^{4} = 256$,which implies $r = \pm 4$.
Case $1$: If $r = 4$,the numbers are $G_{1} = 1 \times 4 = 4$,$G_{2} = 1 \times 4^{2} = 16$,and $G_{3} = 1 \times 4^{3} = 64$.
Case $2$: If $r = -4$,the numbers are $G_{1} = 1 \times (-4) = -4$,$G_{2} = 1 \times (-4)^{2} = 16$,and $G_{3} = 1 \times (-4)^{3} = -64$.
Thus,the three numbers can be $4, 16, 64$ or $-4, 16, -64$.

(A) The given $G.P.$ is $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$
Here,the first term $a = \frac{5}{2}$.
The common ratio $r = \frac{5/4}{5/2} = \frac{5}{4} \times \frac{2}{5} = \frac{1}{2}$.
The $n^{\text{th}}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.
For the $20^{\text{th}}$ term:
$a_{20} = \frac{5}{2} \left(\frac{1}{2}\right)^{20-1} = \frac{5}{2} \left(\frac{1}{2}\right)^{19} = \frac{5}{2^1 \times 2^{19}} = \frac{5}{2^{20}}$.
For the $n^{\text{th}}$ term:
$a_n = \frac{5}{2} \left(\frac{1}{2}\right)^{n-1} = \frac{5}{2^1 \times 2^{n-1}} = \frac{5}{2^{1+n-1}} = \frac{5}{2^n}$.

(A) Given: Common ratio,$r = 2$.
Let $a$ be the first term of the $G.P.$
We know that the $n^{\text{th}}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.
Given $a_8 = 192$,we have $a r^7 = 192$.
Substituting $r = 2$: $a(2)^7 = 192$.
$a(128) = 192$.
$a = \frac{192}{128} = \frac{3}{2}$.
Now,find the $12^{\text{th}}$ term: $a_{12} = a r^{11}$.
$a_{12} = \left(\frac{3}{2}\right) \times (2)^{11} = 3 \times 2^{10}$.
$a_{12} = 3 \times 1024 = 3072$.

(N/A) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
According to the given condition:
$a_{5} = ar^{5-1} = ar^{4} = p$ .........$(1)$
$a_{8} = ar^{8-1} = ar^{7} = q$ .........$(2)$
$a_{11} = ar^{11-1} = ar^{10} = s$ .........$(3)$
Dividing equation $(2)$ by $(1)$,we obtain:
$\frac{ar^{7}}{ar^{4}} = \frac{q}{p} \Rightarrow r^{3} = \frac{q}{p}$ .........$(4)$
Dividing equation $(3)$ by $(2)$,we obtain:
$\frac{ar^{10}}{ar^{7}} = \frac{s}{q} \Rightarrow r^{3} = \frac{s}{q}$ .........$(5)$
Equating the values of $r^{3}$ from $(4)$ and $(5)$:
$\frac{q}{p} = \frac{s}{q}$
$\Rightarrow q^{2} = ps$
Thus,the result is proved.

(A) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
Given,$a = -3$.
The $n^{th}$ term of a $G.P.$ is given by $a_n = a r^{n-1}$.
Therefore,$a_4 = a r^3 = -3 r^3$ and $a_2 = a r = -3 r$.
According to the given condition,$a_4 = (a_2)^2$.
Substituting the values: $-3 r^3 = (-3 r)^2$.
$-3 r^3 = 9 r^2$.
Dividing both sides by $-3 r^2$ (assuming $r \neq 0$),we get $r = -3$.
Now,the $7^{th}$ term is $a_7 = a r^6$.
$a_7 = (-3) \times (-3)^6 = (-3)^1 \times (-3)^6 = (-3)^7$.
$a_7 = -2187$.
Thus,the $7^{th}$ term of the $G.P.$ is $-2187$.

(A) The given sequence is $2, 2\sqrt{2}, 4, \ldots$
Here,the first term $a = 2$ and the common ratio $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$.
Let the $n^{\text{th}}$ term of the sequence be $128$.
The formula for the $n^{\text{th}}$ term of a geometric progression is $a_n = a r^{n-1}$.
Substituting the values,we get:
$2(\sqrt{2})^{n-1} = 128$
Dividing both sides by $2$:
$(\sqrt{2})^{n-1} = 64$
Expressing in terms of base $2$:
$(2^{1/2})^{n-1} = 2^6$
$2^{\frac{n-1}{2}} = 2^6$
Equating the exponents:
$\frac{n-1}{2} = 6$
$n-1 = 12$
$n = 13$
Thus,the $13^{\text{th}}$ term of the sequence is $128$.

(C) The given sequence is $\sqrt{3}, 3, 3\sqrt{3}, \ldots$
This is a geometric progression where the first term $a = \sqrt{3}$ and the common ratio $r = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Let the $n^{\text{th}}$ term of the sequence be $a_n = 729$.
The formula for the $n^{\text{th}}$ term is $a_n = a r^{n-1}$.
Substituting the values,we get $\sqrt{3} \cdot (\sqrt{3})^{n-1} = 729$.
This simplifies to $(\sqrt{3})^n = 729$.
Since $\sqrt{3} = 3^{1/2}$,we have $(3^{1/2})^n = 3^6$.
$3^{n/2} = 3^6$.
Equating the exponents,$\frac{n}{2} = 6$,which gives $n = 12$.
Thus,the $12^{\text{th}}$ term of the sequence is $729$.

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

(C) For three numbers $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.
Here,$a = \frac{2}{7}$,$b = x$,and $c = -\frac{7}{2}$.
Substituting these values into the condition:
$x^2 = \left(\frac{2}{7}\right) \times \left(-\frac{7}{2}\right)$
$x^2 = -1$
In the set of real numbers,the square of a number cannot be negative. Therefore,there are no real values of $x$ that satisfy this condition.
If the question intended for the sequence to be $\frac{2}{7}, x, \frac{7}{2}$ (where $c = \frac{7}{2}$),then $x^2 = \left(\frac{2}{7}\right) \times \left(\frac{7}{2}\right) = 1$,which gives $x = \pm 1$.
Given the options,it is highly probable that the intended sequence was $\frac{2}{7}, x, \frac{7}{2}$.

(A) The given $G.P.$ is $0.15, 0.015, 0.0015, \dots$
Here,the first term $a = 0.15$ and the common ratio $r = \frac{0.015}{0.15} = 0.1$.
The sum of $n$ terms of a $G.P.$ is given by $S_{n} = \frac{a(1-r^{n})}{1-r}$.
For $n = 20$,we have $S_{20} = \frac{0.15[1-(0.1)^{20}]}{1-0.1}$.
$S_{20} = \frac{0.15}{0.9}[1-(0.1)^{20}]$.
$S_{20} = \frac{15}{90}[1-(0.1)^{20}]$.
$S_{20} = \frac{1}{6}[1-(0.1)^{20}]$.

(A) The given $G.P.$ is $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$
Here,the first term $a = \sqrt{7}$ and the common ratio $r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3}$.
The sum of $n$ terms of a $G.P.$ is given by $S_{n} = \frac{a(r^{n}-1)}{r-1}$.
Substituting the values,we get $S_{n} = \frac{\sqrt{7}((\sqrt{3})^{n}-1)}{\sqrt{3}-1}$.
To rationalize the denominator,multiply the numerator and denominator by $(\sqrt{3}+1)$:
$S_{n} = \frac{\sqrt{7}((\sqrt{3})^{n}-1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$
$S_{n} = \frac{\sqrt{7}(\sqrt{3}+1)(3^{\frac{n}{2}}-1)}{3-1}$
$S_{n} = \frac{\sqrt{7}(\sqrt{3}+1)}{2}\left[(3)^{\frac{n}{2}}-1\right]$.

(A) The given $G.P.$ is $1, -a, a^{2}, -a^{3}, \ldots$
Here,the first term $a_{1} = 1$.
The common ratio $r = \frac{-a}{1} = -a$.
The sum of $n$ terms of a $G.P.$ is given by the formula $S_{n} = \frac{a_{1}(1-r^{n})}{1-r}$.
Substituting the values,we get $S_{n} = \frac{1(1-(-a)^{n})}{1-(-a)}$.
Therefore,$S_{n} = \frac{1-(-a)^{n}}{1+a}$.

(A) The given $G.P.$ is $x^{3}, x^{5}, x^{7}, \ldots$
Here,the first term $a = x^{3}$ and the common ratio $r = \frac{x^{5}}{x^{3}} = x^{2}$.
The sum of $n$ terms of a $G.P.$ is given by the formula $S_{n} = \frac{a(1-r^{n})}{1-r}$.
Substituting the values of $a$ and $r$:
$S_{n} = \frac{x^{3}(1-(x^{2})^{n})}{1-x^{2}}$
$S_{n} = \frac{x^{3}(1-x^{2n})}{1-x^{2}}$

We need to evaluate the sum $\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)}$.
Using the linearity property of summation,we have:
$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right) = } \sum\limits_{k = 1}^{11} {2 + } \sum\limits_{k = 1}^{11} {{3^k}} = 2 \times 11 + \sum\limits_{k = 1}^{11} {{3^k}} = 22 + \sum\limits_{k = 1}^{11} {{3^k}} \quad \dots (1)$
The sum $\sum\limits_{k = 1}^{11} {{3^k}}$ is a geometric series with first term $a = 3$,common ratio $r = 3$,and number of terms $n = 11$.
The sum of a geometric series is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values,we get:
$S_{11} = \frac{3(3^{11} - 1)}{3 - 1} = \frac{3}{2}(3^{11} - 1)$.
Substituting this back into equation $(1)$:
$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right) = 22 + \frac{3}{2}(3^{11} - 1)}$.

(A) Let the first three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given,the product of the terms is $1$:
$(\frac{a}{r}) \times a \times (ar) = 1$
$a^3 = 1 \Rightarrow a = 1$.
Given,the sum of the terms is $\frac{39}{10}$:
$\frac{a}{r} + a + ar = \frac{39}{10}$
Substitute $a = 1$:
$\frac{1}{r} + 1 + r = \frac{39}{10}$
$1 + r + r^2 = \frac{39}{10}r$
$10r^2 + 10r + 10 = 39r$
$10r^2 - 29r + 10 = 0$
$10r^2 - 25r - 4r + 10 = 0$
$5r(2r - 5) - 2(2r - 5) = 0$
$(5r - 2)(2r - 5) = 0$
$r = \frac{2}{5}$ or $r = \frac{5}{2}$.
If $r = \frac{5}{2}$,the terms are $\frac{1}{5/2}, 1, \frac{5}{2}$,which are $\frac{2}{5}, 1, \frac{5}{2}$.
If $r = \frac{2}{5}$,the terms are $\frac{1}{2/5}, 1, \frac{2}{5}$,which are $\frac{5}{2}, 1, \frac{2}{5}$.

(C) The given $G.P.$ is $3, 3^{2}, 3^{3}, \dots$
Let $n$ terms of this $G.P.$ be required to obtain the sum $120$.
The formula for the sum of $n$ terms of a $G.P.$ is $S_{n} = \frac{a(r^{n}-1)}{r-1}$ for $r > 1$.
Here,the first term $a = 3$ and the common ratio $r = \frac{3^{2}}{3} = 3$.
Substituting the values into the formula:
$120 = \frac{3(3^{n}-1)}{3-1}$
$120 = \frac{3(3^{n}-1)}{2}$
$120 \times \frac{2}{3} = 3^{n}-1$
$40 \times 2 = 3^{n}-1$
$80 = 3^{n}-1$
$3^{n} = 81$
$3^{n} = 3^{4}$
Therefore,$n = 4$.
Thus,$4$ terms of the given $G.P.$ are required to obtain the sum $120$.

(N/A) Let the $G.P.$ be $a, ar, ar^{2}, ar^{3}, ar^{4}, ar^{5}, \dots$
According to the given condition:
$a + ar + ar^{2} = 16$ --- $(1)$
$ar^{3} + ar^{4} + ar^{5} = 128$ --- $(2)$
From $(1)$,$a(1 + r + r^{2}) = 16$
From $(2)$,$ar^{3}(1 + r + r^{2}) = 128$
Dividing $(2)$ by $(1)$,we get:
$\frac{ar^{3}(1 + r + r^{2})}{a(1 + r + r^{2})} = \frac{128}{16}$
$r^{3} = 8 \Rightarrow r = 2$
Substituting $r = 2$ in $(1)$:
$a(1 + 2 + 4) = 16$ $\Rightarrow 7a = 16$ $\Rightarrow a = \frac{16}{7}$
The sum to $n$ terms $S_{n}$ is given by $S_{n} = \frac{a(r^{n} - 1)}{r - 1}$
$S_{n} = \frac{16}{7} \times \frac{(2^{n} - 1)}{2 - 1} = \frac{16}{7}(2^{n} - 1)$

(A) Given $a=729$ and $a_{7}=64$.
Let $r$ be the common ratio of the $G.P.$
We know that $a_{n}=a r^{n-1}$.
$a_{7}=a r^{6} = 729 r^{6} = 64$.
$r^{6} = \frac{64}{729} = \left(\frac{2}{3}\right)^{6}$.
Thus,$r = \frac{2}{3}$.
The sum of $n$ terms of a $G.P.$ is given by $S_{n}=\frac{a(1-r^{n})}{1-r}$.
$S_{7}=\frac{729(1-(2/3)^{7})}{1-2/3} = \frac{729(1-(2/3)^{7})}{1/3}$.
$S_{7} = 3 \times 729 \times \left(1 - \frac{2^{7}}{3^{7}}\right) = 3^{7} \times \left(\frac{3^{7}-2^{7}}{3^{7}}\right)$.
$S_{7} = 3^{7} - 2^{7} = 2187 - 128 = 2059$.

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
According to the given conditions:
$a + ar = -4$ .......$(1)$
$ar^4 = 4 \times ar^2$
Since $a \neq 0$,we have $r^2 = 4$,which implies $r = 2$ or $r = -2$.
Case $1$: If $r = 2$,then from $(1)$:
$a + 2a = -4$ $\Rightarrow 3a = -4$ $\Rightarrow a = -\frac{4}{3}$.
The $G.P.$ is $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$
Case $2$: If $r = -2$,then from $(1)$:
$a - 2a = -4$ $\Rightarrow -a = -4$ $\Rightarrow a = 4$.
The $G.P.$ is $4, -8, 16, -32, \dots$

(N/A) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
According to the given condition:
$a_{4} = ar^{3} = x$ $(1)$
$a_{10} = ar^{9} = y$ $(2)$
$a_{16} = ar^{15} = z$ $(3)$
Dividing $(2)$ by $(1)$,we obtain:
$\frac{y}{x} = \frac{ar^{9}}{ar^{3}} = r^{6}$
Dividing $(3)$ by $(2)$,we obtain:
$\frac{z}{y} = \frac{ar^{15}}{ar^{9}} = r^{6}$
Since $\frac{y}{x} = \frac{z}{y} = r^{6}$,the ratio between consecutive terms is constant.
Therefore,$x, y, z$ are in $G.P.$

(A) The required sum is the sum of the products of corresponding terms:
Sum $= (2 \times 128) + (4 \times 32) + (8 \times 8) + (16 \times 2) + (32 \times \frac{1}{2})$
Sum $= 256 + 128 + 64 + 32 + 16$
This is a geometric series with first term $a = 256$,common ratio $r = \frac{128}{256} = \frac{1}{2}$,and number of terms $n = 5$.
The sum of $n$ terms of a $G.P.$ is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.
$S_5 = \frac{256(1 - (\frac{1}{2})^5)}{1 - \frac{1}{2}} = \frac{256(1 - \frac{1}{32})}{\frac{1}{2}} = 512 \times \frac{31}{32} = 16 \times 31 = 496$.

(N/A) The given sequences are $a, ar, ar^{2}, \dots, ar^{n-1}$ and $A, AR, AR^{2}, \dots, AR^{n-1}$.
Their corresponding products are $aA, (ar)(AR), (ar^{2})(AR^{2}), \dots, (ar^{n-1})(AR^{n-1})$,which simplifies to $aA, (arR), (arR)^{2}, \dots, (arR)^{n-1}$.
To check if this is a $G.P.$,we find the ratio of consecutive terms:
$\frac{\text{Second term}}{\text{First term}} = \frac{arAR}{aA} = rR$
$\frac{\text{Third term}}{\text{Second term}} = \frac{ar^{2}AR^{2}}{arAR} = rR$
Since the ratio of consecutive terms is constant,the sequence forms a $G.P.$ with a common ratio of $rR$.

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
The terms are $a_{1}=a, a_{2}=a r, a_{3}=a r^{2}, a_{4}=a r^{3}.$
By the given conditions:
$a_{3} = a_{1} + 9 \Rightarrow a r^{2} = a + 9$ ..........$(1)$
$a_{2} = a_{4} + 18 \Rightarrow a r = a r^{3} + 18$ ..........$(2)$
From $(1),$ we have $a(r^{2} - 1) = 9.$ ..........$(3)$
From $(2),$ we have $a r(1 - r^{2}) = 18 \Rightarrow -a r(r^{2} - 1) = 18.$ ..........$(4)$
Dividing $(4)$ by $(3):$
$\frac{-a r(r^{2} - 1)}{a(r^{2} - 1)} = \frac{18}{9}$
$-r = 2 \Rightarrow r = -2.$
Substituting $r = -2$ into $(1):$
$a((-2)^{2} - 1) = 9$
$a(4 - 1) = 9$
$3a = 9 \Rightarrow a = 3.$
The four numbers are $a, ar, ar^{2}, ar^{3}.$
Substituting $a = 3$ and $r = -2:$
$3, 3(-2), 3(-2)^{2}, 3(-2)^{3}$
$3, -6, 12, -24.$

(N/A) Let $A$ be the first term and $R$ be the common ratio of the $G.P.$
According to the given information:
$A R^{p-1} = a$
$A R^{q-1} = b$
$A R^{r-1} = c$
Now,consider the expression $a^{q-r} \cdot b^{r-p} \cdot c^{p-q}$:
$= (A R^{p-1})^{q-r} \cdot (A R^{q-1})^{r-p} \cdot (A R^{r-1})^{p-q}$
$= A^{q-r+r-p+p-q} \cdot R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$
Calculating the exponent of $A$:
$q-r+r-p+p-q = 0$
Calculating the exponent of $R$:
$(pq - pr - q + r) + (qr - pq - r + p) + (rp - rq - p + q) = 0$
Thus,the expression becomes:
$= A^0 \cdot R^0 = 1 \cdot 1 = 1$
Hence,the result is proved.

The first term of the $G.P.$ is $a$ and the $n^{\text{th}}$ term is $b$.
Therefore,the $G.P.$ is $a, ar, ar^{2}, ar^{3}, \dots, ar^{n-1}$,where $r$ is the common ratio.
$b = ar^{n-1}$ .........$(1)$
$P = \text{Product of } n \text{ terms}$
$P = (a)(ar)(ar^{2}) \dots (ar^{n-1})$
$P = (a \times a \times \dots \times a)(r \times r^{2} \times \dots \times r^{n-1})$
$P = a^{n} r^{1 + 2 + \dots + (n-1)}$ .........$(2)$
Here,$1, 2, \dots, (n-1)$ is an $A.P.$
Sum of $n-1$ terms $= \frac{(n-1)}{2} [1 + (n-1)] = \frac{n(n-1)}{2}$.
$P = a^{n} r^{\frac{n(n-1)}{2}}$
Squaring both sides:
$P^{2} = (a^{n} r^{\frac{n(n-1)}{2}})^{2} = a^{2n} r^{n(n-1)}$
$P^{2} = (a^{2} r^{n-1})^{n}$
$P^{2} = (a \cdot ar^{n-1})^{n}$
Using $(1)$,$P^{2} = (ab)^{n}$.
Thus,the result is proved.

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
The sum of the first $n$ terms is given by $S_n = \frac{a(1-r^n)}{1-r}$.
The terms from the $(n+1)^{th}$ to the $(2n)^{th}$ term form a $G.P.$ with the first term $a_{n+1} = ar^n$ and the number of terms equal to $n$.
The sum of these terms is $S' = \frac{a_{n+1}(1-r^n)}{1-r} = \frac{ar^n(1-r^n)}{1-r}$.
The ratio of the sum of the first $n$ terms to the sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ is:
$\text{Ratio} = \frac{\frac{a(1-r^n)}{1-r}}{\frac{ar^n(1-r^n)}{1-r}} = \frac{a(1-r^n)}{1-r} \times \frac{1-r}{ar^n(1-r^n)} = \frac{1}{r^n}$.
Thus,the ratio is $\frac{1}{r^n}$.

Since $a, b, c, d$ are in $G.P.$,we have $b=ar, c=ar^{2}, d=ar^{3}$.
$L.H.S. = (a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})$
$= (a^{2} + a^{2}r^{2} + a^{2}r^{4})(a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6})$
$= a^{2}(1 + r^{2} + r^{4}) \times a^{2}r^{2}(1 + r^{2} + r^{4})$
$= a^{4}r^{2}(1 + r^{2} + r^{4})^{2}$
$R.H.S. = (ab+bc+cd)^{2}$
$= (a(ar) + (ar)(ar^{2}) + (ar^{2})(ar^{3}))^{2}$
$= (a^{2}r + a^{2}r^{3} + a^{2}r^{5})^{2}$
$= (a^{2}r(1 + r^{2} + r^{4}))^{2}$
$= a^{4}r^{2}(1 + r^{2} + r^{4})^{2}$
Since $L.H.S. = R.H.S.$,the identity is proved.

(A) Let $G_{1}$ and $G_{2}$ be two numbers between $3$ and $81$ such that the sequence $3, G_{1}, G_{2}, 81$ forms a $G.P.$
Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
Here,the first term $a = 3$ and the fourth term $a_{4} = 81$.
The formula for the $n^{th}$ term of a $G.P.$ is $a_{n} = a r^{n-1}$.
Thus,$81 = 3 \times r^{4-1} = 3 r^{3}$.
$r^{3} = \frac{81}{3} = 27$.
$r = \sqrt[3]{27} = 3$.
Now,$G_{1} = a r = 3 \times 3 = 9$.
$G_{2} = a r^{2} = 3 \times (3)^{2} = 3 \times 9 = 27$.
Therefore,the two numbers are $9$ and $27$.

(A) The number of bacteria doubles every hour,forming a geometric progression $(G.P.)$ where the initial number of bacteria $a = 30$ and the common ratio $r = 2$.
The number of bacteria after $t$ hours is given by $a_{t+1} = a \times r^{t} = 30 \times 2^{t}$.
At the end of the $2^{\text{nd}}$ hour $(t=2)$: $30 \times 2^{2} = 30 \times 4 = 120$.
At the end of the $4^{\text{th}}$ hour $(t=4)$: $30 \times 2^{4} = 30 \times 16 = 480$.
At the end of the $n^{\text{th}}$ hour $(t=n)$: $30 \times 2^{n}$.

(A) The principal amount deposited is $P = Rs. 500$.
The annual interest rate is $r = 10\% = 0.10$.
The amount $A$ after $n$ years with compound interest is given by the formula $A = P(1 + r)^n$.
Here,$n = 10$ years.
Substituting the values,we get $A = 500(1 + 0.10)^{10}$.
$A = 500(1.1)^{10}$.

Given that $(a^{2}+b^{2}+c^{2}) p^{2}-2(ab+bc+cd) p+(b^{2}+c^{2}+d^{2}) \leq 0$ $(1)$
Expanding the expression,we get:
$(a^{2}p^{2}-2abp+b^{2})+(b^{2}p^{2}-2bcp+c^{2})+(c^{2}p^{2}-2cdp+d^{2}) \leq 0$
This can be written as:
$(ap-b)^{2}+(bp-c)^{2}+(cp-d)^{2} \leq 0$ $(2)$
Since the sum of squares of real numbers is always non-negative,the only way for the sum to be $\leq 0$ is if each term is equal to $0$:
$(ap-b)^{2} = 0, (bp-c)^{2} = 0, (cp-d)^{2} = 0$
This implies $ap=b, bp=c, cp=d$.
Thus,$\frac{b}{a} = p, \frac{c}{b} = p, \frac{d}{c} = p$.
Since the common ratio is constant,$a, b, c, d$ are in $G.P.$

(A) Let the sum of $n$ terms of the $G.P.$ be $S_n = 315$.
The formula for the sum of $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.
Given that the first term $a = 5$ and the common ratio $r = 2$.
Substituting the values into the formula: $315 = \frac{5(2^n - 1)}{2 - 1}$.
$315 = 5(2^n - 1)$.
$2^n - 1 = \frac{315}{5} = 63$.
$2^n = 64 = 2^6$.
Therefore,$n = 6$.
The last term of the $G.P.$ is the $n^{th}$ term,given by $a_n = a \cdot r^{n-1}$.
$a_6 = 5 \cdot 2^{6-1} = 5 \cdot 2^5 = 5 \cdot 32 = 160$.
Thus,the last term is $160$ and the number of terms is $6$.

(C) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
Given $a = 1$.
The third term $a_3 = ar^2 = r^2$ and the fifth term $a_5 = ar^4 = r^4$.
According to the problem,$a_3 + a_5 = 90$.
So,$r^2 + r^4 = 90$,which implies $r^4 + r^2 - 90 = 0$.
Let $x = r^2$. Then $x^2 + x - 90 = 0$.
Factoring the quadratic equation: $(x + 10)(x - 9) = 0$.
This gives $x = -10$ or $x = 9$.
Since $x = r^2$ must be non-negative for real $r$,we have $r^2 = 9$.
Therefore,$r = \pm 3$.

(B) Let the $G.P.$ be $a, ar, ar^2, ar^3, \dots, ar^{2n-1}$.
Number of terms $= 2n$.
Sum of all terms $S_{2n} = \frac{a(r^{2n}-1)}{r-1}$.
Sum of terms at odd places $S_{odd} = a + ar^2 + \dots + ar^{2n-2} = \frac{a((r^2)^n - 1)}{r^2 - 1} = \frac{a(r^{2n}-1)}{r^2-1}$.
Given that $S_{2n} = 5 \times S_{odd}$.
$\frac{a(r^{2n}-1)}{r-1} = 5 \times \frac{a(r^{2n}-1)}{r^2-1}$.
Since $r \neq 1$ and $r^{2n} \neq 1$,we can cancel $\frac{a(r^{2n}-1)}{r-1}$ from both sides.
$1 = \frac{5}{r+1}$.
$r+1 = 5$.
$r = 4$.

(A) Given that $\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx}$.
Applying componendo and dividendo or cross-multiplying:
$(a+bx)(b-cx) = (a-bx)(b+cx)$
$ab - acx + b^2x - bcx^2 = ab + acx - b^2x - bcx^2$
$2b^2x = 2acx$
Since $x \neq 0$,we get $b^2 = ac$,which implies $\frac{b}{a} = \frac{c}{b} \dots (1)$.
Similarly,given $\frac{b+cx}{b-cx} = \frac{c+dx}{c-dx}$.
$(b+cx)(c-dx) = (b-cx)(c+dx)$
$bc - bdx + c^2x - cdx^2 = bc + bdx - c^2x - cdx^2$
$2c^2x = 2bdx$
Since $x \neq 0$,we get $c^2 = bd$,which implies $\frac{c}{b} = \frac{d}{c} \dots (2)$.
From $(1)$ and $(2)$,we have $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.
Therefore,$a, b, c,$ and $d$ are in $G.P.$

Let the $G.P.$ be $a, ar, ar^{2}, \ldots, ar^{n-1}$.
According to the given information:
$S = \frac{a(r^{n}-1)}{r-1}$
$P = a \cdot (ar) \cdot (ar^{2}) \cdots (ar^{n-1}) = a^{n} r^{1+2+\ldots+(n-1)}$
$P = a^{n} r^{\frac{n(n-1)}{2}}$
$R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^{2}} + \cdots + \frac{1}{ar^{n-1}}$
$R = \frac{r^{n-1} + r^{n-2} + \cdots + 1}{ar^{n-1}} = \frac{\frac{1(r^{n}-1)}{r-1}}{ar^{n-1}} = \frac{r^{n}-1}{ar^{n-1}(r-1)}$
Now,consider $P^{2} R^{n}$:
$P^{2} R^{n} = (a^{n} r^{\frac{n(n-1)}{2}})^{2} \times \left( \frac{r^{n}-1}{ar^{n-1}(r-1)} \right)^{n}$
$P^{2} R^{n} = a^{2n} r^{n(n-1)} \times \frac{(r^{n}-1)^{n}}{a^{n} r^{n(n-1)} (r-1)^{n}}$
$P^{2} R^{n} = \frac{a^{n} (r^{n}-1)^{n}}{(r-1)^{n}}$
$P^{2} R^{n} = \left( \frac{a(r^{n}-1)}{r-1} \right)^{n}$
$P^{2} R^{n} = S^{n}$
Hence,the result is proved.

(N/A) It is given that $a, b, c,$ and $d$ are in $G.P.$
$\therefore b^{2}=ac$ ........$(1)$
$c^{2}=bd$ ........$(2)$
$ad=bc$ ........$(3)$
It has to be proved that $(a^{n}+b^{n}), (b^{n}+c^{n}), (c^{n}+d^{n})$ are in $G.P.$,i.e.,
$(b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$
Consider $L.H.S.$
$(b^{n}+c^{n})^{2}=b^{2n}+2b^{n}c^{n}+c^{2n}$
$=(b^{2})^{n}+2b^{n}c^{n}+(c^{2})^{n}$
$=(ac)^{n}+2b^{n}c^{n}+(bd)^{n}$ [ Using $(1)$ and $(2)$ ]
$=a^{n}c^{n}+b^{n}c^{n}+b^{n}c^{n}+b^{n}d^{n}$
$=a^{n}c^{n}+b^{n}c^{n}+a^{n}d^{n}+b^{n}d^{n}$ [ Using $(3)$,since $bc=ad$,so $b^{n}c^{n}=a^{n}d^{n}$ ]
$=c^{n}(a^{n}+b^{n})+d^{n}(a^{n}+b^{n})$
$=(a^{n}+b^{n})(c^{n}+d^{n}) = R.H.S.$
$\therefore (b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$
Thus,$(a^{n}+b^{n}), (b^{n}+c^{n}),$ and $(c^{n}+d^{n})$ are in $G.P.$

(A) The number of letters mailed in each set forms a $G.P.$: $4, 4^2, 4^3, \ldots, 4^8$.
First term $(a) = 4$.
Common ratio $(r) = 4$.
Number of terms $(n) = 8$.
The sum of $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
$S_8 = \frac{4(4^8 - 1)}{4 - 1} = \frac{4(65536 - 1)}{3} = \frac{4(65535)}{3} = 4(21845) = 87380$.
The cost to mail one letter is $50$ paise,which is $Rs. 0.50$.
Total cost $= 87380 \times 0.50 = Rs. 43690$.

(A) The initial cost of the machine is $P = 15625$.
The rate of depreciation is $r = 20\% = 0.2$.
The value of the machine after $n = 5$ years is given by the formula $V = P(1 - r)^n$.
$V = 15625 \times (1 - 0.2)^5$
$V = 15625 \times (0.8)^5$
$V = 15625 \times \left(\frac{4}{5}\right)^5$
$V = 15625 \times \frac{1024}{3125}$
$V = 5 \times 1024 = 5120$
Thus,the estimated value at the end of $5$ years is $Rs. 5120$.

(D) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given the product is $27$,so $\frac{a}{r} \times a \times ar = 27$ $\Rightarrow a^3 = 27$ $\Rightarrow a = 3$.
The sum $S = \frac{3}{r} + 3 + 3r = 3(\frac{1}{r} + r + 1)$.
Case $1$: If $r > 0$,by $AM \geq GM$,$\frac{1}{r} + r \geq 2$. Thus,$S = 3(\frac{1}{r} + r + 1) \geq 3(2 + 1) = 9$.
Case $2$: If $r < 0$,let $r = -k$ where $k > 0$. Then $\frac{1}{r} + r = -(\frac{1}{k} + k) \leq -2$. Thus,$S = 3(\frac{1}{r} + r + 1) \leq 3(-2 + 1) = -3$.
Therefore,$S \in (-\infty, -3] \cup [9, \infty)$.

(B) Let the first term be $a > 0$ and the common ratio be $r > 0$.
Given the sum of the second,third,and fourth terms is $3$:
$ar + ar^2 + ar^3 = 3$ --- $(1)$
Given the sum of the sixth,seventh,and eighth terms is $243$:
$ar^5 + ar^6 + ar^7 = 243$
$r^4(ar + ar^2 + ar^3) = 243$
Substituting $(1)$ into this equation:
$r^4(3) = 243$ $\Rightarrow r^4 = 81$ $\Rightarrow r = 3$ (since $r > 0$).
Substituting $r = 3$ into $(1)$:
$a(3) + a(9) + a(27) = 3$
$39a = 3 \Rightarrow a = \frac{3}{39} = \frac{1}{13}$.
The sum of the first $50$ terms $S_{50}$ is given by:
$S_{50} = \frac{a(r^{50}-1)}{r-1} = \frac{\frac{1}{13}(3^{50}-1)}{3-1} = \frac{1}{26}(3^{50}-1)$.

(C) The given expression is a geometric progression with first term $a = 2^{10}$,common ratio $r = \frac{3}{2}$,and number of terms $n = 11$.
The sum of the geometric progression is given by $S' = a \frac{r^n - 1}{r - 1}$.
Substituting the values: $S' = 2^{10} \frac{(\frac{3}{2})^{11} - 1}{\frac{3}{2} - 1} = 2^{10} \frac{\frac{3^{11}}{2^{11}} - 1}{\frac{1}{2}} = 2^{11} \left( \frac{3^{11} - 2^{11}}{2^{11}} \right) = 3^{11} - 2^{11}$.
Given that $S' = S - 2^{11}$,we have $3^{11} - 2^{11} = S - 2^{11}$.
Therefore,$S = 3^{11}$.

(C) Given the equation: $(a^{2}+b^{2}+c^{2}) p^{2} - 2(ab+bc+cd) p + (b^{2}+c^{2}+d^{2}) = 0$.
Rearranging the terms,we get: $(a^{2}p^{2} - 2abp + b^{2}) + (b^{2}p^{2} - 2bcp + c^{2}) + (c^{2}p^{2} - 2cdp + d^{2}) = 0$.
This can be written as: $(ap - b)^{2} + (bp - c)^{2} + (cp - d)^{2} = 0$.
Since $a, b, c, d, p$ are real numbers,the sum of squares is zero only if each term is zero:
$ap - b = 0 \Rightarrow p = \frac{b}{a}$
$bp - c = 0 \Rightarrow p = \frac{c}{b}$
$cp - d = 0 \Rightarrow p = \frac{d}{c}$
Thus,$\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$.
This implies that $a, b, c, d$ are in $G.P.$

(D) Let $a_{n}$ be the side length of square $A_{n}$.
Given that the side length of $A_{n}$ equals the diagonal of $A_{n+1}$,we have $a_{n} = \sqrt{2} a_{n+1}$,which implies $a_{n+1} = \frac{a_{n}}{\sqrt{2}}$.
This forms a geometric progression for the side lengths with first term $a_{1} = 12$ and common ratio $r = \frac{1}{\sqrt{2}}$.
The side length $a_{n}$ is given by $a_{n} = a_{1} \times r^{n-1} = 12 \times \left(\frac{1}{\sqrt{2}}\right)^{n-1}$.
The area of $A_{n}$ is $(a_{n})^2 = 144 \times \left(\frac{1}{2}\right)^{n-1} = \frac{144}{2^{n-1}}$.
We want the area to be less than $1$,so $\frac{144}{2^{n-1}} < 1$.
This implies $2^{n-1} > 144$.
Since $2^{7} = 128$ and $2^{8} = 256$,we must have $n-1 \geq 8$.
Therefore,$n \geq 9$. The smallest value of $n$ is $9$.

(C) Let the geometric series be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 1$ for an increasing series.
Given $T_2 + T_6 = \frac{25}{2} \Rightarrow ar(1 + r^4) = \frac{25}{2} \dots (1)$
Given $T_3 \cdot T_5 = 25$ $\Rightarrow (ar^2)(ar^4) = 25$ $\Rightarrow a^2r^6 = 25$ $\Rightarrow ar^3 = 5$ (since $a, r > 0$)
From $(1)$,$ar + ar^5 = \frac{25}{2}$. Substituting $a = \frac{5}{r^3}$:
$\frac{5}{r^3} \cdot r + \frac{5}{r^3} \cdot r^5 = \frac{25}{2} \Rightarrow \frac{5}{r^2} + 5r^2 = \frac{25}{2}$
Dividing by $5$: $\frac{1}{r^2} + r^2 = \frac{5}{2} \Rightarrow 2r^4 - 5r^2 + 2 = 0$
$(2r^2 - 1)(r^2 - 2) = 0 \Rightarrow r^2 = 2$ (since $r > 1$)
Then $a = \frac{5}{r^3} = \frac{5}{2\sqrt{2}}$.
The sum of $4^{\text{th}}, 6^{\text{th}}, 8^{\text{th}}$ terms is $ar^3 + ar^5 + ar^7 = ar^3(1 + r^2 + r^4)$.
$= 5(1 + 2 + 4) = 5(7) = 35$.

(D) Let the first four terms be $a, ar, ar^2, ar^3$.
The sum of the first four terms is $S_4 = a(1+r+r^2+r^3) = a\frac{r^4-1}{r-1} = \frac{65}{12} \quad (1)$.
The sum of their reciprocals is $\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} = \frac{r^3+r^2+r+1}{ar^3} = \frac{1}{a} \frac{r^4-1}{r^3(r-1)} = \frac{65}{18} \quad (2)$.
Dividing $(1)$ by $(2)$,we get $\frac{a\frac{r^4-1}{r-1}}{\frac{1}{a}\frac{r^4-1}{r^3(r-1)}} = a^2 r^3 = \frac{65/12}{65/18} = \frac{18}{12} = \frac{3}{2}$.
Given the product of the first three terms is $a \cdot ar \cdot ar^2 = a^3 r^3 = 1$,which implies $ar = 1$,so $a = \frac{1}{r}$.
Substituting $a = \frac{1}{r}$ into $a^2 r^3 = \frac{3}{2}$,we get $(\frac{1}{r})^2 r^3 = r = \frac{3}{2}$.
Thus,$a = \frac{1}{3/2} = \frac{2}{3}$.
The third term $\alpha = ar^2 = \frac{2}{3} \times (\frac{3}{2})^2 = \frac{2}{3} \times \frac{9}{4} = \frac{3}{2}$.
Therefore,$2\alpha = 2 \times \frac{3}{2} = 3$.

(C) The given sequence is a geometric progression with first term $a = -16$ and common ratio $r = -1/2$.
The $n^{\text{th}}$ term is $t_{n} = a r^{n-1} = -16(-1/2)^{n-1}$.
Let the arithmetic mean and geometric mean of $t_{p}$ and $t_{q}$ be $x_{1}$ and $x_{2}$. The equation $4x^{2}-9x+5=0$ has roots $x = 1$ and $x = 5/4$.
Since the arithmetic mean is greater than the geometric mean for distinct terms,we have $AM = 5/4$ and $GM = 1$.
$AM = (t_{p} + t_{q})/2 = 5/4 \Rightarrow t_{p} + t_{q} = 5/2$.
$GM = \sqrt{t_{p} t_{q}} = 1 \Rightarrow t_{p} t_{q} = 1$.
Substituting $t_{p} = -16(-1/2)^{p-1}$ and $t_{q} = -16(-1/2)^{q-1}$:
$t_{p} t_{q} = 256(-1/2)^{p+q-2} = 1 \Rightarrow (-1/2)^{p+q-2} = 1/256 = (1/2)^{8}$.
Since $(-1/2)^{p+q-2} = (1/2)^{8}$,$p+q-2$ must be an even number equal to $8$.
$p+q-2 = 8 \Rightarrow p+q = 10$.

(B) The sum of the infinite $GP$ is given by $\frac{a}{1-r} = 15 \dots (i)$.
The series formed by the squares of the terms is $a^{2}, a^{2}r^{2}, a^{2}r^{4}, \dots$,which is also a $GP$ with first term $a^{2}$ and common ratio $r^{2}$.
The sum of this series is $\frac{a^{2}}{1-r^{2}} = 150$.
We can rewrite this as $\frac{a}{1-r} \cdot \frac{a}{1+r} = 150$.
Substituting $(i)$ into this equation,we get $15 \cdot \frac{a}{1+r} = 150$,which implies $\frac{a}{1+r} = 10 \dots (ii)$.
Dividing $(i)$ by $(ii)$,we get $\frac{1+r}{1-r} = \frac{15}{10} = \frac{3}{2}$.
Solving for $r$: $2 + 2r = 3 - 3r$ $\Rightarrow 5r = 1$ $\Rightarrow r = \frac{1}{5}$.
Substituting $r = \frac{1}{5}$ into $(i)$: $\frac{a}{1 - 1/5} = 15$ $\Rightarrow \frac{a}{4/5} = 15$ $\Rightarrow a = 15 \cdot \frac{4}{5} = 12$.
The series $ar^{2}, ar^{4}, ar^{6}, \dots$ is a $GP$ with first term $A = ar^{2}$ and common ratio $R = r^{2}$.
Sum $= \frac{ar^{2}}{1-r^{2}} = \frac{12 \cdot (1/5)^{2}}{1 - (1/5)^{2}} = \frac{12 \cdot (1/25)}{1 - 1/25} = \frac{12/25}{24/25} = \frac{12}{24} = \frac{1}{2}$.

(D) The given expression is a geometric series: $\sum_{n=1}^{10} \frac{1}{2^n \cdot 3^{11-n}} = \frac{K}{2^{10} \cdot 3^{10}}$.
Multiplying both sides by $2^{10} \cdot 3^{10}$,we get $K = \sum_{n=1}^{10} 2^{10-n} \cdot 3^{n-1} = 3^0 \cdot 2^9 + 3^1 \cdot 2^8 + \ldots + 3^9 \cdot 2^0$.
This is a geometric progression with first term $a = 2^9$,common ratio $r = \frac{3}{2}$,and $n = 10$ terms.
$K = \frac{2^9 ((\frac{3}{2})^{10} - 1)}{\frac{3}{2} - 1} = \frac{2^9 (\frac{3^{10}}{2^{10}} - 1)}{\frac{1}{2}} = 2^{10} \cdot \frac{3^{10} - 2^{10}}{2^{10}} = 3^{10} - 2^{10}$.
We need to find the remainder when $K = 3^{10} - 2^{10}$ is divided by $6$.
$K = 3^{10} - 2^{10} = (3^5 - 2^5)(3^5 + 2^5) = (243 - 32)(243 + 32) = (211)(275)$.
$211 = 6 \times 35 + 1$,so $211 \equiv 1 \pmod{6}$.
$275 = 6 \times 45 + 5$,so $275 \equiv 5 \pmod{6}$.
$K \equiv 1 \times 5 \equiv 5 \pmod{6}$.
Thus,the remainder is $5$.