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

- 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.

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

\(189\)

- 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.

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

\(S_n=a\frac{1-r^n}{1-r}\), 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.

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

\(43\)

- 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.

- 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 \(r=2\).

\(r=2\)

- Focus on whether the vertical changes between successive partial-sum points are shrinking or growing. - Compare the absolute value of each common ratio with \(1\). - For the convergent series, check whether the algebraic infinite sum agrees with the level suggested visually.

a) In panel a), the partial sums rise by smaller and smaller amounts and level off near \(16\). In panel b), the partial sums grow rapidly and do not level off. b) For \(P\), \(|r|=\frac{1}{2}<1\), so the infinite series converges: \(S_\infty=\frac{8}{1-\frac{1}{2}}=16\). This matches the horizontal leveling suggested by panel a). For \(Q\), \(|r|=2>1\), so the infinite series diverges, matching the unbounded growth of its partial sums in panel b).

a) \(P\)'s partial sums approach a finite limit; \(Q\)'s partial sums grow without bound. b) \(P\) converges to \(16\). \(Q\) diverges.

- 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) \(n=14\)