IB Mathematics: Analysis and Approaches Quiz: Geometric Sequences And Series
20 questions · exam conditions
0:00
Geometric Sequences And SeriesQuestion 1 of 20

A geometric sequence has first term u1=sin⁡θu_1 = \sin \theta and common ratio r=sin⁡θr = \sin \theta, for 0<θ<π20 < \theta < \frac{\pi}{2}. The sum to infinity of the sequence is 2. Find the value of cos⁡2θ\cos^2 \theta.

13\frac{1}{3}
49\frac{4}{9}
59\frac{5}{9}
23\frac{2}{3}
← Back to quizzes

IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Geometric Sequences And Series

Practice Geometric Sequences And Series in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Geometric Sequences And Series, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A geometric sequence has first term u1=sin⁡θu_1 = \sin \theta and common ratio r=sin⁡θr = \sin \theta, for 0<θ<π20 < \theta < \frac{\pi}{2}. The sum to infinity of the sequence is 2. Find the value of cos⁡2θ\cos^2 \theta.

  1. 13\frac{1}{3}
  2. 49\frac{4}{9}
  3. 59\frac{5}{9} (correct answer)
  4. 23\frac{2}{3}
Explanation: The sum to infinity of a geometric sequence is given by S∞=u11−rS_\infty = \frac{u_1}{1-r}. We are given u1=sin⁡θu_1 = \sin \theta, r=sin⁡θr = \sin \theta, and S∞=2S_\infty = 2. The condition for convergence, ∣r∣<1|r|<1, is satisfied because for 0<θ<π20 < \theta < \frac{\pi}{2}, we have 0<sin⁡θ<10 < \sin \theta < 1. Substituting the given values into the formula: 2=sin⁡θ1−sin⁡θ2 = \frac{\sin \theta}{1 - \sin \theta}. Now, we solve for sin⁡θ\sin \theta. 2(1−sin⁡θ)=sin⁡θ⇒2−2sin⁡θ=sin⁡θ⇒2=3sin⁡θ⇒sin⁡θ=232(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⁡2θ\cos^2 \theta. Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we have cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta. Substituting the value of sin⁡θ\sin \theta: cos⁡2θ=1−(23)2=1−49=59\cos^2 \theta = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}.

Question 2

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.

  1. 40 m
  2. 50 m
  3. 80 m
  4. 90 m (correct answer)
Explanation: 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: 10×0.8=810 \times 0.8 = 8, 8×0.8=6.48 \times 0.8 = 6.4, and so on. The first term is u1=8u_1 = 8 and the common ratio is r=0.8r = 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∞=u11−r=81−0.8=80.2=40S_\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 2×40=802 \times 40 = 80 m. The total vertical distance is the initial drop plus this amount: 10+80=9010 + 80 = 90 m.

Question 3

A geometric sequence has first term u1=sin⁡θu_1 = \sin \theta and common ratio r=sin⁡θr = \sin \theta, for 0<θ<π20 < \theta < \frac{\pi}{2}. The sum to infinity of the sequence is 2. Find the value of cos⁡2θ\cos^2 \theta.

  1. 13\frac{1}{3}
  2. 49\frac{4}{9}
  3. 59\frac{5}{9} (correct answer)
  4. 23\frac{2}{3}
Explanation: The sum to infinity of a geometric sequence is given by S∞=u11−rS_\infty = \frac{u_1}{1-r}. We are given u1=sin⁡θu_1 = \sin \theta, r=sin⁡θr = \sin \theta, and S∞=2S_\infty = 2. The condition for convergence, ∣r∣<1|r|<1, is satisfied because for 0<θ<π20 < \theta < \frac{\pi}{2}, we have 0<sin⁡θ<10 < \sin \theta < 1. Substituting the given values into the formula: 2=sin⁡θ1−sin⁡θ2 = \frac{\sin \theta}{1 - \sin \theta}. Now, we solve for sin⁡θ\sin \theta. 2(1−sin⁡θ)=sin⁡θ⇒2−2sin⁡θ=sin⁡θ⇒2=3sin⁡θ⇒sin⁡θ=232(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⁡2θ\cos^2 \theta. Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we have cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta. Substituting the value of sin⁡θ\sin \theta: cos⁡2θ=1−(23)2=1−49=59\cos^2 \theta = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}.

Question 4

A geometric series has a first term of aa and a common ratio of rr, where ∣r∣<1|r|<1. The sum to infinity is SS. What is the sum of the terms with odd indices (u1+u3+u5+…u_1 + u_3 + u_5 + \dots) in terms of SS and rr?

  1. S2\frac{S}{2}
  2. S1−r\frac{S}{1-r}
  3. S1+r\frac{S}{1+r} (correct answer)
  4. S(1−r)S(1-r)
Explanation: The original series is u1,u2,u3,…u_1, u_2, u_3, \dots or a,ar,ar2,…a, ar, ar^2, \dots with sum S=a1−rS = \frac{a}{1-r}. The series of terms with odd indices is u1,u3,u5,…u_1, u_3, u_5, \dots which is a,ar2,ar4,…a, ar^2, ar^4, \dots. This is a new geometric series with first term a′=aa' = a and common ratio r′=r2r' = r^2. The sum of this new series is Sodd=a′1−r′=a1−r2S_{odd} = \frac{a'}{1-r'} = \frac{a}{1-r^2}. We want to express this in terms of SS and rr. We can factor the denominator: Sodd=a(1−r)(1+r)S_{odd} = \frac{a}{(1-r)(1+r)}. We recognize that a1−r\frac{a}{1-r} is the sum of the original series, SS. Therefore, we can substitute SS into the expression: Sodd=(a1−r)⋅11+r=S1+rS_{odd} = (\frac{a}{1-r}) \cdot \frac{1}{1+r} = \frac{S}{1+r}.

Question 5

The first three terms of a geometric sequence are xx, x+4x+4 and 3x+43x+4. Given that x>0x>0, find the sum of the first 5 terms of the sequence.

  1. 62
  2. 124 (correct answer)
  3. 128
  4. 248
Explanation: For a geometric sequence, the ratio of consecutive terms is constant. So, x+4x=3x+4x+4\frac{x+4}{x} = \frac{3x+4}{x+4}. This gives (x+4)2=x(3x+4)(x+4)^2 = x(3x+4). Expanding both sides: x2+8x+16=3x2+4xx^2 + 8x + 16 = 3x^2 + 4x. Rearranging gives 2x2−4x−16=02x^2 - 4x - 16 = 0, which simplifies to x2−2x−8=0x^2 - 2x - 8 = 0. Factoring gives (x−4)(x+2)=0(x-4)(x+2) = 0. Since the question states x>0x>0, we must have x=4x=4. The first three terms are 44, 4+4=84+4=8, and 3(4)+4=163(4)+4=16. This is a geometric sequence with first term u1=4u_1 = 4 and common ratio r=2r=2. The sum of the first 5 terms is S5=u1(r5−1)r−1=4(25−1)2−1=4(32−1)=4(31)=124S_5 = \frac{u_1(r^5 - 1)}{r-1} = \frac{4(2^5 - 1)}{2-1} = 4(32 - 1) = 4(31) = 124.

Question 6

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.

  1. 40 m
  2. 50 m
  3. 80 m
  4. 90 m (correct answer)
Explanation: 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: 10×0.8=810 \times 0.8 = 8, 8×0.8=6.48 \times 0.8 = 6.4, and so on. The first term is u1=8u_1 = 8 and the common ratio is r=0.8r = 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∞=u11−r=81−0.8=80.2=40S_\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 2×40=802 \times 40 = 80 m. The total vertical distance is the initial drop plus this amount: 10+80=9010 + 80 = 90 m.

Question 7

An infinite geometric series has a sum of 12 and a common ratio of −12-\frac{1}{2}. Find the sum of the first 4 terms.

  1. 454\frac{45}{4} (correct answer)
  2. 452\frac{45}{2}
  3. 18
  4. 27
Explanation: First, find the first term u1u_1 using the sum to infinity formula S∞=u11−rS_\infty = \frac{u_1}{1-r}. We have 12=u11−(−1/2)=u13/212 = \frac{u_1}{1 - (-1/2)} = \frac{u_1}{3/2}. Solving for u1u_1 gives u1=12×32=18u_1 = 12 \times \frac{3}{2} = 18. Now, use the formula for the sum of the first nn terms, Sn=u1(1−rn)1−rS_n = \frac{u_1(1-r^n)}{1-r}. For n=4n=4, S4=18(1−(−12)4)1−(−12)=18(1−116)32=18(1516)32=18×1516×23=6×1516×2=18016=454S_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}.

Question 8

A geometric series has first term ln⁡x\ln x and common ratio 2. The sum of the first three terms is ln⁡(128)\ln(128). Find the value of xx.

  1. 2\sqrt{2}
  2. 2 (correct answer)
  3. 222\sqrt{2}
  4. 4
Explanation: The first three terms of the sequence are u1=ln⁡xu_1 = \ln x, u2=2ln⁡x=ln⁡(x2)u_2 = 2\ln x = \ln(x^2), and u3=4ln⁡x=ln⁡(x4)u_3 = 4\ln x = \ln(x^4). The sum of the first three terms is S3=ln⁡x+2ln⁡x+4ln⁡x=7ln⁡xS_3 = \ln x + 2\ln x + 4\ln x = 7\ln x. We are given that this sum is equal to ln⁡(128)\ln(128). So, 7ln⁡x=ln⁡(128)7\ln x = \ln(128). Using the logarithm property bln⁡a=ln⁡(ab)b \ln a = \ln(a^b), we have ln⁡(x7)=ln⁡(128)\ln(x^7) = \ln(128). Since 128=27128 = 2^7, this becomes ln⁡(x7)=ln⁡(27)\ln(x^7) = \ln(2^7). Therefore, x7=27x^7 = 2^7, which implies x=2x=2.

Question 9

A geometric series has a first term of aa and a common ratio of rr, where ∣r∣<1|r|<1. The sum to infinity is SS. What is the sum of the terms with odd indices (u1+u3+u5+…u_1 + u_3 + u_5 + \dots) in terms of SS and rr?

  1. S2\frac{S}{2}
  2. S1−r\frac{S}{1-r}
  3. S1+r\frac{S}{1+r} (correct answer)
  4. S(1−r)S(1-r)
Explanation: The original series is u1,u2,u3,…u_1, u_2, u_3, \dots or a,ar,ar2,…a, ar, ar^2, \dots with sum S=a1−rS = \frac{a}{1-r}. The series of terms with odd indices is u1,u3,u5,…u_1, u_3, u_5, \dots which is a,ar2,ar4,…a, ar^2, ar^4, \dots. This is a new geometric series with first term a′=aa' = a and common ratio r′=r2r' = r^2. The sum of this new series is Sodd=a′1−r′=a1−r2S_{odd} = \frac{a'}{1-r'} = \frac{a}{1-r^2}. We want to express this in terms of SS and rr. We can factor the denominator: Sodd=a(1−r)(1+r)S_{odd} = \frac{a}{(1-r)(1+r)}. We recognize that a1−r\frac{a}{1-r} is the sum of the original series, SS. Therefore, we can substitute SS into the expression: Sodd=(a1−r)⋅11+r=S1+rS_{odd} = (\frac{a}{1-r}) \cdot \frac{1}{1+r} = \frac{S}{1+r}.

Question 10

A geometric series has first term ln⁡x\ln x and common ratio 2. The sum of the first three terms is ln⁡(128)\ln(128). Find the value of xx.

  1. 2\sqrt{2}
  2. 2 (correct answer)
  3. 222\sqrt{2}
  4. 4
Explanation: The first three terms of the sequence are u1=ln⁡xu_1 = \ln x, u2=2ln⁡x=ln⁡(x2)u_2 = 2\ln x = \ln(x^2), and u3=4ln⁡x=ln⁡(x4)u_3 = 4\ln x = \ln(x^4). The sum of the first three terms is S3=ln⁡x+2ln⁡x+4ln⁡x=7ln⁡xS_3 = \ln x + 2\ln x + 4\ln x = 7\ln x. We are given that this sum is equal to ln⁡(128)\ln(128). So, 7ln⁡x=ln⁡(128)7\ln x = \ln(128). Using the logarithm property bln⁡a=ln⁡(ab)b \ln a = \ln(a^b), we have ln⁡(x7)=ln⁡(128)\ln(x^7) = \ln(128). Since 128=27128 = 2^7, this becomes ln⁡(x7)=ln⁡(27)\ln(x^7) = \ln(2^7). Therefore, x7=27x^7 = 2^7, which implies x=2x=2.

Question 11

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, u3u_3?

  1. 1
  2. 2 (correct answer)
  3. 4
  4. 8
Explanation: Let the first term be u1u_1 and the common ratio be rr. We are given S∞=u11−r=16S_\infty = \frac{u_1}{1-r} = 16 and S2=u1+u1r=u1(1+r)=12S_2 = u_1 + u_1r = u_1(1+r) = 12. From the first equation, u1=16(1−r)u_1 = 16(1-r). Substitute this into the second equation: 16(1−r)(1+r)=1216(1-r)(1+r) = 12. This simplifies to 16(1−r2)=1216(1-r^2) = 12, so 1−r2=1216=341-r^2 = \frac{12}{16} = \frac{3}{4}. This gives r2=1−34=14r^2 = 1 - \frac{3}{4} = \frac{1}{4}. So, r=±12r = \pm \frac{1}{2}. Since the terms are all positive, the common ratio rr must be positive, so r=12r = \frac{1}{2}. Now we can find u1u_1 using u1(1+r)=12u_1(1+r)=12: u1(1+12)=12⇒u1(32)=12⇒u1=12×23=8u_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 u3=u1r2=8×(12)2=8×14=2u_3 = u_1 r^2 = 8 \times (\frac{1}{2})^2 = 8 \times \frac{1}{4} = 2.

Question 12

Given the geometric series ex+e2x+e3x+…e^x + e^{2x} + e^{3x} + \dots, for which values of xx does this series have a finite sum?

  1. x<0x < 0 (correct answer)
  2. x>0x > 0
  3. −1<x<1-1 < x < 1
  4. All real xx
Explanation: This is a geometric series with first term u1=exu_1 = e^x. The common ratio is r=e2xex=exr = \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|r| < 1. In this case, we need ∣ex∣<1|e^x| < 1. Since the exponential function exe^x is always positive for any real xx, this condition simplifies to 0<ex<10 < e^x < 1. To solve for xx, we take the natural logarithm of all parts of the inequality. As ln⁡(x)\ln(x) is an increasing function, the inequality signs are preserved. ln⁡(ex)<ln⁡(1)\ln(e^x) < \ln(1). This simplifies to x<0x < 0. The lower bound ex>0e^x > 0 is true for all real xx and does not impose a further constraint.

Question 13

For what range of values of xx does the infinite geometric series 1+(2x−3)+(2x−3)2+…1 + (2x-3) + (2x-3)^2 + \dots converge?

  1. x<2x < 2
  2. 1<x<21 < x < 2 (correct answer)
  3. x<1 or x>2x < 1 \text{ or } x > 2
  4. −1<x<1-1 < x < 1
Explanation: An infinite geometric series converges if and only if the absolute value of the common ratio rr is less than 1. In this series, the first term is 1 and the common ratio is r=2x−3r = 2x-3. So, for convergence, we must have ∣2x−3∣<1|2x-3| < 1. This inequality can be written as −1<2x−3<1-1 < 2x-3 < 1. To solve for xx, we add 3 to all parts of the inequality: −1+3<2x<1+3-1+3 < 2x < 1+3, which simplifies to 2<2x<42 < 2x < 4. Finally, we divide all parts by 2: 1<x<21 < x < 2.

Question 14

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 u1u_1?

  1. -4
  2. 4
  3. 12 (correct answer)
  4. 16
Explanation: We are given u2=u1r=−6u_2 = u_1 r = -6 and S∞=u11−r=8S_\infty = \frac{u_1}{1-r} = 8. From the first equation, r=−6/u1r = -6/u_1. Substitute this into the second equation: 8=u11−(−6/u1)=u11+6/u1=u1(u1+6)/u1=u12u1+68 = \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(u1+6)=u128(u_1+6) = u_1^2, which is u12−8u1−48=0u_1^2 - 8u_1 - 48 = 0. Factoring gives (u1−12)(u1+4)=0(u_1 - 12)(u_1 + 4) = 0. So, the possible values for u1u_1 are 12 and -4. We must check if the series converges for these values. Case 1: If u1=12u_1 = 12, then r=−6/12=−1/2r = -6/12 = -1/2. Since ∣r∣<1|r| < 1, this is a valid solution. Case 2: If u1=−4u_1 = -4, then r=−6/(−4)=3/2r = -6/(-4) = 3/2. Since ∣r∣>1|r| > 1, the sum to infinity does not converge. Therefore, the only possible value for the first term is 12.

Question 15

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 rr.

  1. 2\sqrt{2}
  2. 8
  3. 4
  4. 2 (correct answer)
Explanation: 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 aa and common ratio be rr. The first three terms are aa, arar, and ar2ar^2, so their sum is a+ar+ar2=a(1+r+r2)=7a + ar + ar^2 = a(1 + r + r^2) = 7. The fourth, fifth, and sixth terms are ar3ar^3, ar4ar^4, and ar5ar^5, so their sum is ar3+ar4+ar5=ar3(1+r+r2)=56ar^3 + ar^4 + ar^5 = ar^3(1 + r + r^2) = 56. Notice that both equations contain the factor (1+r+r2)(1 + r + r^2). Dividing the second equation by the first: ar3(1+r+r2)a(1+r+r2)=567\frac{ar^3(1 + r + r^2)}{a(1 + r + r^2)} = \frac{56}{7} This simplifies to r3=8r^3 = 8, so r=2r = 2. Looking at the wrong answers: Choice A (2\sqrt{2}) would give r3=22≈2.83r^3 = 2\sqrt{2} \approx 2.83, which is too small. Choice B (8) treats r3=56r^3 = 56 directly, ignoring the relationship between the two sums entirely. Choice C (4) might come from incorrectly solving r2=8r^2 = 8 instead of r3=8r^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.

Question 16

The sum of an infinite geometric series is SS and its first term is u1u_1. A new series is formed by squaring each term of the original series. What is the sum of this new series?

  1. S2S^2
  2. u1S22S−u1\frac{u_1 S^2}{2S-u_1} (correct answer)
  3. u1S2S−u1\frac{u_1 S}{2S-u_1}
  4. u12S\frac{u_1^2}{S}
Explanation: The original series has first term u1u_1 and common ratio rr. Its sum is S=u11−rS = \frac{u_1}{1-r}. The new series formed by squaring each term is u12,(u1r)2,(u1r2)2,…u_1^2, (u_1r)^2, (u_1r^2)^2, \dots. This is a geometric series with first term u12u_1^2 and common ratio r2r^2. Since ∣r∣<1|r|<1 for the original series to converge, we have r2<1r^2 < 1, so the new series also converges. The sum of the new series is Snew=(u1)21−r2=u12(1−r)(1+r)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 rr in terms of SS and u1u_1: 1−r=u1S⇒r=1−u1S1-r = \frac{u_1}{S} \Rightarrow r = 1 - \frac{u_1}{S}. Now we find 1+r1+r: 1+r=1+(1−u1S)=2−u1S=2S−u1S1+r = 1 + (1-\frac{u_1}{S}) = 2 - \frac{u_1}{S} = \frac{2S-u_1}{S}. Substitute these into the formula for SnewS_{new}: Snew=u12(u1S)(2S−u1S)=u12u1(2S−u1)S2=u12⋅S2u1(2S−u1)=u1S22S−u1S_{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}.

Question 17

The third term of a geometric sequence is 12 and the sixth term is 96. Find the sum of the first eight terms.

  1. 255
  2. 381
  3. 510
  4. 765 (correct answer)
Explanation: Let the first term be u1u_1 and the common ratio be rr. We are given u3=u1r2=12u_3 = u_1 r^2 = 12 and u6=u1r5=96u_6 = u_1 r^5 = 96. Dividing the second equation by the first gives u1r5u1r2=9612\frac{u_1 r^5}{u_1 r^2} = \frac{96}{12}, which simplifies to r3=8r^3 = 8, so r=2r=2. Substitute r=2r=2 into the first equation: u1(22)=12⇒4u1=12⇒u1=3u_1(2^2) = 12 \Rightarrow 4u_1 = 12 \Rightarrow u_1 = 3. The sum of the first eight terms is given by the formula Sn=u1(rn−1)r−1S_n = \frac{u_1(r^n - 1)}{r-1}. So, S8=3(28−1)2−1=3(256−1)=3(255)=765S_8 = \frac{3(2^8 - 1)}{2-1} = 3(256 - 1) = 3(255) = 765.

Question 18

The sum of the first nn terms of a sequence is given by the formula Sn=5(2n−1)S_n = 5(2^n - 1). Find the 7th term, u7u_7.

  1. 315
  2. 320 (correct answer)
  3. 635
  4. 640
Explanation: The nn-th term of a sequence can be found using the formula un=Sn−Sn−1u_n = S_n - S_{n-1}. To find the 7th term, u7u_7, we calculate S7−S6S_7 - S_6. S7=5(27−1)=5(128−1)=5(127)=635S_7 = 5(2^7 - 1) = 5(128 - 1) = 5(127) = 635. S6=5(26−1)=5(64−1)=5(63)=315S_6 = 5(2^6 - 1) = 5(64 - 1) = 5(63) = 315. Therefore, u7=S7−S6=635−315=320u_7 = S_7 - S_6 = 635 - 315 = 320. Alternatively, one can identify the sequence. u1=S1=5(21−1)=5u_1 = S_1 = 5(2^1-1)=5. u2=S2−S1=5(22−1)−5=15−5=10u_2 = S_2 - S_1 = 5(2^2-1) - 5 = 15 - 5 = 10. This is a geometric sequence with u1=5u_1=5 and r=2r=2. The 7th term is u7=u1r7−1=5×26=5×64=320u_7 = u_1 r^{7-1} = 5 \times 2^6 = 5 \times 64 = 320.

Question 19

An infinite geometric series has a sum of 12 and a common ratio of −12-\frac{1}{2}. Find the sum of the first 4 terms.

  1. 454\frac{45}{4} (correct answer)
  2. 452\frac{45}{2}
  3. 18
  4. 27
Explanation: First, find the first term u1u_1 using the sum to infinity formula S∞=u11−rS_\infty = \frac{u_1}{1-r}. We have 12=u11−(−1/2)=u13/212 = \frac{u_1}{1 - (-1/2)} = \frac{u_1}{3/2}. Solving for u1u_1 gives u1=12×32=18u_1 = 12 \times \frac{3}{2} = 18. Now, use the formula for the sum of the first nn terms, Sn=u1(1−rn)1−rS_n = \frac{u_1(1-r^n)}{1-r}. For n=4n=4, S4=18(1−(−12)4)1−(−12)=18(1−116)32=18(1516)32=18×1516×23=6×1516×2=18016=454S_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}.

Question 20

Given the geometric series ex+e2x+e3x+…e^x + e^{2x} + e^{3x} + \dots, for which values of xx does this series have a finite sum?

  1. x<0x < 0 (correct answer)
  2. x>0x > 0
  3. −1<x<1-1 < x < 1
  4. All real xx
Explanation: This is a geometric series with first term u1=exu_1 = e^x. The common ratio is r=e2xex=exr = \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|r| < 1. In this case, we need ∣ex∣<1|e^x| < 1. Since the exponential function exe^x is always positive for any real xx, this condition simplifies to 0<ex<10 < e^x < 1. To solve for xx, we take the natural logarithm of all parts of the inequality. As ln⁡(x)\ln(x) is an increasing function, the inequality signs are preserved. ln⁡(ex)<ln⁡(1)\ln(e^x) < \ln(1). This simplifies to x<0x < 0. The lower bound ex>0e^x > 0 is true for all real xx and does not impose a further constraint.