Free math worksheets

- Compare consecutive terms using both subtraction and division. - A constant multiplicative change identifies a geometric pattern. - Count the displayed terms after deciding which type of series it is.

Each term is obtained by multiplying the previous term by \(3\): \(6\div2=3\), \(18\div6=3\), and \(54\div18=3\). Therefore, the sum is a geometric series with first term \(a_1=2\), common ratio \(r=3\), and \(n=4\) terms.

Geometric series; \(a_1=2\), \(r=3\), \(n=4\).

- The subscript on \(a_4\) identifies a single position in the sequence. - The symbol \(S_4\) represents an accumulated total, not another sequence term. - List exactly which terms contribute to the partial sum.

1. The fourth term is \(a_4=24\). 2. The partial sum is \(S_4=3+6+12+24=45\). 3. The symbol \(a_4\) names one term of the sequence, while \(S_4\) names the cumulative total of the first four terms.

\(a_4=24\); \(S_4=45\). The first is one sequence term, while the second is the sum of the first four terms.

- The decision depends on the common ratio, not on the size of the first term. - Compare the absolute value of the ratio with \(1\).

1. An infinite geometric series converges when \(|r|<1\). 2. Here \(\left|\frac{1}{4}\right|<1\), so the series converges.

The series converges because \(|r|=\frac{1}{4}<1\).

- Identify the constant ratio between consecutive terms. - Count how many terms are included before choosing the exponent in the finite-sum formula. - Use a version of the geometric-series formula whose signs fit a ratio greater than \(1\).

This is a geometric series with \(a_1=5\), \(r=3\), and \(n=6\). Using the finite geometric-series formula, \(S_6=5\frac{3^6-1}{3-1}=1820\).

\(1820\)

- The lower sigma bound tells you what exponent appears in the first term. - Determine the number of terms from the inclusive bounds before using the finite geometric-series formula. - Keep the ratio as a fraction while simplifying.

1. The series has first term \(96\), common ratio \(\frac{1}{2}\), and \(6\) terms. 2. Therefore, \(S_6=96\frac{1-(\frac{1}{2})^6}{1-\frac{1}{2}}=189\).

\(a_1=96\), \(r=\frac{1}{2}\), \(n=6\), so \(S_6=96\frac{1-(\frac{1}{2})^6}{1-\frac{1}{2}}=189\).

- List what quantity each term of the geometric series represents in this context. - Use the rebound factor as the common ratio and count only the five upward rebound heights requested. - Round only after evaluating the finite sum.

The rebound heights form a geometric series with \(a_1=12\), \(r=\frac{3}{4}\), and \(n=5\). Thus, \(S_5=12\frac{1-(\frac{3}{4})^5}{1-\frac{3}{4}}=\frac{2343}{64}\approx36.609375\). Rounded to the nearest tenth, the total upward distance is \(36.6\,\text{ft}\).

\(36.6\,\text{ft}\)

- Separate the number of new people in one hour from the cumulative number who have heard the rumor. - For the cumulative total, include the initial person together with every later group. - For the threshold, compare consecutive whole-hour totals near the target.

a) The number who hear the rumor for the first time during hour \(n\) is \(3^n\). During hour \(5\), \(3^5=243\) people hear it for the first time. b) After \(n\) hours, the total is the geometric sum \(1+3+3^2+\cdots+3^n=\frac{3^{n+1}-1}{2}\). For \(n=8\), the total is \(\frac{3^9-1}{2}=9841\). c) After \(10\) hours, the total is \(\frac{3^{11}-1}{2}=88{,}573\), below \(250{,}000\). After \(11\) hours, it is \(\frac{3^{12}-1}{2}=265{,}720\), above \(250{,}000\). Therefore, the first such time is \(11\) full hours.

a) \(243\) people b) \(9841\) people c) \(11\) hours

- Write the original sum and the multiplied sum on separate lines so equal powers line up. - Choose the subtraction order so that most middle terms cancel. - Factor the expression multiplying \(S_n\) before isolating the sum.

1. Multiply the sum by \(r\): \(rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n\). 2. Subtract the second equation from the first: \(S_n-rS_n=a-ar^n\). 3. Factor both sides: \((1-r)S_n=a(1-r^n)\). 4. Since \(r\ne1\), divide by \(1-r\): \(S_n=a\frac{1-r^n}{1-r}\).

\(S_n-rS_n=a-ar^n\), so \((1-r)S_n=a(1-r^n)\) and \(S_n=a\frac{1-r^n}{1-r}\) for \(r\ne1\).

- Include the sign when finding the common ratio; the terms alternate because the ratio is negative. - Count the number of terms carefully before choosing the exponent in the finite-sum formula. - Keep the negative ratio inside parentheses when raising it to a power.

1. The series has \(a_1=64\), common ratio \(r=-\frac{1}{2}\), and \(n=7\) terms. 2. Therefore, \(S_7=64\frac{1-(-\frac{1}{2})^7}{1-(-\frac{1}{2})}=43\).

\(a_1=64\), \(r=-\frac{1}{2}\), \(n=7\), so \(S_7=64\frac{1-(-\frac{1}{2})^7}{1-(-\frac{1}{2})}=43\).

- Express the partial sum as a function of \(n\) before comparing it with the threshold. - After isolating the exponential expression, compare nearby powers of the common ratio. - Verify both the chosen value and the preceding integer to establish that your answer is the smallest one.

The partial sum is \(S_n=3\frac{2^n-1}{2-1}=3(2^n-1)\). We need \(3(2^n-1)>3000\), so \(2^n>1001\). Since \(2^9=512\) and \(2^{10}=1024\), the smallest possible value is \(n=10\). Indeed, \(S_9=1533\) and \(S_{10}=3069\).

\(n=10\)

- Find the common ratio of each series before deciding whether an infinite sum exists. - For an infinite geometric series, the size of \(|r|\) determines whether the partial sums approach a finite value. - Apply the infinite-sum formula only after confirming the convergence condition.

Series \(A\) has common ratio \(r=\frac{1}{2}\). Since \(|r|<1\), it converges, and \(A=\frac{12}{1-\frac{1}{2}}=24\). Series \(B\) has common ratio \(r=-\frac{3}{2}\). Since \(|r|>1\), its terms do not approach \(0\), so the infinite series diverges.

\(A\) converges to \(24\). \(B\) diverges.

- Find the ratio of one term to the term immediately before it, including the sign. - For an infinite geometric series, convergence depends on the absolute value of the common ratio. - Decide whether the convergence condition is satisfied before using an infinite-sum formula.

1. The common ratio is \(r=\frac{-6}{18}=-\frac{1}{3}\). 2. Because \(|r|=\frac{1}{3}<1\), the infinite geometric series converges even though its terms alternate in sign. 3. Its sum is \(S_\infty=\frac{18}{1-(-\frac{1}{3})}=\frac{18}{\frac{4}{3}}=\frac{27}{2}\). 4. Therefore, Karim's claim is incorrect: alternating signs do not by themselves imply divergence.

The claim is incorrect. \(r=-\frac{1}{3}\), the series converges, and \(S_\infty=\frac{27}{2}\).

- Read the sequence of plotted partial sums before deciding which ratio belongs to each panel. - Consecutive differences of partial sums recover the terms of the underlying series. - Compare what the partial sums do as \(n\) increases: approach one value, grow, or oscillate. - Relate both behaviors to the strict convergence condition \(|r|<1\).

1. Panel a) has partial sums \(4,8,12,16,20,24\). Their consecutive differences are all \(4\), so every series term is \(4\) and the common ratio is \(r=1\). 2. For panel a), \(S_n=4n\), so the partial sums grow without bound and the infinite series diverges. 3. Panel b) has partial sums \(4,0,4,0,4,0\). The corresponding terms are \(4,-4,4,-4,\ldots\), so the common ratio is \(r=-1\). 4. For panel b), \(S_n=4\) for odd \(n\) and \(S_n=0\) for even \(n\). The partial sums oscillate and do not approach a single value, so the infinite series diverges. 5. Both examples have \(|r|=1\), showing two different forms of divergence at the boundary of the geometric-series convergence condition \(|r|<1\).

a) Panel a): \(r=1\); panel b): \(r=-1\). b) Panel a): \(S_n=4n\), so the partial sums grow without bound. Panel b): \(S_n=4\) for odd \(n\) and \(S_n=0\) for even \(n\), so the partial sums oscillate. Both series diverge. c) At \(|r|=1\), partial sums need not approach a finite limit: they may grow without bound or oscillate.

- Express all four terms in terms of the first term and the unknown ratio. - The given total creates an equation for the ratio rather than a forward sum calculation. - After factoring the resulting polynomial, use the condition that the ratio is positive and real.

The four terms are \(4\), \(4r\), \(4r^2\), and \(4r^3\), so \(4(1+r+r^2+r^3)=60\). Dividing by \(4\) gives \(r^3+r^2+r-14=0\). Factoring, \((r-2)(r^2+3r+7)=0\). The quadratic factor has no real zeros, so the positive real ratio is \(r=2\).

\(r=2\)

- A sequence term can be recovered from two consecutive partial sums by subtracting them. - Use \(S_0=0\) to recover the first term from the first plotted partial sum. - After recovering several terms, compare consecutive terms to find the common ratio. - Use both the ratio test for a geometric series and the visible partial-sum behavior when deciding convergence.

1. In panel a), the partial sums are \(8,12,14,15\). Using \(a_n=S_n-S_{n-1}\) with \(S_0=0\), the first three terms are \(8,4,2\), so \(r=\frac{1}{2}\). 2. Since \(|r|=\frac{1}{2}<1\), the series converges to \(S_\infty=\frac{8}{1-\frac{1}{2}}=16\). The plotted partial sums increase toward \(16\) by smaller amounts. 3. In panel b), the partial sums are \(1,3,7,15\). The first three terms are \(1,2,4\), so \(r=2\). 4. Since \(|r|=2>1\), the series diverges. The plotted partial sums increase by larger amounts rather than leveling off.

a) Panel a): terms \(8,4,2\), \(r=\frac{1}{2}\). Panel b): terms \(1,2,4\), \(r=2\). b) Panel a) converges to \(16\), and its partial sums level toward that value. Panel b) diverges, and its partial sums grow by increasingly large amounts.

- Use the same first term and ratio to connect the finite partial sum with the infinite sum. - For the percentage, compare the finite sum with the eventual total rather than with the next term. - Rewrite the target percentage as an inequality involving the remaining factor \(r^n\). - A logarithm can locate the threshold, but check neighboring integer values because \(n\) counts terms.

The series has \(a_1=500\) and \(r=0.8\). a) \(S_{10}=500\frac{1-0.8^{10}}{1-0.8}\approx2231.565\). b) Since \(|0.8|<1\), \(S_\infty=\frac{500}{1-0.8}=2500\). c) \(\frac{S_{10}}{S_\infty}=1-0.8^{10}\approx0.8926258\), so about \(89.3\%\) has been accumulated. d) We need \(1-0.8^n\ge0.95\), so \(0.8^n\le0.05\). This gives \(n\ge\frac{\ln(0.05)}{\ln(0.8)}\approx13.43\). Checking consecutive integers, \(n=13\) gives about \(94.5\%\), while \(n=14\) gives about \(95.6\%\). Thus the smallest value is \(n=14\).

a) \(S_{10}\approx2231.565\) b) \(S_\infty=2500\) c) \(89.3\%\) d) \(n=14\)