$bold{Geometric;Series}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

a.) This is a geometric series with $t_1=6$, $r=dfrac{18}{6}=3$

The sum of first seven terms is

$$
S_7=6cdot dfrac{3^7-1}{3-1}=6558
$$

b.) This is a geometric series with $t_1=100$ and $r=dfrac{50}{100}=dfrac{1}{2}=0.5$

The sum of the first seven terms is

$$
S_7=100cdot dfrac{0.5^7-1}{0.5-1}=198.4375
$$

c.) This is a geometric series with $t_1=8$ and $r=dfrac{-24}{8}=-3$

The sum of the first seven terms is

$$
S_7=8cdot dfrac{(-3)^7-1}{(-3)-1}=4376
$$

d.) This is a geometric series with $t_1=dfrac{1}{3}$ and $r=dfrac{1/6}{1/3}=0.5$

The sum of the first seven terms is

$$
S_7=dfrac{1}{3}cdot dfrac{(0.5)^7-1}{(0.5)-1}=dfrac{127}{192}
$$

a.) 6558

b.) 198.4375

c.) 4376

d.) $dfrac{127}{192}$

$bold{Geometric;Series}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

This is a geometric series with $t_1=11$ and $r=4$. We need to find the first 6 terms,

$$
S_6=11cdot dfrac{4^6-1}{4-1}=15;015
$$

$$
S_6=15;015
$$

$bold{Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

a.) This is a geometric series with $t_1=6$, $r=dfrac{30}{6}=5$

The $n^{th}$ term is

$t_n=6cdot 5^{n-1}$

Sixth term: $t_6=6cdot 5^{6-1}=18750$

The sum of first six terms is

$$
S_6=6cdot dfrac{5^6-1}{5-1}=23;436
$$

b.) This is a geometric series with $t_1=-11$ and $r=dfrac{-33}{-11}=3$

The $n^{th}$ term is

$t_n=(-11)cdot 3^{n-1}$

Sixth term: $t_6=(-11)cdot 3^{6-1}=-2673$

The sum of first six terms is

$$
S_6=(-11)cdot dfrac{3^6-1}{3-1}=-4004
$$

c.) This is a geometric series with $t_1=21;000;000$ and $r=dfrac{4;200;000}{21;000;000}=0.2$

The $n^{th}$ term is

$t_n=(21;000;000)cdot (0.2)^{n-1}$

Sixth term: $t_6=21;000;000cdot (0.2)^{6-1}=6720$

The sum of first six terms is

$$
S_6=(21;000;000)cdot dfrac{0.2^6-1}{0.2-1}=26;248;320
$$

d.) This is a geometric series with $t_1=dfrac{4}{5}$ and $r=dfrac{8/15}{4/5}=dfrac{2}{3}$

The $n^{th}$ term is

$t_n=left(dfrac{4}{5}right)cdot left(dfrac{2}{3}right)^{n-1}$

Sixth term: $t_6=dfrac{4}{5}cdot left(dfrac{2}{3}right)^{6-1}=dfrac{4}{5}cdot dfrac{2^5}{3^5}=dfrac{128}{1215}$

The sum of first six terms is

$$
S_6=left(dfrac{4}{5}right)cdot dfrac{(2/3)^6-1}{(2/3)-1}=dfrac{532}{243}
$$

e.) This is a geometric series with $t_1=3.4$ and $r=-dfrac{7.14}{3.4}=-2.1$

The $n^{th}$ term is

$t_n=left(3.4right)cdot left(-2.1right)^{n-1}$

Sixth term: $t_-6=3.4cdot(-2.1)^5approx-138.8594$

The sum of first six terms is

$$
S_6=left(3.4right)cdot dfrac{(-2.1)^6-1}{(-2.1)-1}approx -92.9693
$$

f.) This is a geometric series with $t_1=1$ and $r=dfrac{3x^2}{1}=3x^2$.

The $n^{th}$ term is

$t_n=1cdot(3x^2)^{n-1}=(3x^2)^{n-1}$

Sixth term: $t_6=(3x^2)^{6-1}=3^5x^{10}=243x^{10}$

The sum of first six terms is

$$
S_6=1cdot dfrac{(3x^2)^6-1}{3x^2-1}=dfrac{729x^{12}-1}{3x^2-1}
$$

a.) 6558

b.) 198.4375

c.) 4376

d.) $dfrac{127}{192}$

$bold{Identifying;Series}$ For a series containing $t_1+t_2+t_3+t_4+…$

geometric $implies dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=dfrac{t_n}{t_{n-1}}=r$

arithmetic $implies t_2-t_1=t_3-t_2=t_4-t_3=t_n-t_{n-1}=d$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

a.) Since a constant difference of $d=15-10=5$ exists between consecutive terms, the series is arithmetic.

b.) The ratio between consecutive terms is $r=dfrac{21}{7}=3$, thus it is a geometric series with $t_1=7$.

The sum of first 8 terms is

$S_8=7cdot dfrac{3^8-1}{3-1}=22;960$

c.) The ratio between consecutive terms is $r=-dfrac{512}{2048}=-dfrac{1}{4}$, thus it is a geometric series with $t_1=2048$.

The sum of first 8 terms is

$S_8=2048cdot dfrac{left(dfrac{1}{4}right)^8-1}{dfrac{1}{4}-1}=dfrac{13;107}{8}$

e.) The ratio between consecutive terms is $r=dfrac{1.21}{1.1}=1.1$, thus it is a geometric series with $t_1=1.1$

The sum of first 8 terms is

$S_8=1.1cdot dfrac{1.1^8-1}{1.1-1}=12.579$

f.) Since a constant difference of $d=63-81=-18$ exists between consecutive terms, the series is arithmetic.

a.) arithmetic

b.) 22 960

c.) $dfrac{13;107}{8}$

d.) neither arithmetic nor geometric

e.) 12.579

f.) arithmetic

$bold{Identifying;Series}$ For a series containing $t_1+t_2+t_3+t_4+…$

geometric $implies dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=dfrac{t_n}{t_{n-1}}=r$

arithmetic $implies t_2-t_1=t_3-t_2=t_4-t_3=t_n-t_{n-1}=d$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

a.) In this case, $t_1=13$ , $r=5$, $n=7$

Thus sum of the seven terms is

$$
S_7=13cdot dfrac{5^7-1}{5-1}=253;903
$$

b.) We are given $t_1=11$ and $t_7=704$

From the general term of arithmetic sequence

$t_n=t_1cdot r^{n-1}implies 704=11cdot r^{7-1}implies r=pm 2$

for $r=2implies S_7=11cdot dfrac{2^7-1}{2-1}=1397$

for $r=-2implies S_7=11cdot dfrac{(-2)^7-1}{(-2)-1}=473$

c.) We are given $t_2=30$ and $t_1=120$, so we can get $r=dfrac{30}{120}=dfrac{1}{4}$.

We shall evaluate $S_7$

$$
S_7=120cdot dfrac{left(dfrac{1}{4}right)^7-1}{dfrac{1}{4}-1}=159.9902
$$

d.) Observe that each term is 3 times its preceding term, thus, $r=3$. Since $t_3=18$

We can use the general term for geometric sequence to find the first term $t_1$.

$t_n=t_1cdot r^{n-1}implies 18=t_1cdot r^{3-1}implies t_1=2$

Now, we can use the formula for geometric series to evaluate $S_7$

$$
S_7=2dfrac{3^7-1}{3-1}=2186
$$

e.) By inspection, we can notice that $r=dfrac{2}{3}$. Similar to part(d), we will use $t_8=1024$ to calculate $t_1$

$1024=t_1cdot left(dfrac{2}{3}right)^{8-1}implies t_1=17496$

Now, we can evaluate $S_7$

$$
S_7=17496cdot dfrac{(2/3)^7-1}{(2/3)-1}=49;416
$$

f.) We’re given $t_5=5$ and $t_8=-40$, using the general term for geometric sequence, we calculate $r$ and $t_1$ by setting up two equations with two unknowns.

$-40=t_1cdot r^{8-1}$ and $-8=t_1cdot r^{5-1}$

Equate the $t_1$ from the first equation to the second equation.

$dfrac{-40}{r^7}=dfrac{-8}{r^4}implies r^3=-8implies r=-2$

So $t_1=dfrac{-40}{(-2)^7}=dfrac{5}{16}$

Now, we can evaluate $S_7$

$$
S_7=dfrac{5}{16}cdot dfrac{(-2)^7-1}{(-2)-1}=dfrac{215}{16}
$$

a.) 253 903

b.) 1397 or 473

c.) 159.9902

d.) 2186

e.) 49 416

f.) $dfrac{215}{16}$

$bold{Identifying;Series}$ For a series containing $t_1+t_2+t_3+t_4+…$

geometric $implies dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=dfrac{t_n}{t_{n-1}}=r$

arithmetic $implies t_2-t_1=t_3-t_2=t_4-t_3=t_n-t_{n-1}=d$

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

a.) This is a geometric series with $t_1=1$ and $r=dfrac{6}{1}=6$

General Term: $t_n=1cdot 6^{n-1}=6^{n-1}$

To know what value of $n$ that satisfies $t_n=279;936$, we can use the formula for general term

$279;936=1cdot 6^{n-1}$

$6^7=6^{n-1}implies n=8$

The sum of the series containing 9 terms is

$S_9=1cdot dfrac{6^8-1}{6-1}=335;923$

b.) This is a geometric series with $t_1=960$ and $r=dfrac{480}{960}=0.5$

General Term: $t_n=960cdot (0.5)^{n-1}$

To know what value of $n$ that satisfies $t_n=15$, we can use the formula for general term

$15=960cdot left(dfrac{1}{2}right)^{n-1}$

$dfrac{1}{2^{n-1}}=dfrac{15}{960}implies dfrac{1}{2^{n-1}}=dfrac{1}{2^6}implies n=7$

The sum of the series containing 7 terms is

$S_7=960cdot dfrac{(0.5)^7-1}{0.5-1}=1905$

c.) This is a geometric series with $t_1=17$ and $r=dfrac{-51}{17}=-3$

General Term: $t_n=17cdot (-3)^{n-1}$

To know what value of $n$ that satisfies $t_n=334;611$, we can use the formula for general term

$334;611=17cdot left(-3right)^{n-1}$

$dfrac{334;611}{17}=(-3)^{n-1}$

$implies 19683=(-3)^{n-1}implies (-3)^{9}=(-3)^{n-1}implies n =10$

The sum of the series containing 10 terms is

$S_{10}=17cdot dfrac{(-3)^{10}-1}{(-3)-1}=-250;954$

d.) This is a geometric series with $t_1=24;000$ and $r=dfrac{3;600}{24;000}=0.15$

General Term: $t_n=24;000cdot (0.15)^{n-1}$

To know what value of $n$ that satisfies $t_n=1.8225$, we can use the formula for general term

$1.8225=24;000cdot left(0.15right)^{n-1}$

$0.000;075;937;5=0.15^{n-1}$

$0.15^5=0.15^{n-1}implies n = 6$

The sum of the series containing 6 terms is

$S_{6}=24;000cdot dfrac{(0.15)^{6}-1}{(0.15)-1}=28;234.9725$

e.) This is a geometric series with $t_1=-6$ and $r=-dfrac{24}{6}=-4$

General Term: $t_n=-6cdot (-4)^{n-1}$

To know what value of $n$ that satisfies $t_n=98304$, we can use the formula for general term

$98;304=-6left(-4right)^{n-1}$

$dfrac{98;304}{-6}=(46)^{n-1}implies -16;384=(-4)^{n-1}$

$(-4)^7=(-4)^{n-1}implies n = 8$

The sum of the series containing 8 terms is

$S_{8}=(-6)cdot dfrac{(-4)^{8}-1}{(-4)-1}=78;642$

e.) This is a geometric series with $t_1=4$ and $r=dfrac{2}{4}=0.5$

General Term: $t_n=4cdot (0.5)^{n-1}$

To know what value of $n$ that satisfies $t_n=dfrac{1}{1024}$, we can use the formula for general term

$dfrac{1}{1024}=4left( dfrac{1}{2}right)^{n-1}$

$dfrac{1/1024}{4}=left(dfrac{1}{2}right)^{n-1}implies dfrac{1}{4096}=dfrac{1}{2^{n-1}}$

$dfrac{1}{2^{12}}=dfrac{1}{2^{n-1}}implies n = 13$

The sum of the series containing 13 terms is

$S_{13}=(4)cdot dfrac{(0.5)^{13}-1}{(0.5)-1}=dfrac{8191}{1024}$

a.) 335 923

b.) 1905

c.) $-250;954$

d.) 28 234.9725

e.) 78 642

f.) $dfrac{8191}{1024}$

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

In this case, the maximum height that the ball can achieve is $60%$ of its previous height. The distance traveled for each bounce is the sum of the distance in going up and going down, implying that in each bounce, the ball travels twice its maximum height. Thus, we can model the total distance traveled as a function of $n^{th}$ bounce by a geometric series. Initially, the ball is $3m$ above the ground, so the first bounce has a maximum height of $3times0.6=1.8$. We then have a geometric series with $t_1=1.8$ and $r=0.6$. However, note that the total distance is twice this series (going up and going down) plus the initial 3m where the ball is originally at.

Thus, the distance covered on the 4th bounce is

$D_4=3+2left[ 1.8cdot dfrac{(0.6)^4-1}{0.6-1}right]=10.8336;m$

At the end of 4th bounce, the ball has covered a total of $boxed{bold{10.8336;m}}$.

We can confirm our answer by heuristic approach

$D_4=3+2[3(0.6)+3(0.6)^2+3(0.6)^3+3(0.6)^4]=10.8336$ m

$10.8336$ m

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

When $r=1$, the formula for the geometric series becomes undefined. However, notice that when $r=1$, any term on the series is equal to its previous term, and all terms become identical. This can be written as

$S_n=t_1+t_1+t_1+t_1+…$

$n=2implies t_1+t_1=2t_1$

$n=3implies t_1+t_1+t_1=3t_1$

$n=4implies t_1+t_1+t_1+t_1=4t_1$

Therefore, the sum of the first $n$ terms is $S_n=ncdot t_1$

$$
S_n=ncdot t_1
$$

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

Observe that the number of new segment is twice that of the previous one, suggesting that $r=2$. Initially, 1 segment is drawn, indicating $t_1=1$. Thus, the total number of segments in the $n^{th}$ stage can be modeled as a geometric series. Here, we are interested in the 20th stage, using the formula for the sum of geometric series

$S_{20}=1cdotdfrac{2^{20}-1}{2-1}=1;048;575$

Therefore, the $20^{th}$ stage must have $boxed{bold{1;048;575}}$ line segments.

$1;048;575$ line segments

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

Observe that a right triangle can be formed in each stage with hypotenuse of $t_{n-1}$ and legs of length $t_n$.

Remember that the relationship between hypotenuse and leg is given by the Pythagorean relation

$(t_{n-1})^2=(t_n)^2+(t_n)^2implies (t_{n-1})^2=2(t_n)^2=t_n=dfrac{1}{sqrt{2}}cdot (t_{n-1})^{1/2}$

This is a recursive form of a geometric sequence with $r=dfrac{1}{sqrt{2}}$.

Since the length of initial square=1, we can write the length of square added in each stage $t_n$ as

$t_n=1cdot left(dfrac{1}{sqrt{2}}right)^{n-1}=(sqrt{0.5})^{n-1}$

We shall then express the area added in each stage for every square in the previous stage as a function of $n$.

Notice that in each stage, 2 squares are added with side $t_n$ and an isosceles right triangle with leg $t_n$.

Remember that

$A_{square}=(side)^2$ ; $A_{triangle}=dfrac{1}{2}(base)(height)$

Thus, the additional area in each stage for every square in the previous stage is the sum of the areas of two squares and one triangle which can be expressed as

$A_n=2(t_n)^2+dfrac{1}{2}(t_n)(t_n)=2.5t_n^2$

The general term for the length of side can be substituted to get

$A_n=2.5left((sqrt{0.5})^{n-1}right)^2=2.5cdot 0.5^{n-1}$

This corresponds to the area being added on the $n^{th}$ stage for every square added on the $(n-1)^{th}$ stage.

Now we want to know how many new squares are added in each stage. Observe that two squares are added for
every square in the previous stage, thus, it is a geometric sequence with $t_1=1$ and $r=2$

The number of squares added on the $n^{th}$ stage is

$S_n=1cdot 2^{n-1}=2^{n-1}$.

The area added in the $n^{th}$ stage (we will call $T_n$) is the product of the number of squares in the $(n-1)$ stage times the area that results from each added square.

In this case, this is the product of $A_n$ and $S_{n-1}$

$T_{n}=A_ncdot S_{n-1}=(2.5cdot 0.5^{n-1})cdot 2^{n-2}=2.5cdot dfrac{2^{n-2}}{2^{n-1}}=2.5cdot 2^{-1}=1.25$

Thus, the area added in each stage is $1.25$ $m^2$ and is not a function of $n$.

The total area on the $n^{th}$ stage $Z_{n}$ is the sum of the area of the first square and the additional area resulting from the 9 remaining stages

$Z_n=1;m^2+(9 ;text{stages})(1.25; m^2; text{per stage})=12.25;m^2$

Therefore, at the $10^{th}$ stage, the total area is $boxed{bold{12.25;m^2}}$.

$$
12.25;m^2
$$

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

There are 5 employees in the first level and the number of employees in each level is 3 times that of the previous level. Thus, we can model the total number of employees in the first $n$ levels as geometric series with $t_1=5$ and $r=3$.

Using the formula for the sum of geometric series, we have

$S_{7}=5cdotdfrac{3^7-1}{3-1}=5;465$

Thus, there are $boxed{bold{5,465;employees}}$ in this company.

$5;465$ employees

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

If the order of terms in geometric series is reversed, the last term becomes the first term, and the common ratio of the new series is the reciprocal of the original series. In this case, the original series has $r_{o}=dfrac{5}{3}$. Thus, the new reversed series has $r=dfrac{3}{5}$ with $t_1=5;859;375$.

Therefore, if there are $10$ terms in the series, the sum is

$S_{10}=5;859;375dfrac{(3/5)^{10}-1}{(3/5)-1}=14;559;864$

$$
14;559;864
$$

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

The series is subject to the following constraints:

(a) The total must not be more than $$2;000;000$

(b) The first term $t_1$ must be a positive integer

(c) If terms are arranged in increasing order, it must form a geometric series with $2<r<10$

We shall find the possible values of $n, r,$ and $t_1$ that will satisfy the conditions above using the formula for the sum of geometric series:

$S_n=t_1cdot dfrac{r^n-1}{r-1}$

$$
2;000;000geq t_1cdot dfrac{r^{n}-1}{r-1}
$$

There are many possible combinations of $n,r$ and $t_1$ that would satisfy the conditions.

For instance, if $r=4$ and $n=6$, we can find $t_1$ as

$t_1leq dfrac{2;000;000}{dfrac{4^6-1}{4-1}}$

$$
t_1 leq 1465.20
$$

$bold{Arithmetic}$

$S_n=dfrac{n}{2}(t_1+t_n)$

(1) The difference in consecutive terms is constant

(2) If we write the terms in forward and reverse order, the sum of corresponding pairs is constant.

$bold{Geometric}$

$S_n=dfrac{t_{n+1}-t_1}{r-1}, rneq 1$

(1) The ratio between consecutive terms is constant.

(2) If we write the terms in forward and reverse order, the product of corresponding pairs is constant.

$bold{Similarities}$

(1) Both equations contains first term and another term at a particular $n$.

(2) Both equations involve division operation.

$bold{Differences}$

(1) The denominator in arithmetic series is $2$ while for geometric is $r-1$.

(2) The arithmetic series formula involves addition operation while that of geometric series involves substraction.

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

We are given with $S_3=372$ and $t_1=12$, we shall find $r$

$372=12cdot dfrac{r^3-1}{r-1}$

$31=dfrac{r^3-1}{r-1}$

$31(r-1)=r^3-1$

Remember the factor for the difference of two cubes

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$31(r-1)=(r-1)(r^2+r+1^2)implies r=1$

$31=r^2+r+1$

$r^2+r-30=0$

Solve this quadratic equation by factoring

$(r+6)(r-5)=0$

$r=-6$ or $r=5$ or $r=1$.

We will see which value of $r$ would maximize $t_5$.

$r=-6implies t_5=12(-6)^{5-1}=15;552$

$r=5implies 12(5)^{5-1}=7500$

$r=1implies 12cdot (5)^{1-1}=12$

Therefore, the maximum possible value of the fifth term is $boxed{bold{15;552}}$

$$
15;552
$$

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

Given that $t_1=23$ and $t_3=92$, we can use the general term of geometric sequence to solve for $r$

$$
92=23cdot r^{3-1}implies r^2=dfrac{92}{23}implies r^2=4implies r=pm sqrt{2}
$$

We need to find the value of $n$ such that $S_n=62;813$. We can use the formula for the sum of geometric series.

Case 1: $r=2$

62 813 = $23cdot dfrac{(2)^n-1}{(2)-1}$

$$
2731=dfrac{(2)^n-1}{2-1}implies 2^n=2732
$$

Remember that $n$ should take discrete integer value and $2^{10}=1024$ and $2^{11}=2048$. Thus, no value of $n$ can satisfy this.

Case 2: $r=-2$

$62;813=23cdot dfrac{(-2)^n-1}{(-2)-1}$

$-8193=(-2)^{n}-1implies (-2)^n=-8192$

$n=13$

Hence $r=-2$ and there are $boxed{bold{13}}$ terms in the series.

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

In this geometric series

$1+x+x^2+…+x^{14}$

The first term is $t_1=1$ and common ratio $r=x$.

The sum of the 15 terms is

$S_{15}=1+x+x^2+…+x^{14}=1cdot dfrac{x^{15}-1}{x-1}$

Therefore

$x^{15}-1=(x-1)(1+x+x^2+…+x^{14})$

$$
x^{15}-1=(x-1)(1+x+x^2+…+x^{14})
$$

$bold{General;Term;of;Geometric;Sequence:}$ The $n^{th}$ term of a geometric sequence with first term $t_1$ and common ratio $r$ is

$t_n=t_1cdot r^{n-1}$

$bold{Sum;of;Geometric;Series:}$ The sum of first $n$ terms in a geometric series with first term $t_1$ and common ratio $r$ is

$$
S_n=t_1cdot dfrac{r^n-1}{r-1}
$$

b.) This appears to be a geometric series with $t_1=dfrac{1}{2}$ and $r=dfrac{1/4}{1/2}=0.5$

We shall use the formula for the sum of geometric series as

$$
S_n=0.5cdot dfrac{0.5^n-1}{0.5-1}=-(0.5^n-1)=1-0.5^n
$$

c.)

$n=2implies S_n=1-(0.5)^2=0.75$

$n=5implies S_n=1-(0.5)^5=0.9922$

$n=20implies S_n=1-(0.5)^{20}=0.99999$

From the obtained expression for $S_n$, notice that as $n$ increases, $S_n$ approaches $1$. We would expect that the sum of this infinite series is 1.

a.) see figure

b.) $S_n=1-0.5^n$

c.) $S_{infty}=1$