Free math worksheets

- Compare each term with the term immediately before it to identify the common ratio. - A finite series ends after a stated last term; an ellipsis indicates that the pattern continues. - For an infinite geometric series, focus on the size of the common ratio rather than trying to add terms.

a) The series has four terms, so it is finite. Its common ratio is \(\frac{1}{2}\). Every finite series has a finite sum. b) The series is infinite with common ratio \(-\frac{1}{2}\). Because \(\left|-\frac{1}{2}\right|<1\), it has a finite sum. c) The series is infinite with common ratio \(2\). Because \(|2|>1\), it does not have a finite sum.

a) Finite; \(r=\frac{1}{2}\); finite sum. b) Infinite; \(r=-\frac{1}{2}\); finite sum. c) Infinite; \(r=2\); no finite sum.

- Use the same multiplicative change from one term to the next. - Keep the sign of each term consistent with multiplying by a positive ratio. - A finite series expression adds the requested sequence terms in order.

Each term is obtained by multiplying the preceding term by \(\frac{1}{4}\). The first four terms are \(-16,-4,-1,-\frac{1}{4}\). Therefore \(S_4=-16-4-1-\frac{1}{4}\).

First four terms: \(-16,-4,-1,-\frac{1}{4}\) \(S_4=-16-4-1-\frac{1}{4}\)

- Find the ratio of consecutive terms. - Check the convergence condition. - Apply the infinite geometric-series formula. - Rationalize the final denominator.

1. The first term is \(a_1=4\), and the common ratio is \(q=\frac{2\sqrt2}{4}=\frac{\sqrt2}{2}\). 2. Since \(|q|<1\), the series converges. 3. Its sum is \(S=\frac{4}{1-\sqrt2/2}=\frac{8}{2-\sqrt2}\). 4. Rationalize the denominator: \(S=\frac{8(2+\sqrt2)}{(2-\sqrt2)(2+\sqrt2)}\) \(=\frac{8(2+\sqrt2)}{2}=8+4\sqrt2\).

\(8+4\sqrt2\).

- Write a sum formula for the full series and for the tail beginning with the third term. - Divide the two equations to eliminate \(a_1\). - Use positivity to choose the correct square root.

1. The total sum is \(\frac{a_1}{1-q}=12\). 2. The sum beginning with the third term is \(\frac{a_1q^2}{1-q}=3\). 3. Divide the second equation by the first: \(q^2=\frac3{12}=\frac14\). Since all terms are positive and the series is decreasing, \(q=\frac12\). 4. Substitute into the total-sum equation: \(12=\frac{a_1}{1-1/2}=2a_1\), so \(a_1=6\).

\(a_1=6\) and \(q=\frac12\).

- Use the infinite geometric-series sum formula. - Express the second term as \(a_1q\). - Solve the resulting equations and then apply the convergence condition.

1. The sum condition is \(\frac{a_1}{1-q}=16\), so \(a_1=16(1-q)\). 2. Since \(a_2=a_1q\), \(a_1-a_2=a_1(1-q)=4\). 3. Substitute \(a_1=16(1-q)\): \(16(1-q)^2=4\). Thus, \((1-q)^2=\frac14\), so \(q=\frac12\) or \(q=\frac32\). 4. Convergence requires \(|q|<1\), so \(q=\frac12\). 5. Then \(a_1=16\left(1-\frac12\right)=8\).

\(a_1=8\) and \(q=\frac12\).

- Relate the infinite-series sum formula to the first term. - Express \(a_2\) as \(a_1q\). - Find \(q\) before using the sum of the first two terms.

1. Combine \(S=2.5a_1\) with the infinite-series formula: \(\frac{a_1}{1-q}=2.5a_1\). Since \(a_1\ne0\), divide by \(a_1\): \(\frac1{1-q}=2.5\). 2. Solve for \(q\): \(1=2.5(1-q)\), so \(2.5q=1.5\) and \(q=0.6\). 3. Since \(a_2=a_1q\), \(a_1+a_2=a_1(1+q)=16\). Therefore, \(1.6a_1=16\), so \(a_1=10\).

\(a_1=10\) and \(q=0.6\).

- Separate the nonrepeating digit from the repeating block. - Identify the first term and the decimal-place shift between copies of the block. - Add the terminating part after finding the geometric-series sum.

1. Separate the nonrepeating and repeating parts: \(0.4\overline{18}=0.4+(0.018+0.00018+0.0000018+\cdots)\). 2. The geometric series has first term \(a_1=0.018\) and common ratio \(q=0.01\). Its sum is \(\frac{0.018}{1-0.01}=\frac{0.018}{0.99}=\frac1{55}\). 3. Therefore, \(0.4\overline{18}=\frac25+\frac1{55}=\frac{22}{55}+\frac1{55}=\frac{23}{55}\).

\(0.4\overline{18}=0.4+(0.018+0.00018+0.0000018+\cdots)=\frac{23}{55}\).

- Check the condition that must be true before the infinite geometric-series formula can be used. - Consider what happens to the individual terms as the series continues. - A numerical result from a formula is not meaningful if the hypotheses of that formula are not satisfied.

The series is geometric with common ratio \(3\). The formula \(\frac{a}{1-r}\) gives the sum of an infinite geometric series only when \(|r|<1\). Here \(|3|>1\), so the terms do not approach \(0\) and the partial sums grow without bound. Therefore the series diverges and has no finite sum.

The infinite geometric-series formula does not apply because \(|r|=3>1\). The series diverges and has no finite sum.

- Use the sum formula for a convergent infinite geometric series. - Solve that equation for \(q\). - Then use the finite geometric-series formula and reduce the fraction.

1. Use the infinite geometric-series formula: \(12=\frac{8}{1-q}\). Thus, \(1-q=\frac23\), so \(q=\frac13\). 2. Apply the finite geometric-series formula: \(S_5=8\frac{1-(1/3)^5}{1-1/3}\) \(=8\frac{1-1/243}{2/3}=8\cdot\frac{242}{243}\cdot\frac32=\frac{968}{81}\).

\(q=\frac13\) and \(S_5=\frac{968}{81}\).

- Find the ratio of consecutive terms. - Check the convergence condition for a geometric series. - Find both the infinite sum and the five-term partial sum. - Express the partial sum as a percentage of the total.

1. The series is geometric with first term \(a_1=3\) and common ratio \(q=\frac{1.5}{3}=0.5\). Since \(|q|<1\), the series converges. 2. Its sum is \(S=\frac{3}{1-0.5}=6\). 3. The first five terms sum to \(S_5=3\frac{1-(0.5)^5}{1-0.5}=6\cdot(1-0.03125)=5.8125\). 4. The percentage is \(\frac{S_5}{S}\cdot100\%=\frac{5.8125}{6}\cdot100\%=96.875\%\).

The series converges to \(6\). The first five terms sum to \(5.8125\), which is \(96.875\%\) of the total.

- Use the first and fifth terms to solve for the common ratio. - Apply the positivity condition when choosing a root. - Use the finite geometric-series formula for part b).

1. Since \(b_1=3\) and \(b_5=48\), \(48=3q^4\). Thus, \(q^4=16\). Because every term must be positive, \(q=2\). 2. The missing terms are \(b_2=6\), \(b_3=12\), and \(b_4=24\). 3. The first ten terms sum to \(S_{10}=3\frac{2^{10}-1}{2-1}=3\cdot1023=3069\).

a) \(6, 12, 24\). b) \(3069\).

- Express the arithmetic terms using one variable. - Use the relationship among three consecutive geometric terms. - Find the common ratio and check whether its absolute value is less than \(1\). - Then apply the infinite geometric-series formula.

1. Write the arithmetic terms as \(a, a-4, a-8\). Then the geometric terms are \(a+1, a-2, a-3\). 2. Use the middle-term property: \((a-2)^2=(a+1)(a-3)\). Simplifying gives \(a=3.5\). Therefore, \(a_1=3.5\), \(a_2=-0.5\), and \(a_3=-4.5\). 3. The geometric sequence begins \(4.5, 1.5, 0.5\), so \(q=\frac13\). Since \(|q|<1\), the associated infinite series converges. 4. Its sum is \(S=\frac{4.5}{1-1/3}=6.75\).

a) \(a_1=3.5\), \(a_2=-0.5\), and \(a_3=-4.5\). b) The infinite geometric series converges to \(6.75\).

- Express the first three terms of both sequences using \(d\) and \(q\). - Use the condition on the second terms to eliminate \(d\). - Apply the convergence condition for an infinite geometric series.

1. Let the arithmetic common difference be \(d\). Then \(a_2=1+d\), \(a_3=1+2d\), \(b_2=q\), and \(b_3=q^2\). 2. Since \(a_2=b_2+0.125\), \(d=q-0.875\). 3. Use \(a_3=b_3\): \(1+2(q-0.875)=q^2\). Thus, \(q^2-2q+0.75=0\). The solutions are \(q=1.5\) and \(q=0.5\). 4. An infinite geometric series converges only when \(|q|<1\), so only \(q=0.5\) is valid. 5. Its sum is \(S=\frac{1}{1-0.5}=2\).

a) \(q^2-2q+0.75=0\), with solutions \(q=1.5\) and \(q=0.5\). b) The series converges for \(q=0.5\), and its sum is \(2\).

- Write the first two terms to identify the common ratio. - Apply \(|q|<1\). - Simplify the infinite-series formula as a function of \(x\). - Verify that the solution in part c) lies in the convergence domain.

1. The first term is \(a_1=\frac2x\), and the common ratio is \(q=\frac3x\). 2. The series converges when \(\left|\frac3x\right|<1\), which is equivalent to \(|x|>3\). 3. On this domain, \(S(x)=\frac{2/x}{1-3/x}=\frac{2}{x-3}\). 4. Set the sum equal to \(3\): \(\frac{2}{x-3}=3\). Thus, \(2=3x-9\), so \(x=\frac{11}{3}\). Since \(\left|\frac{11}{3}\right|>3\), this value is in the convergence domain.

a) \(|x|>3\). b) \(S(x)=\frac{2}{x-3}\). c) \(x=\frac{11}{3}\).

- Write one equation using the infinite geometric-series sum formula. - Use \(a_2=a_1q\) to write a second equation. - Look for a difference-of-squares expression. - Check each value of \(q\) against the convergence condition.

1. The infinite-series sum gives \(\frac{a_1}{1-q}=12.5\), so \(a_1=12.5(1-q)\). 2. Since \(a_2=a_1q\), the first-two-terms condition is \(a_1(1+q)=12\). 3. Substitute \(a_1=12.5(1-q)\): \(12.5(1-q)(1+q)=12\). Thus, \(12.5(1-q^2)=12\), so \(1-q^2=0.96\) and \(q^2=0.04\). Therefore, \(q=0.2\) or \(q=-0.2\). Both satisfy \(|q|<1\). 4. When \(q=0.2\), \(a_1=12.5\cdot(1-0.2)=10\). When \(q=-0.2\), \(a_1=12.5\cdot(1+0.2)=15\).

\(a_1=10\) and \(q=0.2\), or \(a_1=15\) and \(q=-0.2\).

- Separate the whole-number part from the repeating part. - Determine how many decimal places each copy of the repeating block shifts. - Apply the infinite geometric-series sum formula. - Express the pattern for a repeating block of length \(k\).

1. The number can be written as \(2.141414\ldots=2+0.\overline{14}\). The repeating part is \(0.14+0.0014+0.000014+\cdots\). Therefore, \(a_1=0.14\) and \(q=0.01=\frac1{100}\). 2. Its sum is \(\frac{0.14}{1-0.01}=\frac{0.14}{0.99}=\frac{14}{99}\). Thus, \(2+\frac{14}{99}=\frac{198}{99}+\frac{14}{99}=\frac{212}{99}\). The numerator and denominator have no common factor, so the fraction is in simplest form. 3. If the repeating block has \(k\) digits, each successive copy shifts \(k\) decimal places to the right. Therefore, \(q=10^{-k}\). For a three-digit block, \(q=10^{-3}=0.001\).

a) \(2.141414\ldots=2+(0.14+0.0014+0.000014+\cdots)\), with \(a_1=0.14\) and \(q=0.01\). b) \(\frac{212}{99}\). c) For a repeating block of \(k\) digits, \(q=10^{-k}\). For three digits, \(q=0.001\).

- A two-digit repeating block can be represented using a denominator of \(99\) before simplification. - Simplify the fraction for \(0.\overline{15}\) completely. - Write \(0.\overline{9}\) as \(0.9+0.09+0.009+\cdots\).

1. Multiply the numerator and denominator by \(3\): \(\frac8{33}=\frac{24}{99}\). Therefore, the corresponding repeating decimal is \(0.\overline{24}\). 2. For \(0.\overline{15}\), the first term is \(0.15\) and the common ratio is \(0.01\). Its value is \(\frac{0.15}{1-0.01}=\frac{15}{99}=\frac5{33}\). The simplified denominator is \(33\), not \(99\), so the claim is false. 3. The decimal \(0.\overline{9}\) is the geometric series \(0.9+0.09+0.009+\cdots\), with first term \(0.9\) and common ratio \(0.1\). Therefore, \(0.\overline{9}=\frac{0.9}{1-0.1}=\frac{0.9}{0.9}=1\).

a) \(0.\overline{24}\). b) The claim is false because \(0.\overline{15}=\frac5{33}\). c) \(0.\overline{9}=\frac{0.9}{1-0.1}=1\).

- Separate the series into two interwoven geometric series. - Pay attention to the sign of the common ratio in the second series. - Verify that the absolute value of each common ratio is less than \(1\).

1. Separate the terms into two geometric series: \(\left(\frac3{10}+\frac3{100}+\frac3{1000}+\cdots\right)+\left(\frac25-\frac2{25}+\frac2{125}-\frac2{625}+\cdots\right)\). 2. The first series has first term \(\frac3{10}\) and common ratio \(\frac1{10}\), so \(S_1=\frac{3/10}{1-1/10}=\frac13\). 3. The second series has first term \(\frac25\) and common ratio \(-\frac15\), so \(S_2=\frac{2/5}{1-(-1/5)}=\frac13\). 4. Therefore, \(S=S_1+S_2=\frac13+\frac13=\frac23\).

\(\frac23\)

- Use the same idea as the geometric-series representation of \(0.\overline{9}\). - Identify the first place at which the repeating \(9\) appears. - Find the exact sum of the repeating tail before adding \(0.12\).

1. Separate the terminating part from the repeating nines: \(x=0.12+0.009+0.0009+0.00009+\cdots\). 2. The geometric series has first term \(0.009\) and common ratio \(0.1\). Its sum is \(\frac{0.009}{1-0.1}=\frac{0.009}{0.9}=0.01\). 3. Therefore, \(x=0.12+0.01=0.13\). 4. The repeating tail equals exactly \(0.01\), so adding it to \(0.12\) produces the terminating decimal \(0.13\).

1) \(x=0.12+(0.009+0.0009+0.00009+\cdots)\). 2) \(x=0.13\). 3) The repeating tail has exact sum \(0.01\), and \(0.12+0.01=0.13\).

- Relate the height of an equilateral triangle to its side length. - Determine how the area changes when the side length is multiplied by a constant factor. - Treat the triangle areas as an infinite geometric series.

1. The area of an equilateral triangle with side length \(a\) is \(A=\frac{\sqrt3}{4}a^2\). For the first triangle, \(A_1=\frac{\sqrt3}{4}(6)^2=9\sqrt3\,\text{cm}^2\). 2. The height of an equilateral triangle is \(\frac{\sqrt3}{2}a\), so \(a_{n+1}=\frac{\sqrt3}{2}a_n\). 3. Areas scale by the square of the side-length factor. Therefore, the common ratio of the areas is \(q=\left(\frac{\sqrt3}{2}\right)^2=\frac34\). 4. Since \(|q|<1\), the area sum is \(S=\frac{9\sqrt3}{1-3/4}=36\sqrt3\,\text{cm}^2\).

\(36\sqrt3\,\text{cm}^2\)

- Determine the scale factor for the side lengths. - Recall that areas of similar figures scale by the square of the length factor. - Subtract the original area from the total area sum for part c).

1. The original perimeter is \(P_1=3a\). Connecting side midpoints halves each side length, so the perimeters form a geometric series with common ratio \(\frac12\). Thus, \(S_P=\frac{3a}{1-1/2}=6a\). 2. The original area is \(A_1=\frac{\sqrt3}{4}a^2\). Halving the side length multiplies area by \(\left(\frac12\right)^2=\frac14\), so \(S_A=\frac{A_1}{1-1/4}=\frac43A_1=\frac{\sqrt3}{3}a^2\). 3. Excluding the original triangle, the area sum is \(S_{\mathrm{inside}}=S_A-A_1=\frac43A_1-A_1=\frac13A_1\). Equivalently, \(S_{\mathrm{inside}}=\frac{\sqrt3}{12}a^2\).

a) \(6a\). b) \(\frac{\sqrt3}{3}a^2\). c) \(S_{\mathrm{inside}}=\frac13A_1=\frac{\sqrt3}{12}a^2\).

- Relate the apothem of a regular hexagon to its circumradius. - Square the radius scale factor to obtain the area scale factor. - Use the infinite geometric-series sum formula.

1. A regular hexagon inscribed in a circle of radius \(r_n\) has side length \(r_n\). Its apothem, which is the radius of its incircle, is \(r_{n+1}=r_n\cos 30^\circ=\frac{\sqrt3}{2}r_n\). 2. Circle areas scale by the square of the radius factor, so \(q=\left(\frac{\sqrt3}{2}\right)^2=\frac34\). 3. The first circle has area \(A_1=\pi R^2\). Therefore, \(S=\frac{\pi R^2}{1-3/4}=4\pi R^2\).

a) \(q=\frac34\). b) \(4\pi R^2\).

- Use the Pythagorean theorem to compare consecutive side lengths. - Areas scale by the square of the side-length factor. - The perimeters and areas have different common ratios.

1. If a square has side length \(a_n\), then a side of the next square is the hypotenuse of a right triangle with legs \(\frac{a_n}{2}\). Thus, \(a_{n+1}=\sqrt{\left(\frac{a_n}{2}\right)^2+\left(\frac{a_n}{2}\right)^2}=\frac{a_n}{\sqrt2}\). 2. The areas have first term \(a^2\) and common ratio \(q_A=\left(\frac1{\sqrt2}\right)^2=\frac12\). Therefore, \(S_A=\frac{a^2}{1-1/2}=2a^2\). 3. The perimeters have first term \(4a\) and common ratio \(q_P=\frac1{\sqrt2}\). Therefore, \(S_P=\frac{4a}{1-1/\sqrt2}=4a(2+\sqrt2)\).

The area sum is \(2a^2\), and the perimeter sum is \(4a(2+\sqrt2)\).

- Use the relationship between the inradius and circumradius of an equilateral triangle. - A length scale factor applies directly to circumference but must be squared for area. - Write a separate geometric series for each quantity.

1. For an equilateral triangle, the inradius is half the circumradius. Therefore, consecutive circle radii satisfy \(R_{n+1}=\frac12R_n\). 2. The circumferences form a geometric series with first term \(2\pi R\) and common ratio \(\frac12\). Thus, \(S_C=\frac{2\pi R}{1-1/2}=4\pi R\). 3. The areas form a geometric series with first term \(\pi R^2\) and common ratio \(\left(\frac12\right)^2=\frac14\). Thus, \(S_A=\frac{\pi R^2}{1-1/4}=\frac43\pi R^2\).

The area sum is \(\frac43\pi R^2\), and the circumference sum is \(4\pi R\).

- Treat the horizontal and vertical displacements separately. - Compare displacements that occur two segments apart. - Track both the magnitude and sign of each component.

1. The segment lengths are \(12, 6, 3, \frac32, \ldots\), a geometric sequence with common ratio \(\frac12\). Therefore, the total length is \(L=\frac{12}{1-1/2}=24\). 2. The horizontal displacements are \(12, -3, \frac34, -\frac3{16}, \ldots\), a geometric series with first term \(12\) and common ratio \(-\frac14\). Thus, \(x_\infty=\frac{12}{1-(-1/4)}=\frac{48}{5}=9.6\). 3. The vertical displacements are \(6, -\frac32, \frac38, -\frac3{32}, \ldots\), with the same common ratio. Thus, \(y_\infty=\frac6{1-(-1/4)}=\frac{24}{5}=4.8\). Therefore, \(P_\infty=(9.6, 4.8)\).

a) \(24\). b) \(P_\infty=(9.6, 4.8)\).

- Read the starting distance and angle from the diagram and focus on the first right triangle. - Compare two consecutive perpendicular path lengths using the same trigonometric relationship. - Once the constant scale factor is known, identify the infinite series represented by the path.

1. From the diagram, \(\overline{SP_0}=10\) and the angle between the rays is \(30^\circ\). In right triangle \(SP_0P_1\), \(\overline{P_0P_1}=10\sin30^\circ=5\). 2. Also, \(\overline{SP_1}=10\cos30^\circ=5\sqrt3\). Therefore \(\overline{P_1P_2}=5\sqrt3\sin30^\circ=\frac{5\sqrt3}{2}\). 3. Repeating the same right-triangle relationship shows that each perpendicular path segment is \(\cos30^\circ=\frac{\sqrt3}{2}\) times the previous one. Thus the path lengths form a geometric series with first term \(5\) and common ratio \(\frac{\sqrt3}{2}\). 4. Since \(\frac{\sqrt3}{2}<1\), \(L=\frac{5}{1-\sqrt3/2}=20+10\sqrt3\approx37.32\).

\(L=20+10\sqrt3\approx37.32\).

- Identify the common ratio from the summand. - Apply \(|q|<1\). - Find a formula for \(S(x)\) before setting it equal to \(4\). - Verify that the resulting \(x\) is in the convergence interval.

1. This is a geometric series with first term \(1\) and common ratio \(q=\frac{x-1}{2}\). 2. Convergence requires \(\left|\frac{x-1}{2}\right|<1\). Thus, \(-1<x<3\). 3. On this interval, \(S(x)=\frac1{1-(x-1)/2}=\frac2{3-x}\). 4. Set the sum equal to \(4\): \(\frac2{3-x}=4\). Then \(2=12-4x\), so \(x=2.5\). This value lies in the convergence interval.

a) \(x\in(-1, 3)\). b) \(x=2.5\).

- Express the three terms around the middle term. - Use the product to determine the middle term. - Apply the finite geometric-series sum formula in part b).

1. The first three terms are \(b_1, b_1q, b_1q^2\). Their product is \(b_1(b_1q)(b_1q^2)=(b_1q)^3=1000\). Since the terms are positive, \(b_1q=10\). 2. The first three terms can therefore be written as \(\frac{10}{q}, 10, 10q\). Their sum gives \(\frac{10}{q}+10+10q=35\). Multiplying by \(q\) and simplifying gives \(2q^2-5q+2=0\). Thus, \(q=2\) or \(q=\frac12\). 3. For the increasing sequence, \(q=2\), so \(b_1=\frac{10}{2}=5\). 4. The sum of the first five terms is \(S_5=5\frac{2^5-1}{2-1}=155\).

a) \(q=2\) or \(q=\frac12\). b) \(b_1=5\) and \(S_5=155\).

- Express the first three terms using \(a_0\) and \(q\). - Solve the resulting quadratic equation. - Apply the convergence condition before using the infinite-series sum formula.

1. The first-three-terms condition is \(10(1+q+q^2)=17.5\). Therefore, \(q^2+q-0.75=0\). Factoring gives \((q-0.5)(q+1.5)=0\), so \(q=0.5\) or \(q=-1.5\). 2. An infinite geometric series converges only when \(|q|<1\). Therefore, only \(q=0.5\) is valid. 3. For \(q=0.5\), \(S=\frac{10}{1-0.5}=20\).

1) \(q=0.5\) or \(q=-1.5\). 2) The series converges for \(q=0.5\), and its sum is \(20\).

- Relate the side length of an inscribed regular hexagon to the circle’s radius. - Find the inradius-to-circumradius ratio of a regular hexagon. - Apply the infinite geometric-series sum formula.

1. A regular hexagon inscribed in a circle has side length equal to the circle’s radius. Therefore, the first hexagon has perimeter \(P_1=6R\). 2. The inradius of a regular hexagon is \(\frac{\sqrt3}{2}\) times its circumradius. Thus, each new circle has radius \(\frac{\sqrt3}{2}\) times the preceding circle’s radius. 3. The hexagon perimeters form a geometric series with common ratio \(\frac{\sqrt3}{2}\). Therefore, \(S_P=\frac{6R}{1-\sqrt3/2}=12R(2+\sqrt3)\).

\(12R(2+\sqrt3)\)

- Write the original sum and a second copy in which every term has been multiplied by the common ratio. - Align equal powers before subtracting the two equations so that most terms cancel. - After the cancellation, isolate \(S_n\) before substituting values from the numerical series.

1. Write \(S_n=a+ar+ar^2+\cdots+ar^{n-1}\). 2. Multiply by \(r\): \(rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n\). 3. Subtract: \(S_n-rS_n=a-ar^n\). 4. Factor and solve for \(S_n\): \((1-r)S_n=a(1-r^n)\), so \(S_n=\frac{a(1-r^n)}{1-r}\). 5. For the given series, \(a=3\), \(r=2\), and \(n=9\). Thus \(S_9=\frac{3(1-2^9)}{1-2}=1533\).

\(S_n=\frac{a(1-r^n)}{1-r}\), and the given sum is \(1533\).

- Look at the plotted points with odd values of \(n\) separately from those with even values of \(n\). - Estimate a horizontal level that both groups of points appear to approach. - For the exact check, identify the first term and common ratio and then decide whether the infinite geometric-series formula applies.

a) The odd-indexed partial sums decrease: \(8,6,5.5,5.375\). The even-indexed partial sums increase: \(4,5,5.25,5.3125\). The two groups approach the same value from opposite sides. b) The points approach about \(5.3\). c) The first term is \(8\) and the common ratio is \(-\frac{1}{2}\). Since \(\left|-\frac{1}{2}\right|<1\), the infinite sum exists and is \(\frac{8}{1-(-1/2)}=\frac{16}{3}\).

a) Odd-indexed partial sums decrease while even-indexed partial sums increase; both approach the same limit from opposite sides. b) Approximately \(5.3\). c) Exactly \(\frac{16}{3}\).

- Treat the first drop separately from the later bounces. - Each rebound height contributes distance in two directions before the next bounce begins. - The sequence of maximum rebound heights forms an infinite geometric series.

1. The initial drop contributes \(10\,\text{ft}\). 2. The rebound heights form the geometric series \(6+3.6+2.16+\cdots\), with first term \(6\) and ratio \(\frac{3}{5}\). 3. The sum of all rebound heights is \(\frac{6}{1-3/5}=15\,\text{ft}\). 4. Each rebound height is traveled once upward and once downward, so the rebound portion contributes \(2(15)=30\,\text{ft}\). 5. The total distance is \(10+30=40\,\text{ft}\).

\(40\,\text{ft}\).

- Treat the terms from index \(m\) through index \(n\) as their own finite geometric series. - Determine the first term of that shorter series and how many terms it contains. - Keep the exponent attached to the original sequence index when identifying the segment's first term.

1. The first term of the segment is \(ar^{m-1}\), and the segment contains \(n-m+1\) terms. 2. Applying the finite geometric-series formula to this segment gives \(\sum_{k=m}^{n}a_k=\frac{ar^{m-1}(1-r^{n-m+1})}{1-r}\). 3. Here \(a=3\), \(r=-\frac{1}{2}\), \(m=4\), and \(n=8\). 4. Thus the sum is \(\frac{3(-1/2)^3(1-(-1/2)^5)}{1+1/2}=-\frac{33}{128}\).

\(\sum_{k=m}^{n}a_k=\frac{ar^{m-1}(1-r^{n-m+1})}{1-r}\), and the given segment sums to \(-\frac{33}{128}\).

- The last displayed term shows how the exponent depends on the unknown number of terms. - Express the entire finite sum with the geometric-series formula before trying to identify \(n\). - After simplifying the sum equation, compare powers with the same base.

1. The first term is \(2\) and the common ratio is \(3\). 2. The finite sum is \(S_n=\frac{2(3^n-1)}{3-1}=3^n-1\). 3. Set \(3^n-1=6560\), giving \(3^n=6561\). 4. Since \(6561=3^8\), \(n=8\).

\(n=8\).

- Find the first terms of the difference sequence. - Express the original sequence as its first term plus a sum of differences. - Use the finite geometric-series formula. - Evaluate the exponential term as \(n\to\infty\).

1. The first two differences are \(d_1=15-10=5\) and \(d_2=17.5-15=2.5\). Thus, the common ratio of the difference sequence is \(q=0.5\), and \(d_n=5(0.5)^{n-1}\). 2. Recover \(a_n\) by summing the first \(n-1\) differences: \(a_n=10+\sum_{k=1}^{n-1}5(0.5)^{k-1}\). 3. Using the finite geometric-series formula, \(a_n=10+5\frac{1-(0.5)^{n-1}}{1-0.5}\) \(=20-10(0.5)^{n-1}\). 4. Since \((0.5)^{n-1}\to0\), \(\lim_{n\to\infty}a_n=20\).

a) \(a_n=20-10(0.5)^{n-1}\). b) The sequence converges to \(20\).

- Express \(a_1\) in terms of \(q\) using the infinite-series sum. - Use equal consecutive differences for the modified arithmetic sequence. - Substitute to reduce the system to one equation in \(q\).

1. From the infinite geometric-series formula, \(\frac{a_1}{1-q}=24\), so \(a_1=24(1-q)\). 2. The arithmetic terms are \(a_1, a_1q+2, a_1q^2+1\). Equal consecutive differences give \((a_1q+2)-a_1=(a_1q^2+1)-(a_1q+2)\). 3. Simplifying yields \(a_1(q-1)^2=3\). Substitute \(a_1=24(1-q)\): \(24(1-q)^3=3\). Thus, \((1-q)^3=\frac18\), so \(1-q=\frac12\) and \(q=\frac12\). 4. Therefore, \(a_1=24\left(1-\frac12\right)=12\).

\(a_1=12\) and \(q=\frac12\).

- Separate the total sum into the sums of the odd-indexed and even-indexed terms. - Relate each even-indexed term to the odd-indexed term immediately before it. - Use the signs of the two partial sums to determine the sign of \(q\).

1. The total sum is the sum of the odd-indexed and even-indexed terms: \(S=S_{\mathrm{odd}}+S_{\mathrm{even}}\). Therefore, \(6=9+S_{\mathrm{even}}\), so \(S_{\mathrm{even}}=-3\). 2. Each even-indexed term is \(q\) times the preceding odd-indexed term, so \(S_{\mathrm{even}}=qS_{\mathrm{odd}}\). Thus, \(-3=9q\), giving \(q=-\frac13\). 3. Use the total-sum formula: \(6=\frac{a_1}{1-(-1/3)}=\frac{a_1}{4/3}\). Hence, \(a_1=8\). 4. For part b), \(S<S_{\mathrm{odd}}\), so the even-indexed terms must have a negative sum. Since \(S_{\mathrm{even}}=qS_{\mathrm{odd}}\) and \(S_{\mathrm{odd}}>0\), it follows that \(q<0\).

a) \(q=-\frac13\) and \(a_1=8\). b) Because \(S<S_{\mathrm{odd}}\), the even-indexed terms have a negative sum. Since \(S_{\mathrm{even}}=qS_{\mathrm{odd}}\) and \(S_{\mathrm{odd}}>0\), \(q\) must be negative.

- Use the diagram to relate each square to the circle immediately inside it. - Compare the diagonal of one square with the diameter of the preceding circle to find the scale change between square areas. - Sum the two geometric area sequences separately before comparing them.

1. Let the side length of the \(n\)th square be \(s_n\). Its area is \(A_{Q,n}=s_n^2\). 2. The circle inscribed in that square has radius \(r_n=\frac{s_n}{2}\), so \(A_{C,n}=\frac{\pi}{4}s_n^2\). 3. The next square has diagonal \(s_n\), equal to the diameter of the preceding circle. Therefore its area is \(A_{Q,n+1}=\frac12s_n^2=\frac12A_{Q,n}\). Thus both area sequences are geometric with common ratio \(\frac12\). 4. The square-area sum is \(S_Q=\frac{s^2}{1-1/2}=2s^2\), and the circle-area sum is \(S_C=\frac{(\pi/4)s^2}{1-1/2}=\frac{\pi}{2}s^2\). 5. Hence \(\frac{S_C}{S_Q}=\frac{(\pi/2)s^2}{2s^2}=\frac{\pi}{4}\). The factor \(s^2\) cancels, so the ratio is independent of the original side length.

\(\frac{S_C}{S_Q}=\frac{\pi}{4}\).

- Relate the cube’s space diagonal to the diameter of its circumscribed sphere. - Relate the cube’s side length to the diameter of its inscribed sphere. - Cube the radius scale factor to obtain the volume scale factor.

1. The first sphere has volume \(V_1=\frac43\pi R^3\). 2. If the cube has side length \(a\), its space diagonal equals the original sphere’s diameter: \(a\sqrt3=2R\). Thus, \(a=\frac{2R}{\sqrt3}\). 3. The sphere inscribed in the cube has diameter \(a\), so its radius is \(R_2=\frac{a}{2}=\frac{R}{\sqrt3}\). 4. Consecutive radii have ratio \(\frac1{\sqrt3}\), so consecutive volumes have ratio \(q=\left(\frac1{\sqrt3}\right)^3=\frac1{3\sqrt3}\). 5. Therefore, \(S_V=\frac{(4/3)\pi R^3}{1-1/(3\sqrt3)}=\frac{2(9+\sqrt3)}{13}\pi R^3\).

\(\frac{2(9+\sqrt3)}{13}\pi R^3\)

- Find the inradius of the first equilateral triangle. - Relate the circumradius of an equilateral triangle to its side length. - Square the side-length scale factor to obtain the area ratio.

1. The first triangle has area \(A_{T,1}=\frac{\sqrt3}{4}s^2\). Its inradius is \(r_1=\frac{\sqrt3}{6}s\), so the first circle has area \(A_{C,1}=\pi r_1^2=\frac\pi{12}s^2\). 2. The circumradius of the next equilateral triangle is \(r_1\). Since an equilateral triangle with circumradius \(R\) has side length \(R\sqrt3\), the next side length is \(s_2=r_1\sqrt3=\frac12s\). Thus, corresponding areas are multiplied by \(q=\left(\frac12\right)^2=\frac14\). 3. Therefore, the sum of the triangle areas is \(S_T=\frac{(\sqrt3/4)s^2}{1-1/4}=\frac{\sqrt3}{3}s^2\). The sum of the circle areas is \(S_C=\frac{(\pi/12)s^2}{1-1/4}=\frac\pi9s^2\).

a) \(\frac{\sqrt3}{3}s^2\). b) \(\frac\pi9s^2\).

- Use the two side segments to form a right triangle whose hypotenuse is a side of the new square. - Compare the area of the second square with the area of the first. - Use the same ratio for every stage of the construction.

1. The first square has area \(A_1=a^2\). 2. A side of \(Q_2\) is the hypotenuse of a right triangle with legs \(\frac14a\) and \(\frac34a\). Therefore, \(s_2^2=\left(\frac14a\right)^2+\left(\frac34a\right)^2=\frac58a^2\). Hence, \(A_2=\frac58a^2\). 3. The construction is repeated at the same scale, so the square areas form a geometric series with common ratio \(\frac58\). 4. Thus, \(S=\frac{a^2}{1-5/8}=\frac83a^2\).

\(\frac83a^2\)

- Use the sum formula and the equation \(a_2=a_1q\). - Eliminate \(a_1\) to obtain an equation in \(q\). - For part b), apply the condition that the absolute value of the common ratio must be less than \(1\).

1. The conditions for part a) are \(\frac{a_1}{1-q}=12\) and \(a_1q=\frac83\). 2. From the first equation, \(a_1=12(1-q)\). Substitute into the second: \(12q(1-q)=\frac83\). This simplifies to \(9q^2-9q+2=0\), so \(q=\frac13\) or \(q=\frac23\). 3. The corresponding first terms are \(a_1=12\left(1-\frac13\right)=8\) and \(a_1=12\left(1-\frac23\right)=4\). 4. In part b), the common ratio is \(r=\frac{x-2}{3}\). Convergence requires \(\left|\frac{x-2}{3}\right|<1\). Thus, \(|x-2|<3\), which gives \(-1<x<5\).

a) \((a_1, q)=\left(8, \frac13\right)\) or \((a_1, q)=\left(4, \frac23\right)\). b) \(x\in(-1, 5)\).

- Write the sum formula for the original series. - List the first few products of consecutive terms and identify the new common ratio. - Eliminate \(a_1\) using the original sum equation. - Check that each resulting ratio gives convergent series.

1. The original series satisfies \(\frac{a_1}{1-q}=10\), so \(a_1=10(1-q)\). 2. The terms of the new series are \(a_1a_2=a_1^2q,\quad a_2a_3=a_1^2q^3,\quad a_3a_4=a_1^2q^5,\ldots\). This is geometric with first term \(a_1^2q\) and common ratio \(q^2\). Therefore, \(\frac{a_1^2q}{1-q^2}=\frac{50}{3}\). 3. Substitute \(a_1=10(1-q)\): \(\frac{100(1-q)^2q}{1-q^2}=\frac{50}{3}\). Since \(1-q^2=(1-q)(1+q)\), this simplifies to \(6q^2-5q+1=0\). Thus, \(q=\frac12\) or \(q=\frac13\). 4. The corresponding first terms are \(a_1=10\left(1-\frac12\right)=5\) and \(a_1=10\left(1-\frac13\right)=\frac{20}{3}\).

\(a_1=5\) and \(q=\frac12\), or \(a_1=\frac{20}{3}\) and \(q=\frac13\).

- Write a sum formula for each series. - Compare the square of the original sum with the sum of the squared terms. - Determine the common ratio of the squared-term series.

1. The two sums satisfy \(\frac{a_1}{1-q}=9\) and \(\frac{a_1^2}{1-q^2}=40.5\). 2. Squaring the first equation gives \(\frac{a_1^2}{(1-q)^2}=81\). Divide this value by the sum of the squared-term series: \(\frac{1-q^2}{(1-q)^2}=\frac{81}{40.5}=2\). Since \(1-q^2=(1-q)(1+q)\), \(\frac{1+q}{1-q}=2\). Therefore, \(1+q=2-2q\), so \(q=\frac13\). 3. Then \(a_1=9\left(1-\frac13\right)=6\). 4. If the original series converges, then \(|q|<1\). The squared-term series has common ratio \(q^2\), and \(0\le q^2<1\). Therefore, the squared-term series also converges.

a) \(a_1=6\) and \(q=\frac13\). b) The squared-term series has common ratio \(q^2\). Since \(|q|<1\) implies \(q^2<1\), it converges.

- Instead of comparing two long sums directly, describe the part of the infinite series that remains after the first \(n\) terms. - The remainder is itself a geometric series beginning with the next omitted term. - After isolating a power of the common ratio in an inequality, logarithms can be used to solve for the exponent.

1. The common ratio is \(0.6\), so the infinite sum is \(\frac{20}{1-0.6}=50\). 2. The amount remaining after the first \(n\) terms is \(50(0.6)^n\). 3. Require \(50(0.6)^n<0.01\), so \((0.6)^n<0.0002\). 4. Taking logarithms gives \(n>\frac{\ln(0.0002)}{\ln(0.6)}\approx16.67\). 5. The smallest positive integer that works is \(17\).

\(n=17\).