Free Math Worksheets

- 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, \(0.4\overline{18}=\frac25+\frac1{55}=\frac{22}{55}+\frac1{55}=\frac{23}{55}\).

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

- 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) \(P_\infty=(9.6, 4.8)\).

- Find the first two perpendicular lengths using right-triangle trigonometry. - Determine whether the ratio of consecutive lengths is constant. - Apply the infinite geometric-series sum formula.

1. 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, the next perpendicular segment has length \(\overline{P_1P_2}=5\sqrt3\sin30^\circ=\frac{5\sqrt3}{2}\). 3. Each perpendicular segment is \(\cos30^\circ=\frac{\sqrt3}{2}\) times the previous one. Thus, the lengths form a geometric series with first term \(5\) and common ratio \(\frac{\sqrt3}{2}\). 4. Since the common ratio is less than \(1\), \(L=\frac5{1-\sqrt3/2}=20+10\sqrt3\). Numerically, \(L\approx37.32\).

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

- 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, \(S_P=\frac{6R}{1-\sqrt3/2}=12R(2+\sqrt3)\).

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

- 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.

- Express the first square and circle areas in terms of \(s\). - Use the diagonal of an inscribed square to compare consecutive square areas. - Find both infinite geometric-series sums before taking their ratio.

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 its area is \(A_{C,n}=\pi\left(\frac{s_n}{2}\right)^2=\frac\pi4s_n^2\). 3. The next square has diagonal equal to the circle diameter, which is \(s_n\). Therefore, its area is \(A_{Q,n+1}=\frac{s_n^2}{2}=\frac12A_{Q,n}\). Both the square areas and the circle areas form geometric series with common ratio \(\frac12\). 4. The sum of the square areas is \(S_Q=\frac{s^2}{1-1/2}=2s^2\). The sum of the circle areas is \(S_C=\frac{(\pi/4)s^2}{1-1/2}=\frac\pi2s^2\). 5. Thus, \(\frac{S_C}{S_Q}=\frac{(\pi/2)s^2}{2s^2}=\frac\pi4\). The factor \(s^2\) cancels, so the ratio is independent of the original side length.

\(\frac{S_C}{S_Q}=\frac\pi4\).

- 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.