Geometric Sequences & Series - Algebra 3

1. A ball is dropped from a height of 8 meters. On each bounce it rises to $60\%$ of the height from which it just fell. What total vertical distance does the ball travel before coming to rest?
   Answer: $32$ m Whenever you see a bouncing ball or any repeated-motion distance problem, think geometric series. The key is to track every segment: the ball drops, then rises and falls repeatedly until the bounces become negligible. The initial drop is $8$ meters. After that, each bounce reaches $60\%$ of the previous height, so the rebound heights are $8(0.6), 8(0.6)^2, 8(0.6)^3, \dots$. For each rebound height, the ball travels up that distance and then down that same distance, so each contributes twice its height. The total distance is therefore $$8 + 2\left(8(0.6) + 8(0.6)^2 + 8(0.6)^3 + \cdots\right).$$ The infinite sum in parentheses is a geometric series with first term $8(0.6)=4.8$ and ratio $0.6$. Its sum is $\frac{4.8}{1-0.6}=12$. Twice that is $24$, and adding the initial $8$ gives $$8 + 24 = 32\text{ m}.$$ So $32$ m is correct. The wrong answers show common traps. $12.8$ m is just the first drop plus the first rise, ignoring all later bounces. $20$ m is the sum of only the falling distances $8+4.8+2.88+\cdots$, forgetting that every upward bounce also counts. $24$ m correctly sums all the up-and-down bounce distances after the first drop but forgets to include the initial $8$-meter drop. Your takeaway: physically list out the motion first — down, up, down, up — then form a geometric series. Include the initial drop, and remember to double every rebound because each bounce has both an up and a down segment.
2. A geometric series has first term $48$ and common ratio $\frac{1}{2}$. What is the least number of terms for which the sum exceeds $90$?
   Answer: $5$ The partial sum is $S_n=48\frac{1-(1/2)^n}{1-1/2}=96\left(1-2^{-n}\right)$. For $n=4$, $S_4=96(1-1/16)=90$, which does not exceed 90. For $n=5$, $S_5=96(1-1/32)=93$, so the least value is 5. The value 4 corresponds to reaching exactly 90 rather than exceeding it.
3. A geometric sequence has first term $5$, common ratio $2$, and last term $160$. How many terms are in the sequence?
   Answer: $6$ The last term satisfies $5(2)^{n-1}=160$, so $2^{n-1}=32=2^5$. Thus $n-1=5$, so $n=6$. The value 5 comes from forgetting to add 1 after solving the exponent; 7 and8 come from off-by-one errors in the exponent.
4. How many terms are in the geometric sequence $3,\;6,\;12,\;\ldots,\;768$?
   Answer: $9$ Whenever you see a geometric sequence, start with the general term formula: $a_n = a_1 r^{n-1}$. Here $a_1 = 3$, and each term doubles, so $r=2$. You need the term number $n$ such that $a_n = 768$. Set up the equation: $$3(2^{n-1}) = 768$$ Divide by 3: $$2^{n-1} = 256$$ Since $256 = 2^8$, you get $n-1 = 8$, so $n = 9$. The sequence is $3, 6, 12, 24, 48, 96, 192, 384, 768$, which has exactly 9 terms. The choice $8$ is the exponent $n-1$ — it counts the number of doublings after the first term, not the number of terms. The choice $7$ likely comes from counting only the terms strictly between $3$ and $768$, leaving out both endpoints. The choice $10$ comes from adding an extra term, such as continuing to $1536$ or counting the first term twice. The key habit: in a geometric sequence, the exponent on $r$ is always one less than the term number. So when you solve and get an exponent of 8, remember to add 1 back to find 9 terms. That off-by-one trap is exactly what this question tests.
5. Find the sum of the first 8 terms of the geometric sequence $3,\;-6,\;12,\;-24,\;\ldots$.
   Answer: $-255$ Whenever you see a geometric sequence, first identify the common ratio by dividing any term by the previous one. Here $-6/3=-2$, so each term is multiplied by $-2$. The sum of the first $n$ terms is $$S_n=a_1\frac{1-r^n}{1-r}.$$ With $a_1=3$, $r=-2$, and $n=8$, $$S_8=3\frac{1-(-2)^8}{1-(-2)} =3\frac{1-256}{3} =-255.$$ So the 8-term sum is $-255$. Notice $(-2)^8$ is positive $256$, not $-128$; using the 7th power gives the sum of only 7 terms, which is $129$. That is the trap behind the $129$ choice: it stops one term too soon. The choice $255$ has the right magnitude but the wrong sign; because the 8th term is $-384$, adding it to the 7-term sum $129$ gives $-255$, not $255$. The choice $-129$ is likely the result of taking the 7-term sum $129$ and flipping its sign, or mixing up the formula's numerator and denominator. To stay safe, use the formula with $n=8$, and check whether an even number of alternating terms should make the total negative.
6. The second term of an infinite geometric series is $-15$ and the third term is $7.5$. What is the sum of the series?
   Answer: $20$ Whenever you see an infinite geometric series, the key is to find the common ratio first, then the first term. Here, the third term divided by the second term gives the ratio: $r=\frac{7.5}{-15}=-0.5$ Since the second term is $ar=-15$, we have $a(-0.5)=-15 \quad\Rightarrow\quad a=30$ Now apply the infinite sum formula: $S=\frac{a}{1-r}=\frac{30}{1-(-0.5)}=\frac{30}{1.5}=20$ So the sum is indeed $20$. Why the others miss the mark: - $-60$ comes from using $r=0.5$ instead of $-0.5$, which ignores the alternating sign. That would give $a=-30$ and sum $-60$. - $-10$ treats the second term $-15$ as if it were the first term: $-15/(1-(-0.5))=-10$. But you must divide by the ratio to recover the actual first term. - $5$ treats the third term $7.5$ as the first term: $7.5/(1-(-0.5))=5$. This skips the step of finding $a$ entirely. A good habit: always compute $r$ first, then use a given term to find $a$ before applying $S=\frac{a}{1-r}$. Also watch the sign of $r$ — a negative ratio means the series alternates, and ignoring that sign is a common trap.
7. An infinite geometric series has sum $12$ and common ratio $-\frac13$. What is its second term?
   Answer: $-\frac{16}{3}$ Whenever an infinite geometric series gives you its sum and common ratio, your first move is to recover the first term using the formula $S = \frac{a}{1-r}$. Here $S=12$ and $r=-\frac13$, so $$a = S(1-r) = 12\left(1+\frac13\right)=16.$$ That $16$ is the first term, not the answer. The second term is the first term multiplied by the common ratio: $$ar = 16 \cdot \left(-\frac13\right) = -\frac{16}{3}.$$ Now see why the other choices miss the mark. $-4$ comes from multiplying the sum $12$ by the ratio $-\frac13$; that conflates the total sum with the first term. $\frac{16}{9}$ is actually the third term, $ar^2 = 16 \cdot \frac19$; it confuses the term position with the exponent and also drops the negative sign. $16$ is the first term you computed correctly, but the question asks for the second term, not the first. On exam day, write the sum formula first, solve for $a$, then take one more step to $ar$. Remember: the infinite sum formula gives you the first term, but the question may ask for a different term. Always check the requested term number and use $ar^{n-1}$.
8. The $n$th partial sum of a geometric series is $S_n=3(2^n-1)$. What is the fourth term $a_4$?
   Answer: $24$ When you are given a formula for the partial sum $S_n$, you are being tested on the relationship between a running total and a single term. The key move is to remember that the $n$th term is the difference between consecutive partial sums: $$a_n = S_n - S_{n-1}$$ For this geometric series, $$S_4 = 3(2^4-1)=3(16-1)=45$$ and $$S_3 = 3(2^3-1)=3(8-1)=21$$ So the fourth term is $$a_4 = S_4 - S_3 = 45 - 21 = 24.$$ The choice $45$ is the trap of reporting the partial sum $S_4$ itself, not the fourth term. The choice $21$ is $S_3$, the sum of the first three terms, not the fourth term. The choice $48$ comes from incorrectly using $a_4 = 3(2^4)=48$, treating the exponent as $n$ instead of $n-1$ for a term; even in the sum formula, the exponent inside is $n$, but the individual term is $3 \cdot 2^{3}$. A quick study tip: whenever a question gives you partial sums, immediately write $a_n = S_n - S_{n-1}$. That single formula converts a potentially tricky series question into a simple subtraction problem.
9. An infinite geometric series has sum $12$ and common ratio $-\frac{1}{3}$. What is the first term?
   Answer: $16$ Since $S_\infty=\frac{a_1}{1-r}$, we have $12=\frac{a_1}{1-(-1/3)}=\frac{a_1}{4/3}$, so $a_1=16$. The value $-16$ has the wrong sign; $-\frac{16}{3}$ is actually the second term, $16(-1/3)=-16/3$; $-4$ comes from computing $S_\infty r=12(-1/3)=-4$.
10. For what values of $x$ does the infinite geometric series $\displaystyle \sum_{n=0}^{\infty}(2x-1)^n$ converge?
    Answer: $0<x<1$ The ratio is $2x-1$. A geometric series converges only when $|2x-1|<1$, so $-1<2x-1<1$, which gives $0<x<1$. The interval $0\le x\le1$ includes endpoints where $|r|=1$; $x<1$ omits the lower bound; $0\le x<1$ incorrectly includes $x=0$, where $r=-1$.
11. For a geometric series, the sum of the first $2$ terms is $8$, the sum of the first $4$ terms is $80$, and the first term is positive. What is the common ratio?
    Answer: $3$ Let $a$ be the first term and $r$ the ratio. Then $a(1+r)=8$ and $a(1+r+r^2+r^3)=80$. Dividing gives $1+r^2=10$, so $r^2=9$, meaning $r=\pm3$. If $r=-3$, then $a=8/(1-3)=-4$, contradicting the positivity of the first term, so $r=3$. The value $-3$ ignores that positivity condition; $10$ comes from treating $80/8$ as the ratio; $3$ or $-3$ forgets to check the first-term sign.
12. In a geometric sequence, $a_2=36$ and $a_5=\frac{4}{3}$. What is $a_{10}$?
    Answer: $\frac{4}{729}$ Since $\frac{a_5}{a_2}=r^3=\frac{4/3}{36}=\frac{1}{27}$, we get $r=\frac13$. Then $a_1=\frac{a_2}{r}=108$, so $a_{10}=108\left(\frac13\right)^9=\frac{4}{729}$. The value $\frac{4}{2187}$ comes from treating the given $a_2=36$ as $a_1$ and applying $r^9$; $\frac{4}{243}$ is $a_9$, from using exponent 8 instead of 9 (or $r^4$ from $a_5$); $\frac{4}{81}$ comes from computing $a_5r^3$ instead of $a_5r^5$.
13. In a geometric sequence, $a_4=40$ and $a_6=160$. If the sum of the first $n$ terms is $2555$, what is $n$?
    Answer: $9$ Since $a_6/a_4=r^2=4$, $r=\pm 2$. If $r=-2$, then $a_1=-5$ and $S_n=-\frac{5}{3}(1-(-2)^n)$; odd $n$ gives a negative sum, while even $n$ would require $2^n=1534$, impossible. Thus $r=2$, so $a_1=5$, and $5(2^n-1)=2555$. Therefore $2^n=512$, so $n=9$. The value 10 may come from solving $2^{n-1}=512$; 8 and 11 are common index/exponent slips.
14. In a geometric sequence, $a_3=12$ and $a_6=96$. What is the sum of the first $8$ terms?
    Answer: $765$ Since $a_6/a_3=r^3=8$, $r=2$. Then $a_1=a_3/r^2=12/4=3$, so $S_8=3(2^8-1)=3(255)=765$. The value $381$ is $S_7$, using one fewer term; $1530$ comes from using $a_1=6$ instead of 3; $3060$ comes from using $a_1=12$ instead of 3.
15. Evaluate $\displaystyle \sum_{k=1}^{10}5(-2)^{k-1}$.
    Answer: $-1705$ Here $a_1=5$, $r=-2$, and $n=10$. Thus $S_{10}=5\frac{1-(-2)^{10}}{1-(-2)}=5\frac{1-1024}{3}=-1705$. The value $-341$ omits the leading factor 5; $341$ omits that factor and also loses the sign; $1705$ keeps the factor but loses the sign.
16. For which values of $x$ does the infinite series $\sum_{n=1}^{\infty}\left(\frac{3}{x-1}\right)^{n-1}$ converge, and what is its sum?
    Answer: $x<-2$ or $x>4$; $\frac{x-1}{x-4}$ Whenever you see an infinite series whose terms are powers of a fixed expression, think geometric series: it converges exactly when $|r|<1$, and its sum is $\frac{a}{1-r}$. Here the first term is $\left(\frac{3}{x-1}\right)^0=1$, and the common ratio is $r=\frac{3}{x-1}$. So convergence requires $$\left|\frac{3}{x-1}\right|<1 \quad \Rightarrow \quad |x-1|>3.$$ This means $x-1>3$ or $x-1<-3$, giving the interval $x<-2$ or $x>4$. For those $x$, the sum is $$\frac{1}{1-\frac{3}{x-1}}=\frac{x-1}{x-4}.$$ So the correct choice is the one with interval $x<-2$ or $x>4$ and sum $\frac{x-1}{x-4}$. The choice with interval $-4<x<2$ misplaces the center of the inequality: the condition is about distance from $1$, not from $-1$, and it must be greater than $3$. The choice with sum $\frac{x-1}{x+4}$ uses the correct interval but adds the ratio in the denominator instead of subtracting it: $1-r$, not $1+r$. The choice $-2<x<4$ reverses the inequality; that interval actually makes $|r|>1$, so the series diverges there. Study tip: for any geometric series, write down $a$ and $r$ first. Then solve $|r|<1$ carefully, and simplify $\frac{a}{1-r}$ separately. That keeps the interval and the sum from getting tangled together.
17. The repeating decimal $0.272727\ldots$ can be written as an infinite geometric series. Which fraction equals this decimal?
    Answer: $\frac{3}{11}$ Whenever you see a repeating decimal, think of it as an infinite geometric series whose common ratio is a power of $\frac{1}{10}$. Here, $$0.272727\ldots = 0.27 + 0.0027 + 0.000027 + \cdots$$ This is a geometric series with first term $a=\frac{27}{100}$ and common ratio $r=\frac{1}{100}$. Its sum is $$\frac{a}{1-r}=\frac{\frac{27}{100}}{1-\frac{1}{100}} =\frac{\frac{27}{100}}{\frac{99}{100}} =\frac{27}{99} =\frac{3}{11}.$$ So $\frac{3}{11}$ is the correct fraction. The choice $\frac{27}{100}$ is just the first two decimal places, not the full repeating value — it equals $0.27$, not $0.272727\ldots$. The choice $\frac{5}{18}$ equals $0.2777\ldots$, which repeats a single $7$, not the two-digit block $27$. And $\frac{1}{4}$ equals $0.25$, which is close but definitely not the repeating pattern. A quick study tip: for a decimal with a repeating block of length $n$, write the block over $n$ nines and simplify. Since $27$ repeats, $\frac{27}{99}=\frac{3}{11}$. This shortcut saves time and helps you check your geometric-series work.
18. In a geometric sequence, $a_2=6$ and $a_5=48$. What is $a_7$?
    Answer: $192$ Whenever you see a geometric sequence, isolate the common ratio $r$ first. Between $a_2=6$ and $a_5=48$, there are 3 steps, so $6r^3=48 \quad\Rightarrow\quad r^3=8 \quad\Rightarrow\quad r=2.$ Now $a_7$ is 2 steps after $a_5$, so multiply by $r^2$: $a_7=48\cdot 2^2=48\cdot4=192.$ That is the correct value. The choice $96$ is one step too soon — it is actually $a_6=48\cdot2$, not $a_7$. The choice $384$ comes from multiplying by $2^3=8$ instead of $2^2=4$; that would land on $a_8$, or it happens if you count terms from $a_5$ to $a_7$ inclusively rather than counting the gaps between them. The choice $48$ simply repeats $a_5$, ignoring the common ratio entirely. Remember: in a geometric sequence, the exponent is the number of gaps, not the number of terms listed. To go from $a_5$ to $a_7$ is 2 gaps, so use $r^2$. Always confirm how many steps you are moving before multiplying.
19. If $x-2,\;2x+1,\;4x+5$ are consecutive terms of a geometric sequence, what is the value of $x$?
    Answer: $-\frac{11}{7}$ When you see consecutive terms of a geometric sequence, remember that the ratio between each pair of consecutive terms is constant. So here, $\frac{2x+1}{x-2} = \frac{4x+5}{2x+1}$. Cross-multiplying gives $(2x+1)^2 = (x-2)(4x+5)$. Expand carefully: $4x^2+4x+1 = 4x^2-3x-10$. Cancel the $4x^2$ terms: $4x+1 = -3x-10$, so $7x = -11$, and $x = -\frac{11}{7}$. This checks: the terms become $-\frac{25}{7}, -\frac{15}{7}, -\frac{9}{7}$, each multiplied by $\frac35$. The other choices come from common traps. $-\frac{11}{3}$ is what you get if you expand $(2x+1)^2$ incorrectly as $4x^2+1$, dropping the middle term $4x$. $-1$ makes the sequence arithmetic, not geometric: $-3, -1, 1$ have a common difference, not a common ratio. $-\frac{7}{11}$ is the reciprocal of the correct answer, suggesting the equation $7x=-11$ was solved by flipping the fraction instead of dividing by 7. On exam day, set up the ratio equality first, then expand carefully—especially when squaring binomials. A quick check of the three terms with your answer confirms whether the common ratio is constant.
20. A geometric sequence has common ratio $3$ and $a_4=54$. Which of the following is an explicit rule for $a_n$?
    Answer: $a_n=2(3)^{n-1}$ The first term is $a_1=a_4/r^3=54/27=2$, so the explicit rule is $a_n=2(3)^{n-1}$. The form $6(3)^{n-1}$ comes from dividing by $r^2$ rather than $r^3$; $18(3)^{n-1}$ comes from dividing by r only; $54(3)^{n-1}$ treats $a_4$ as though it were $a_1$.

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