Geometric Sequences & Series - IB Mathematics Analysis and Approaches

1. In a convergent geometric series of positive terms, the sum is 16 and the sum of the first two terms is 12. What is the third term, $u_3$?
   Answer: 2 Let the first term be $u_1$ and the common ratio be $r$. We are given $S_\infty = \frac{u_1}{1-r} = 16$ and $S_2 = u_1 + u_1r = u_1(1+r) = 12$. From the first equation, $u_1 = 16(1-r)$. Substitute this into the second equation: $16(1-r)(1+r) = 12$. This simplifies to $16(1-r^2) = 12$, so $1-r^2 = \frac{12}{16} = \frac{3}{4}$. This gives $r^2 = 1 - \frac{3}{4} = \frac{1}{4}$. So, $r = \pm \frac{1}{2}$. Since the terms are all positive, the common ratio $r$ must be positive, so $r = \frac{1}{2}$. Now we can find $u_1$ using $u_1(1+r)=12$: $u_1(1+\frac{1}{2}) = 12 \Rightarrow u_1(\frac{3}{2}) = 12 \Rightarrow u_1 = 12 \times \frac{2}{3} = 8$. The third term is $u_3 = u_1 r^2 = 8 \times (\frac{1}{2})^2 = 8 \times \frac{1}{4} = 2$.
2. A geometric sequence has first term $u_1 = \sin \theta$ and common ratio $r = \sin \theta$, for $0 < \theta < \frac{\pi}{2}$. The sum to infinity of the sequence is 2. Find the value of $\cos^2 \theta$.
   Answer: $\frac{5}{9}$ The sum to infinity of a geometric sequence is given by $S_\infty = \frac{u_1}{1-r}$. We are given $u_1 = \sin \theta$, $r = \sin \theta$, and $S_\infty = 2$. The condition for convergence, $|r|<1$, is satisfied because for $0 < \theta < \frac{\pi}{2}$, we have $0 < \sin \theta < 1$. Substituting the given values into the formula: $2 = \frac{\sin \theta}{1 - \sin \theta}$. Now, we solve for $\sin \theta$. $2(1 - \sin \theta) = \sin \theta \Rightarrow 2 - 2\sin \theta = \sin \theta \Rightarrow 2 = 3\sin \theta \Rightarrow \sin \theta = \frac{2}{3}$. The question asks for $\cos^2 \theta$. Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we have $\cos^2 \theta = 1 - \sin^2 \theta$. Substituting the value of $\sin \theta$: $\cos^2 \theta = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}$.
3. An infinite geometric series has a sum of 12 and a common ratio of $-\frac{1}{2}$. Find the sum of the first 4 terms.
   Answer: $\frac{45}{4}$ First, find the first term $u_1$ using the sum to infinity formula $S_\infty = \frac{u_1}{1-r}$. We have $12 = \frac{u_1}{1 - (-1/2)} = \frac{u_1}{3/2}$. Solving for $u_1$ gives $u_1 = 12 \times \frac{3}{2} = 18$. Now, use the formula for the sum of the first $n$ terms, $S_n = \frac{u_1(1-r^n)}{1-r}$. For $n=4$, $S_4 = \frac{18(1 - (-\frac{1}{2})^4)}{1 - (-\frac{1}{2})} = \frac{18(1 - \frac{1}{16})}{\frac{3}{2}} = \frac{18(\frac{15}{16})}{\frac{3}{2}} = 18 \times \frac{15}{16} \times \frac{2}{3} = 6 \times \frac{15}{16} \times 2 = \frac{180}{16} = \frac{45}{4}$.
4. Given the geometric series $e^x + e^{2x} + e^{3x} + \dots$, for which values of $x$ does this series have a finite sum?
   Answer: $x < 0$ This is a geometric series with first term $u_1 = e^x$. The common ratio is $r = \frac{e^{2x}}{e^x} = e^x$. An infinite geometric series converges (has a finite sum) if and only if the absolute value of its common ratio is less than 1, i.e., $|r| < 1$. In this case, we need $|e^x| < 1$. Since the exponential function $e^x$ is always positive for any real $x$, this condition simplifies to $0 < e^x < 1$. To solve for $x$, we take the natural logarithm of all parts of the inequality. As $\ln(x)$ is an increasing function, the inequality signs are preserved. $\ln(e^x) < \ln(1)$. This simplifies to $x < 0$. The lower bound $e^x > 0$ is true for all real $x$ and does not impose a further constraint.
5. The sum of the first $n$ terms of a sequence is given by $S_n = 3^n - 1$. Which statement is true for the sequence $u_n$?
   Answer: The sequence is geometric with common ratio 3. When you're given a sum formula for the first n terms of a sequence, you need to find the individual terms to determine the sequence's properties. The key insight is that $u_n = S_n - S_{n-1}$ for $n \geq 2$, and $u_1 = S_1$. Let's find the terms systematically. For $n = 1$: $u_1 = S_1 = 3^1 - 1 = 2$. For $n \geq 2$: $u_n = S_n - S_{n-1} = (3^n - 1) - (3^{n-1} - 1) = 3^n - 3^{n-1} = 3^{n-1}(3-1) = 2 \cdot 3^{n-1}$. This gives us $u_1 = 2$, $u_2 = 2 \cdot 3^1 = 6$, $u_3 = 2 \cdot 3^2 = 18$, $u_4 = 2 \cdot 3^3 = 54$, etc. Notice that each term is 3 times the previous term: $\frac{u_2}{u_1} = \frac{6}{2} = 3$, $\frac{u_3}{u_2} = \frac{18}{6} = 3$. This confirms the sequence is geometric with common ratio 3, making D correct. A is wrong because the differences aren't constant (6-2=4, 18-6=12, etc.), so it's not arithmetic. B is incorrect since we've established it is geometric. C incorrectly states the common ratio as 2 when we calculated it to be 3. Study tip: When given $S_n$, always use $u_n = S_n - S_{n-1}$ to find the general term. Check if consecutive terms have a constant ratio (geometric) or constant difference (arithmetic) to classify the sequence type.
6. The sum of an infinite geometric series is $S$ and its first term is $u_1$. A new series is formed by squaring each term of the original series. What is the sum of this new series?
   Answer: $\frac{u_1 S^2}{2S-u_1}$ The original series has first term $u_1$ and common ratio $r$. Its sum is $S = \frac{u_1}{1-r}$. The new series formed by squaring each term is $u_1^2, (u_1r)^2, (u_1r^2)^2, \dots$. This is a geometric series with first term $u_1^2$ and common ratio $r^2$. Since $|r|<1$ for the original series to converge, we have $r^2 < 1$, so the new series also converges. The sum of the new series is $S_{new} = \frac{(u_1)^2}{1-r^2} = \frac{u_1^2}{(1-r)(1+r)}$. From the original sum formula, we can express $r$ in terms of $S$ and $u_1$: $1-r = \frac{u_1}{S} \Rightarrow r = 1 - \frac{u_1}{S}$. Now we find $1+r$: $1+r = 1 + (1-\frac{u_1}{S}) = 2 - \frac{u_1}{S} = \frac{2S-u_1}{S}$. Substitute these into the formula for $S_{new}$: $S_{new} = \frac{u_1^2}{(\frac{u_1}{S})(\frac{2S-u_1}{S})} = \frac{u_1^2}{\frac{u_1(2S-u_1)}{S^2}} = u_1^2 \cdot \frac{S^2}{u_1(2S-u_1)} = \frac{u_1 S^2}{2S-u_1}$.
7. A ball is dropped from a height of 10m. After hitting the ground, it rebounds to 80% of its previous height. This process continues until the ball comes to rest. Calculate the total vertical distance travelled by the ball.
   Answer: 90 m The total vertical distance is the initial drop plus the sum of all subsequent upward and downward movements. Initial drop = 10 m. The heights of the rebounds form a geometric sequence: $10 \times 0.8 = 8$, $8 \times 0.8 = 6.4$, and so on. The first term is $u_1 = 8$ and the common ratio is $r = 0.8$. The total distance travelled after the first drop is the sum of the upward and downward paths. The sum of the heights of the rebounds is an infinite geometric series: $S_\infty = \frac{u_1}{1-r} = \frac{8}{1-0.8} = \frac{8}{0.2} = 40$ m. This sum represents the total distance travelled upwards. The ball must also travel this same distance downwards. So, the total distance after the initial drop is $2 \times 40 = 80$ m. The total vertical distance is the initial drop plus this amount: $10 + 80 = 90$ m.
8. The third term of a geometric sequence is 12 and the sixth term is 96. Find the sum of the first eight terms.
   Answer: 765 Let the first term be $u_1$ and the common ratio be $r$. We are given $u_3 = u_1 r^2 = 12$ and $u_6 = u_1 r^5 = 96$. Dividing the second equation by the first gives $\frac{u_1 r^5}{u_1 r^2} = \frac{96}{12}$, which simplifies to $r^3 = 8$, so $r=2$. Substitute $r=2$ into the first equation: $u_1(2^2) = 12 \Rightarrow 4u_1 = 12 \Rightarrow u_1 = 3$. The sum of the first eight terms is given by the formula $S_n = \frac{u_1(r^n - 1)}{r-1}$. So, $S_8 = \frac{3(2^8 - 1)}{2-1} = 3(256 - 1) = 3(255) = 765$.
9. For what range of values of $x$ does the infinite geometric series $1 + (2x-3) + (2x-3)^2 + \dots$ converge?
   Answer: $1 < x < 2$ An infinite geometric series converges if and only if the absolute value of the common ratio $r$ is less than 1. In this series, the first term is 1 and the common ratio is $r = 2x-3$. So, for convergence, we must have $|2x-3| < 1$. This inequality can be written as $-1 < 2x-3 < 1$. To solve for $x$, we add 3 to all parts of the inequality: $-1+3 < 2x < 1+3$, which simplifies to $2 < 2x < 4$. Finally, we divide all parts by 2: $1 < x < 2$.
10. The sum of the first $n$ terms of a sequence is given by the formula $S_n = 5(2^n - 1)$. Find the 7th term, $u_7$.
    Answer: 320 The $n$-th term of a sequence can be found using the formula $u_n = S_n - S_{n-1}$. To find the 7th term, $u_7$, we calculate $S_7 - S_6$. $S_7 = 5(2^7 - 1) = 5(128 - 1) = 5(127) = 635$. $S_6 = 5(2^6 - 1) = 5(64 - 1) = 5(63) = 315$. Therefore, $u_7 = S_7 - S_6 = 635 - 315 = 320$. Alternatively, one can identify the sequence. $u_1 = S_1 = 5(2^1-1)=5$. $u_2 = S_2 - S_1 = 5(2^2-1) - 5 = 15 - 5 = 10$. This is a geometric sequence with $u_1=5$ and $r=2$. The 7th term is $u_7 = u_1 r^{7-1} = 5 \times 2^6 = 5 \times 64 = 320$.
11. A geometric series has first term $\ln x$ and common ratio 2. The sum of the first three terms is $\ln(128)$. Find the value of $x$.
    Answer: 2 The first three terms of the sequence are $u_1 = \ln x$, $u_2 = 2\ln x = \ln(x^2)$, and $u_3 = 4\ln x = \ln(x^4)$. The sum of the first three terms is $S_3 = \ln x + 2\ln x + 4\ln x = 7\ln x$. We are given that this sum is equal to $\ln(128)$. So, $7\ln x = \ln(128)$. Using the logarithm property $b \ln a = \ln(a^b)$, we have $\ln(x^7) = \ln(128)$. Since $128 = 2^7$, this becomes $\ln(x^7) = \ln(2^7)$. Therefore, $x^7 = 2^7$, which implies $x=2$.
12. The product of the first three terms of a geometric sequence is 216. The sum of these three terms is 26. If the common ratio $r > 1$, find the first term.
    Answer: 2 Let the three terms be $\frac{a}{r}$, $a$, and $ar$. The product of these terms is $(\frac{a}{r})(a)(ar) = a^3$. We are given that the product is 216, so $a^3 = 216$, which gives $a = 6$. The middle term is 6. The sum of the terms is $\frac{a}{r} + a + ar = 26$. Substituting $a=6$, we get $\frac{6}{r} + 6 + 6r = 26$. Subtracting 6 from both sides gives $\frac{6}{r} + 6r = 20$. Multiplying the entire equation by $r$ gives $6 + 6r^2 = 20r$, which can be rearranged into the quadratic equation $6r^2 - 20r + 6 = 0$. Dividing by 2 gives $3r^2 - 10r + 3 = 0$. Factoring gives $(3r-1)(r-3) = 0$. The possible values for $r$ are $r = \frac{1}{3}$ or $r = 3$. Since the question states $r > 1$, we must have $r=3$. The question asks for the first term, which is $\frac{a}{r} = \frac{6}{3} = 2$.
13. A geometric series has a first term of $a$ and a common ratio of $r$, where $|r|<1$. The sum to infinity is $S$. What is the sum of the terms with odd indices ($u_1 + u_3 + u_5 + \dots$) in terms of $S$ and $r$?
    Answer: $\frac{S}{1+r}$ The original series is $u_1, u_2, u_3, \dots$ or $a, ar, ar^2, \dots$ with sum $S = \frac{a}{1-r}$. The series of terms with odd indices is $u_1, u_3, u_5, \dots$ which is $a, ar^2, ar^4, \dots$. This is a new geometric series with first term $a' = a$ and common ratio $r' = r^2$. The sum of this new series is $S_{odd} = \frac{a'}{1-r'} = \frac{a}{1-r^2}$. We want to express this in terms of $S$ and $r$. We can factor the denominator: $S_{odd} = \frac{a}{(1-r)(1+r)}$. We recognize that $\frac{a}{1-r}$ is the sum of the original series, $S$. Therefore, we can substitute $S$ into the expression: $S_{odd} = (\frac{a}{1-r}) \cdot \frac{1}{1+r} = \frac{S}{1+r}$.
14. In a geometric sequence of positive terms, each term is equal to the sum of the two subsequent terms. Find the common ratio.
    Answer: $\frac{\sqrt{5}-1}{2}$ Let the terms of the sequence be $u_n$ and the common ratio be $r$. The condition is $u_n = u_{n+1} + u_{n+2}$ for any $n$. We can write this in terms of the first term and ratio, for example, using $n=1$: $u_1 = u_2 + u_3 \Rightarrow u_1 = u_1r + u_1r^2$. Since $u_1$ cannot be zero (otherwise all terms are zero), we can divide by $u_1$ to get $1 = r + r^2$. This is a quadratic equation: $r^2 + r - 1 = 0$. Using the quadratic formula, $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$. The sequence has positive terms, which implies $u_1 > 0$ and the common ratio $r$ must also be positive. Therefore, we must take the positive root: $r = \frac{-1 + \sqrt{5}}{2}$.
15. The first three terms of a geometric sequence are $x$, $x+4$ and $3x+4$. Given that $x>0$, find the sum of the first 5 terms of the sequence.
    Answer: 124 For a geometric sequence, the ratio of consecutive terms is constant. So, $\frac{x+4}{x} = \frac{3x+4}{x+4}$. This gives $(x+4)^2 = x(3x+4)$. Expanding both sides: $x^2 + 8x + 16 = 3x^2 + 4x$. Rearranging gives $2x^2 - 4x - 16 = 0$, which simplifies to $x^2 - 2x - 8 = 0$. Factoring gives $(x-4)(x+2) = 0$. Since the question states $x>0$, we must have $x=4$. The first three terms are $4$, $4+4=8$, and $3(4)+4=16$. This is a geometric sequence with first term $u_1 = 4$ and common ratio $r=2$. The sum of the first 5 terms is $S_5 = \frac{u_1(r^5 - 1)}{r-1} = \frac{4(2^5 - 1)}{2-1} = 4(32 - 1) = 4(31) = 124$.
16. The second term of a geometric sequence is -6 and the sum to infinity is 8. Which of the following is a possible value for the first term $u_1$?
    Answer: 12 We are given $u_2 = u_1 r = -6$ and $S_\infty = \frac{u_1}{1-r} = 8$. From the first equation, $r = -6/u_1$. Substitute this into the second equation: $8 = \frac{u_1}{1 - (-6/u_1)} = \frac{u_1}{1 + 6/u_1} = \frac{u_1}{(u_1+6)/u_1} = \frac{u_1^2}{u_1+6}$. This leads to the quadratic equation $8(u_1+6) = u_1^2$, which is $u_1^2 - 8u_1 - 48 = 0$. Factoring gives $(u_1 - 12)(u_1 + 4) = 0$. So, the possible values for $u_1$ are 12 and -4. We must check if the series converges for these values. Case 1: If $u_1 = 12$, then $r = -6/12 = -1/2$. Since $|r| < 1$, this is a valid solution. Case 2: If $u_1 = -4$, then $r = -6/(-4) = 3/2$. Since $|r| > 1$, the sum to infinity does not converge. Therefore, the only possible value for the first term is 12.
17. For a geometric sequence, the sum of the first three terms is 7, and the sum of the fourth, fifth, and sixth terms is 56. Find the common ratio $r$.
    Answer: 2 When you encounter geometric sequence problems involving sums of consecutive terms, set up equations using the general term formula and properties of geometric series. Let the first term be $a$ and common ratio be $r$. The first three terms are $a$, $ar$, and $ar^2$, so their sum is $a + ar + ar^2 = a(1 + r + r^2) = 7$. The fourth, fifth, and sixth terms are $ar^3$, $ar^4$, and $ar^5$, so their sum is $ar^3 + ar^4 + ar^5 = ar^3(1 + r + r^2) = 56$. Notice that both equations contain the factor $(1 + r + r^2)$. Dividing the second equation by the first: $\frac{ar^3(1 + r + r^2)}{a(1 + r + r^2)} = \frac{56}{7}$ This simplifies to $r^3 = 8$, so $r = 2$. Looking at the wrong answers: Choice A ($\sqrt{2}$) would give $r^3 = 2\sqrt{2} \approx 2.83$, which is too small. Choice B (8) treats $r^3 = 56$ directly, ignoring the relationship between the two sums entirely. Choice C (4) might come from incorrectly solving $r^2 = 8$ instead of $r^3 = 8$, or from misunderstanding the ratio between consecutive groups of terms. The key insight is recognizing that when you have sums of consecutive terms in a geometric sequence, you can often eliminate variables by finding ratios between these sums. Always look for common factors that will cancel out, leaving you with a simpler equation involving only the common ratio.

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