- Do consecutive terms have a constant ratio? - What condition on \(|r|\) makes an infinite geometric series converge? - Does this ratio satisfy that condition?

1. The common ratio is \(r=\frac{5}{4}\). 2. Because \(|r|\ge1\), the terms do not produce a convergent geometric series. 3. The series diverges.

The series diverges.

- Find the ratio between consecutive terms, including its sign. - Recall that a convergent geometric series requires the ratio’s absolute value to be strictly less than \(1\). - Check whether the alternating partial sums settle toward one value.

1. The common ratio is \(r=-1\). 2. Because \(|r|\ge1\), the terms do not produce a convergent geometric series. 3. The series diverges.

The common ratio is \(r=-1\), so \(|r|=1\) and the series diverges.

- Simplify \(1^n\) to identify what every term equals. - Consider what happens when the same nonzero amount is added to each successive partial sum. - Verify the geometric convergence condition before attempting to use the infinite-sum formula.

1. The common ratio is \(r=1\). 2. Because \(|r|\ge1\), the terms do not produce a convergent geometric series. 3. The series diverges.

Since \(r=1\), the convergence condition \(|r|<1\) fails. The infinite geometric sum formula is not valid, and the series diverges.

- Divide each term after the first by its preceding term. - A geometric pattern needs the same multiplier at every step.

1. Consecutive ratios are \(10/5=2\), \(20/10=2\), and \(40/20=2\). 2. The repeated ratio is constant, so the terms form a geometric pattern with \(r=2\).

Yes. The common ratio is \(r=2\).

- Compare multiplication between terms with addition between terms. Which matters here? - Test whether one multiplier works for every pair of neighboring terms in each list.

1. In a), the three consecutive ratios all equal \(3\). 2. In b), the first two ratios equal \(3\), but the last is \(50/18=25/9\). 3. Therefore only a) is geometric.

Only a) is geometric, with common ratio \(3\). In b), the last ratio is \(25/9\), not \(3\).

- Check whether the given ratio permits a finite infinite-series sum. - Which number belongs in the numerator of the given formula?

1. Since \(|r|=\frac12<1\), the infinite geometric sum exists. 2. Substitute into the formula: \(S=\frac{1}{1-\frac12}=2\).

\(S=2\).

- Because the index starts at \(0\), substitute \(n=0\) to identify the first term. - Read the common ratio from the base raised to the power \(n\), then test whether its absolute value is less than \(1\). - For a convergent geometric series, use \(\frac{a_1}{1-r}\) with the first term and ratio you identified.

1. The common ratio is \(r=\frac{1}{3}\), and \(|r|<1\). 2. The first term is \(5\). 3. The sum is \(\frac{5}{1-(\frac{1}{3})}=\frac{15}{2}\).

The first term is \(5\), the common ratio is \(r=\frac{1}{3}\), and the series converges to \(\frac{15}{2}\).

- Keep the negative sign when identifying the common ratio; it controls the alternating pattern. - The convergence criterion depends on \(|r|\), not on whether \(r\) is positive or negative. - Once convergence is established, substitute the first term and signed ratio into \(\frac{a_1}{1-r}\).

1. The common ratio is \(r=- \frac{1}{2}\), and \(|r|<1\). 2. The first term is \(8\). 3. The sum is \(\frac{8}{1-(- \frac{1}{2})}=\frac{16}{3}\).

Since \(|r|=\frac{1}{2}<1\), the series converges to \(\frac{16}{3}\).

- Identify the coefficient outside the power as the first term because the exponent begins at \(0\). - Check the magnitude of the common ratio before using any infinite-sum formula. - Simplify \(1-r\) first, then divide the fractional first term by that result.

1. The common ratio is \(r=\frac{3}{5}\), and \(|r|<1\). 2. The first term is \(\frac{7}{4}\). 3. The sum is \(\frac{\frac{7}{4}}{1-(\frac{3}{5})}=\frac{35}{8}\).

The series converges to \(\frac{35}{8}\).

- Distinguish the negative first term from the positive common ratio. - Verify \(|r|<1\) before applying the geometric-series sum formula. - The sign of the final sum should be consistent with all terms of the series being negative.

1. The common ratio is \(r=\frac{1}{4}\), and \(|r|<1\). 2. The first term is \(-6\). 3. The sum is \(\frac{-6}{1-(\frac{1}{4})}=-8\).

The common ratio is \(r=\frac{1}{4}\), and the series converges to \(-8\).

- Do consecutive terms have a constant ratio? - Is the absolute value of that ratio less than \(1\)? - Which first term and common ratio belong in the sum formula?

1. The common ratio is \(r=-\frac23\), so the series is geometric and \(|r|<1\). 2. The first term is \(9\). 3. The sum is \(\frac{9}{1-(-\frac23)}=\frac{27}{5}\).

The series is geometric with common ratio \(r=-\frac23\), and it converges to \(\frac{27}{5}\).

- What formula relates the first term, common ratio, and sum of a convergent geometric series? - Substitute the known first term and sum, then solve for \(r\). - Does the resulting ratio satisfy \(|r|<1\)?

1. The sum condition gives \(\frac{6}{1-r}=15\). 2. Thus \(6=15-15r\), so \(15r=9\). 3. \(r=\frac35\), which satisfies \(|r|<1\).

\(r=\frac35\).

- What formula relates the first term, common ratio, and sum of a convergent geometric series? - Substitute the known ratio and sum, then solve for the first term. - Check that the given ratio satisfies the convergence condition.

1. Let the first term be \(a\). 2. \(\frac{a}{1-(-\frac14)}=20\). 3. \(\frac{4a}{5}=20\), so \(a=25\).

The first term is \(25\).

- What are the first term and common ratio in terms of \(k\)? - Substitute them into the convergent geometric-series sum formula. - Solve the resulting equation for \(k\).

1. The sum is \(\frac{k}{1-\frac25}=\frac{5k}{3}\). 2. Set \(\frac{5k}{3}=18\). 3. \(k=\frac{54}{5}\).

\(k=\frac{54}{5}\).

- How is each intensity reading related to the previous one? - What are the first term and common ratio of the resulting series? - Does the ratio satisfy the condition for an infinite geometric sum?

1. The intensity readings form a geometric series with first term \(80\) and ratio \(\frac38\). 2. Since \(\left|\frac38\right|<1\), the total is \(\frac{80}{1-\frac38}=128\).

The sum of the intensity readings is \(128\) units.

- How is each later adjustment related to the previous one? - What are the first term and common ratio of the resulting series? - Does the ratio satisfy the condition for an infinite geometric sum?

1. The adjustments are \(30,30(0.12),30(0.12)^2,\ldots\). 2. The common ratio is \(0.12=\frac3{25}\). 3. The total is \(\frac{30}{1-0.12}=\frac{375}{11}\approx34.09\).

The total adjustment is \(\frac{375}{11}\approx34.09\) brightness units.

- Assign positive and negative signs according to direction. - What common ratio relates consecutive signed displacements? - Does that ratio satisfy the condition for an infinite geometric sum?

1. The signed displacements form \(24-12+6-\cdots\), a geometric series with ratio \(-\frac12\). 2. The sum is \(\frac{24}{1-(-\frac12)}=16\).

The limiting displacement is \(16\) pixels to the right.

- How is the area added in each phase related to the previous phase? - What are the first term and common ratio of the resulting series? - Does the ratio satisfy the condition for an infinite geometric sum?

1. The added areas form a geometric series with first term \(160\) and ratio \(0.75=\frac34\). 2. The total is \(\frac{160}{1-\frac34}=640\,\text{ft}^2\).

A total of \(640\,\text{ft}^2\) of seating area would be added.

- Identify the common ratio of each geometric series. - Evaluate the two convergent sums separately. - Preserve the subtraction between the two series.

1. The first geometric series has sum \(\frac{7}{1-\frac12}=14\). 2. The second has sum \(\frac{2}{1-\frac13}=3\). 3. The difference is \(14-3=11\).

\(11\).

- What is the common ratio of the series? - Does its absolute value satisfy the convergence condition? - When is the infinite geometric-series sum formula valid?

1. The common ratio is \(-\frac43\). 2. Its magnitude is \(\frac43>1\), so the terms do not approach \(0\). 3. The infinite-sum expression is not valid, and the series diverges.

The series diverges; the infinite geometric-sum formula cannot be used.

- Identify the first term and common ratio of each series. - Evaluate each convergent geometric sum. - Compare the two exact values.

1. \(A=\frac{5}{1-\frac13}=\frac{15}{2}\). 2. \(B=\frac{4}{1-\frac25}=\frac{20}{3}\). 3. \(\frac{15}{2}>\frac{20}{3}\).

\(A=\frac{15}{2}\) and \(B=\frac{20}{3}\); series \(A\) has the greater sum.

- What place-value contribution does each repeated block \(27\) make? - What constant ratio relates one contribution to the next? - Use the infinite geometric-series formula to simplify the value.

1. \(0.272727\ldots=\frac{27}{100}+\frac{27}{100^2}+\frac{27}{100^3}+\cdots\). 2. The first term is \(\frac{27}{100}\) and the ratio is \(\frac1{100}\). 3. The sum is \(\frac{\frac{27}{100}}{1-\frac1{100}}=\frac{27}{99}=\frac3{11}\).

\(0.272727\ldots=\frac{27}{100}+\frac{27}{100^2}+\frac{27}{100^3}+\cdots=\frac3{11}\).

- Expand several layers of the nested expression. - Track how many factors of the repeated fraction multiply each copy of the constant. - Identify the first term and the factor between successive expanded terms.

1. Expanding the nesting gives \(2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\cdots\). 2. This is geometric with first term \(2\) and ratio \(\frac13\). 3. Its sum is \(\frac{2}{1-1/3}=3\).

The nested expression equals \(3\).

- How can you generate the next term when the ratio is known? - Write the terms before adding any of them. - A partial sum includes every term through the stated index.

1. Multiply each term by \(\frac12\) to get \(3,\frac32,\frac34,\frac38\). 2. \(S_3=3+\frac32+\frac34=\frac{21}{4}\). 3. \(S_4=\frac{21}{4}+\frac38=\frac{45}{8}\).

The first four terms are \(3,\frac32,\frac34,\frac38\). \(S_3=\frac{21}{4}\) and \(S_4=\frac{45}{8}\).

- Include the sign when finding the common ratio. - Compare the absolute value of the ratio with one. - Remember the negative sign in the denominator of the sum formula.

1. Each term is multiplied by \(-\frac12\), so \(r=-\frac12\). 2. Because \(|r|=\frac12<1\), the series converges even though the signs alternate. 3. The sum is \(\frac{4}{1-(-1/2)}=\frac83\).

The series converges to \(\frac83\).

- The remaining series starts at the first term you have not included. - Its ratio is unchanged from the original geometric pattern. - Apply the sum formula to this tail rather than the whole series.

1. The fourth term is \(\frac13\) and the ratio is \(\frac13\). 2. The remaining terms form a geometric series beginning at the fourth term. 3. Its sum is \(\frac{1/3}{1-1/3}=\frac12\).

The fourth term is \(\frac13\). The tail beginning with that term sums to \(\frac12\).

- Each vertical change between neighboring partial sums is one series term. - Compare the signed sizes of consecutive vertical changes. - Think about what a negative multiplier does to a term's sign.

1. The first successive changes are \(S_1-S_0=1\), \(S_2-S_1=-\frac12\), and \(S_3-S_2=\frac14\). 2. Each change is \(-\frac12\) times the preceding change, so the series has ratio \(r=-\frac12\). 3. A negative ratio alternates the direction of consecutive additions, producing the plotted up-and-down pattern.

The first three changes are \(1\), \(-\frac12\), and \(\frac14\). Their common ratio is \(-\frac12\); its negative sign makes the changes alternate up and down.

- How many places to the right does the next nonzero digit occur? - Each one-place shift changes the value by the same factor in base \(5\). - Compare two consecutive nonzero place values, not two adjacent digits.

1. The nonzero digits occur in the first, third, and fifth fractional places, with contributions \(\frac15,\frac1{125},\frac1{3125}\). 2. Each contribution is \(\frac1{25}\) of the previous one, so the geometric ratio is \(r=\frac1{25}\).

The first three place values are \(\frac15,\frac1{125},\frac1{3125}\), with common ratio \(r=\frac1{25}\).

- How much does the first digit after the point contribute? - What changes when the same digit moves one place farther right? - Check that the resulting geometric ratio permits an infinite sum.

1. The place values give \(0.\overline{2}_5=\frac25+\frac2{25}+\frac2{125}+\cdots\). 2. This geometric series has first term \(\frac25\) and ratio \(\frac15\). 3. Its sum is \(\frac{2/5}{1-1/5}=\frac12\).

\(0.\overline{2}_5=\frac25+\frac2{25}+\frac2{125}+\cdots=\frac12\) in base \(10\).

- Substitute the lower index to find the first term of the series. - What is the common ratio, and does its absolute value satisfy the convergence condition? - Which first term belongs in the geometric-series sum formula?

1. The common ratio is \(r=\frac{2}{5}\), and \(|r|<1\). 2. The first term is \(\frac{16}{25}\). 3. The sum is \(\frac{\frac{16}{25}}{1-(\frac{2}{5})}=\frac{16}{15}\).

The first term is \(\frac{16}{25}\), and the series converges to \(\frac{16}{15}\).

- Substitute the lower index to find the first term of the tail. - What is the constant ratio between consecutive terms? - Does the ratio satisfy the convergence condition, and which first term belongs in the sum formula?

1. The common ratio is \(r=- \frac{1}{3}\), and \(|r|<1\). 2. The first term is \(- \frac{10}{3}\). 3. The sum is \(\frac{- \frac{10}{3}}{1-(- \frac{1}{3})}=- \frac{5}{2}\).

The first term is \(-\frac{10}{3}\), the common ratio is \(r=-\frac{1}{3}\), and the series converges to \(-\frac{5}{2}\).

- Substitute the lower index to find the first term of the tail. - What is the common ratio, and does its absolute value satisfy the convergence condition? - Which first term belongs in the geometric-series sum formula?

1. The common ratio is \(r=\frac{1}{2}\), and \(|r|<1\). 2. The first term is \(\frac{3}{32}\). 3. The sum is \(\frac{\frac{3}{32}}{1-(\frac{1}{2})}=\frac{3}{16}\).

The series converges to \(\frac{3}{16}\).

- Evaluate the term at the lower index carefully; an odd power keeps the ratio’s negative sign. - The tail has the same common ratio as the underlying geometric sequence, so test its absolute value. - Use the term at \(n=3\) as the first term in the infinite geometric-series formula.

1. The common ratio is \(r=- \frac{3}{4}\), and \(|r|<1\). 2. The first term is \(- \frac{81}{16}\). 3. The sum is \(\frac{- \frac{81}{16}}{1-(- \frac{3}{4})}=- \frac{81}{28}\).

The first term is \(-\frac{81}{16}\), and the series converges to \(-\frac{81}{28}\).

- What expression is the common ratio? - What inequality must the absolute value of that ratio satisfy? - Solve the resulting compound inequality for \(c\).

1. The common ratio is \(\frac{c-1}{3}\). 2. Convergence requires \(\left|\frac{c-1}{3}\right|<1\). 3. Thus \(-3<c-1<3\), so \(-2<c<4\).

\(-2<c<4\).

- How can the first two terms be recovered from \(S_1\) and \(S_2\)? - What ratio do those two terms determine? - Does that ratio satisfy the convergence condition, and what sum formula then applies?

1. The first term is \(a_1=S_1=9\). 2. The second term is \(a_2=S_2-S_1=3\). 3. The common ratio is \(r=\frac{a_2}{a_1}=\frac13\). 4. The infinite sum is \(\frac{9}{1-\frac13}=\frac{27}{2}\).

The common ratio is \(\frac13\), and the sum is \(\frac{27}{2}\).

- Separate the initial drop from the repeated rebound distances. - What geometric series represents the rebound heights? - Why is each rebound height counted twice in the total distance?

1. The initial downward distance is \(10\,\text{ft}\). 2. The rebound heights are \(8,8\left(\frac45\right),8\left(\frac45\right)^2,\ldots\), whose sum is \(\frac{8}{1-\frac45}=40\). 3. Each rebound height is traveled once upward and once downward. 4. The total distance is \(10+2\cdot40=90\,\text{ft}\).

The total vertical distance is \(90\,\text{ft}\).

- First determine the size of the first correction and how one correction distance compares with the next. - The requested corrections are two indices apart, so compare the 1st correction with the 3rd rather than with the 2nd. - Decide whether the odd-numbered correction distances form their own geometric series.

1. The first correction is \(\frac{2}{3}\cdot18=12\,\text{cm}\). 2. After each correction, the remaining error is multiplied by \(\frac{1}{3}\), so each correction distance is \(\frac{1}{3}\) of the preceding correction distance. 3. Consecutive odd-numbered corrections are two steps apart, so their common ratio is \(\left(\frac{1}{3}\right)^2=\frac{1}{9}\). 4. The odd-numbered correction distances therefore form the geometric series \(12+12\left(\frac{1}{9}\right)+12\left(\frac{1}{9}\right)^2+\cdots\). Its sum is \(\frac{12}{1-\frac{1}{9}}=\frac{27}{2}\,\text{cm}\).

The total distance moved during the odd-numbered corrections is \(\frac{27}{2}\,\text{cm}\).

- List the first few even-indexed terms. - What constant ratio relates one selected term to the next selected term? - Which first term and ratio belong in the infinite geometric-series sum formula?

1. The even-indexed terms are \(a_2=4\), \(a_4=\frac49\), \(a_6=\frac4{81},\ldots\). 2. The ratio between successive even-indexed terms is \(\frac19\). 3. The sum is \(\frac{4}{1-\frac19}=\frac92\).

The sum is \(\frac92\).

- List the first few odd-indexed terms. - How many original index steps separate one selected term from the next? - What ratio does that produce for the selected geometric series?

1. The odd-indexed terms are \(5,\frac54,\frac5{16},\ldots\). 2. The ratio between successive odd-indexed terms is \(\left(-\frac12\right)^2=\frac14\). 3. The sum is \(\frac{5}{1-\frac14}=\frac{20}{3}\).

The sum is \(\frac{20}{3}\).

- Can the expression be split into two separate series? - What are the first term and common ratio of each component series? - Evaluate each convergent geometric series before combining the results.

1. Both component geometric series converge, so the sum can be separated. 2. \(\sum_{n=0}^{\infty}3\left(\frac12\right)^n=\frac{3}{1-\frac12}=6\). 3. \(\sum_{n=0}^{\infty}2\left(\frac14\right)^n=\frac{2}{1-\frac14}=\frac83\). 4. The requested sum is \(6-\frac83=\frac{10}{3}\).

\(\frac{10}{3}\).

- Simplify each displayed pair to identify the grouped geometric series. - What are the first term and ratio of the grouped series? - Then evaluate the original geometric series directly and compare the results.

1. The grouped terms are \(2,\frac12,\frac18,\ldots\), with ratio \(\frac14\). 2. The grouped sum is \(\frac{2}{1-\frac14}=\frac83\). 3. The original series has first term \(4\) and ratio \(-\frac12\), so its sum is \(\frac{4}{1+\frac12}=\frac83\).

Both methods give \(\frac83\).

- What are the first term and common ratio? - Which finite geometric-series formula gives the sum of the first \(n\) terms? - What happens to the ratio raised to the \(n\)th power as \(n\to\infty\)?

1. The first \(n\) terms have \(S_n=3\cdot\frac{1-\left(\frac25\right)^n}{1-\frac25}=5\left(1-\left(\frac25\right)^n\right)\). 2. Since \(\left(\frac25\right)^n\to0\), \(S_n\to5\).

\(S_n=5\left(1-\left(\frac25\right)^n\right)\), and the infinite sum is \(5\).

- Express both the infinite total and the stated finite total using the same first term and ratio. - The first term is common to both sides and can be removed. - Use the sign restriction to select between the algebraic possibilities.

1. The infinite sum is \(\frac{a}{1-r}\), and the first two terms sum to \(a(1+r)\). 2. The condition gives \(\frac{a}{1-r}=4a(1+r)\). 3. Cancel \(a>0\) and simplify to obtain \(1=4(1-r^2)\), so \(r^2=\frac34\). 4. Since \(0<r<1\), \(r=\frac{\sqrt3}{2}\).

\(r=\frac{\sqrt3}{2}\).

- Write one infinite-sum equation for each common ratio. - Eliminate the shared first term by dividing the equations. - Factor the difference of squares before solving for the ratio.

1. The two sums satisfy \(a/(1-r)=12\) and \(a/(1-r^2)=9\). 2. Dividing the second equation by the first gives \((1-r)/(1-r^2)=9/12=3/4\). 3. Since \(1-r^2=(1-r)(1+r)\), this becomes \(1/(1+r)=3/4\), so \(r=1/3\). 4. Substituting into \(a/(1-r)=12\) gives \(a=12(2/3)=8\).

\(r=\frac13\) and \(a=8\).

- First impose the condition that makes an infinite geometric total possible. - Only after finding that domain should you write the infinite sum. - Track the sign of the denominator when solving the inequality.

1. Convergence requires \(|2x-1|<1\), which gives \(0<x<1\). 2. On this interval, the sum is \(\frac{1}{1-(2x-1)}=\frac{1}{2(1-x)}\). 3. Since \(1-x>0\), the inequality \(\frac{1}{2(1-x)}>3\) gives \(1>6(1-x)\), so \(x>\frac56\). 4. Combining the conditions gives \(\frac56<x<1\).

\(x\in\left(\frac56,1\right)\).

- Determine how both growth and discounting affect one payment's present value relative to the previous one. - Use the combined factor as the geometric ratio. - The first present value occurs one year from now, not today.

1. The present value of the first payment is \(2000/1.10\). 2. From one year to the next, the payment grows by \(1.04\) while the discount denominator grows by \(1.10\), so present values have ratio \(1.04/1.10=52/55\). 3. Since \(52/55<1\), the present values form a convergent geometric series. 4. Their sum is \(\$\frac{100000}{3}\), or approximately \(\$33{,}333.33\).

The present value is \(\$\frac{100000}{3}\), approximately \(\$33{,}333.33\).

- Treat the repeating digits as one place-value block. - Find the target-base value of that block and locate its first occurrence. - Determine the factor by which each later block is shifted.

1. Each repeated block \(13_5\) has base-10 value \(1\cdot5+3=8\). 2. The first block occupies the \(5^{-2}\) place, and successive blocks are multiplied by \(5^{-2}=1/25\). 3. Thus the value is \(\frac{8}{25}+\frac{8}{25^2}+\frac{8}{25^3}+\cdots\). 4. The sum is \(\frac{8/25}{1-1/25}=\frac13\).

\(0.\overline{13}_5=\frac13\).

- Determine separately how the number of tiles and the area of each tile change by stage. - Combine those two changes to find the factor for total area from one stage to the next. - Sum the stage totals rather than the areas of individual tiles one at a time.

1. Stage \(n\) has \(4^n\) tiles, each with area \(25\left(\frac{1}{10}\right)^n\,\text{cm}^2\). 2. The total area at stage \(n\) is \(25\left(\frac{4}{10}\right)^n=25\left(\frac25\right)^n\,\text{cm}^2\). 3. The stage totals form a convergent geometric series with ratio \(\frac25\). 4. The total area is \(\frac{25}{1-2/5}=\frac{125}{3}\,\text{cm}^2\).

The total area is \(\frac{125}{3}\,\text{cm}^2\).

- Count how many equal ratio steps connect the two given endpoints. - Use the endpoint relationship to determine that ratio. - Once the inserted values are known, treat the continued sequence as an infinite series.

1. With three equal ratio steps from \(4\) to \(\frac12\), \(4r^3=\frac12\), so \(r^3=\frac18\) and \(r=\frac12\). 2. The two inserted means are \(4r=2\) and \(4r^2=1\). 3. The infinite series is \(4+2+1+\frac12+\cdots\). 4. Its sum is \(\frac{4}{1-1/2}=8\).

The geometric means are \(2\) and \(1\), and the infinite sum is \(8\).

- Express the omitted portion using the first term that was not included. - How does the size of that omitted portion change when one more term is kept? - Check consecutive possible stopping points to identify the least one.

1. The first omitted term after \(N\) terms is \(10\left(-\frac13\right)^N\). 2. The tail is geometric, so its absolute value is \(\left|\frac{10(-1/3)^N}{1+1/3}\right|=\frac{7.5}{3^N}\). 3. The condition is \(\frac{7.5}{3^N}<0.001\), or \(3^N>7500\). 4. Since \(3^8=6561\) and \(3^9=19683\), the least such integer is \(N=9\).

\(N=9\).

- Recover terms by subtracting consecutive partial sums. - Factor the resulting expression to expose a fixed repeated multiplier. - Check the result against the limit of the partial sums.

1. The first term is \(a_1=S_1=15(1+1/3)=20\). 2. For \(n\ge2\), \(a_n=S_n-S_{n-1}=15\left[\left(-\frac13\right)^{n-1}-\left(-\frac13\right)^n\right]=20\left(-\frac13\right)^{n-1}\). 3. Thus the terms are geometric with first term \(20\) and ratio \(-1/3\). 4. Since \(|r|<1\), the sum is \(20/(1+1/3)=15\), matching \(\lim S_n\).

The first term is \(20\), the common ratio is \(-\frac13\), and the sum is \(15\).

- Express the next term as a multiple of the current term. - Factor each averaged term as a constant multiple of the original term. - Compare the two complete series sums.

1. Since \(a_{n+1}=ra_n\), \(b_n=\frac{1+r}{2}a_n\). 2. Hence \(\sum b_n=\frac{1+r}{2}\sum a_n=6(1+r)\). 3. The equation \(6(1+r)=9\) gives \(r=1/2\). 4. The original first term is \(a=12(1-r)=6\).

The common ratio is \(\frac12\), and the first term is \(6\).

- Express each specified term using the same first term and ratio. - Divide the equations to isolate a power of the ratio. - Check that the recovered ratio satisfies the convergence condition.

1. The equations are \(ar=6\) and \(ar^4=\frac34\). 2. Dividing gives \(r^3=\frac{1}{8}\), so \(r=\frac12\). 3. Then \(a=\frac{6}{r}=12\). 4. The infinite sum is \(\frac{12}{1-1/2}=24\).

The first term is \(12\), the common ratio is \(\frac12\), and the sum is \(24\).

- Identify the common ratio of the reciprocal terms. - Compare the absolute values of a nonzero number and its reciprocal. - Apply the geometric convergence condition to both ratios.

1. The original geometric series can converge only if \(|r|<1\). 2. The reciprocal terms are \(\frac{1}{a_n}=\frac{1}{a}\left(\frac{1}{r}\right)^{n-1}\), so the reciprocal series has ratio \(\frac{1}{r}\). 3. If \(|r|<1\), then \(\left|\frac{1}{r}\right|>1\), so the reciprocal series diverges. 4. Conversely, if the reciprocal series converges, then \(\left|\frac{1}{r}\right|<1\), which implies \(|r|>1\), so the original series diverges.

The two series cannot both converge.

- Relate the middle term to the product of its two neighbors. - After finding the parameter, check the ratio between both adjacent pairs. - Use the first of the three terms as the first term of the requested series.

1. Consecutive geometric terms satisfy \(x^2=(x+4)(x-2)\). 2. Expanding gives \(x^2=x^2+2x-8\), so \(x=4\). 3. The three terms are \(8,4,2\), giving ratio \(r=\frac12\). 4. The infinite sum is \(\frac{8}{1-1/2}=16\).

\(x=4\), the common ratio is \(\frac12\), and the series sum is \(16\).

- Find the first product explicitly. - Track how both factors change when the index increases by one. - Combine the two ratio changes into the ratio of the product series.

1. The first product is \(a_1a_2=3\left(-\frac32\right)=-\frac92\). 2. Moving from \(a_na_{n+1}\) to \(a_{n+1}a_{n+2}\) multiplies both factors by \(-\frac12\), so the product ratio is \(\frac14\). 3. The product series is geometric and converges because \(\frac14<1\). 4. Its sum is \(\frac{-9/2}{1-1/4}=-6\).

The series converges to \(-6\).

- Track how the initial quantity and repeated factor depend on the parameter. - Simplify the total before studying the parameter. - Distinguish a limit in the parameter from a limit in the term index.

1. The series is geometric with first term \(1-c\) and ratio \(c\). 2. Since \(|c|<1\), its sum is \(\frac{1-c}{1-c}=1\). 3. The \(N\)th partial sum is \(S_N=1-c^{N+1}\). 4. For each fixed \(N\), \(S_N\to0\) as \(c\to1^-\). Thus no fixed number of initial terms captures most of the total when \(c\) is close to \(1\); the contribution is spread across increasingly many terms even though the infinite sum remains \(1\).

The series sums to \(1\) for every \(|c|<1\). As \(c\to1^-\), each fixed partial sum tends to \(0\), so increasingly many terms are needed to approach the unchanged infinite total.

- Relate each remaining tail to the term where that tail begins. - Compare how successive tails change. - Reduce the new sequence of tails to a repeated multiplicative pattern.

1. The \(k\)th tail is \(R_k=ar^{k-1}/(1-r)\). 2. Therefore \((R_k)\) itself is geometric with first term \(a/(1-r)\) and common ratio \(r\). 3. Since the original series converges and \(a\ne0\), \(|r|<1\), so the tail series also converges. 4. Its sum is \(\frac{a/(1-r)}{1-r}=\frac{a}{(1-r)^2}\).

\(\sum_{k=1}^{\infty}R_k=\frac{a}{(1-r)^2}\).

- How are the first term, second term, and common ratio related? - Combine that relation with the infinite geometric-series sum formula. - After solving, which candidate ratio satisfies \(|r|<1\)?

1. The common ratio is \(r=-\frac{6}{a}\). 2. The sum equation is \(\frac{a}{1+\frac{6}{a}}=8\), so \(\frac{a^2}{a+6}=8\). 3. Thus \(a^2-8a-48=0\), giving \(a=12\) or \(a=-4\). 4. If \(a=12\), then \(r=-\frac12\), which converges; if \(a=-4\), then \(r=\frac32\), which diverges. 5. Therefore \(a=12\) and \(r=-\frac12\).

\(a=12\) and \(r=-\frac12\).

- How is the fourth term related to the first term in a geometric sequence? - How does that same factor relate the fourth-term tail to the full series? - Use the ratio of the two given sums to solve for \(r\).

1. If the full sum is \(S\), the tail beginning with the fourth term equals \(r^3S\). 2. Thus \(r^3(30)=\frac{15}{4}\). 3. \(r^3=\frac18\), so \(r=\frac12\).

\(r=\frac12\).

- How are the listed index classes related to the original common ratio? - What can be learned by comparing the two indexed totals before finding the first term? - How can either indexed total recover the remaining unknown once the ratio is known?

1. The first indexed subseries has sum \(\frac{a}{1-r^3}=12\). 2. The second indexed subseries has sum \(\frac{ar}{1-r^3}=-6\). 3. Dividing the second equation by the first gives \(r=-\frac12\). 4. Substitution gives \(a=12\left(1-\left(-\frac12\right)^3\right)=\frac{27}{2}\). 5. The original sum is \(\frac{a}{1-r}=\frac{27/2}{3/2}=9\).

\(a=\frac{27}{2}\), \(r=-\frac12\), and the original series has sum \(9\).

- How does squaring every term change the first term and the repeated multiplier? - Can one total be used to eliminate one unknown from the other condition? - Which candidate solution is compatible with the stated sign and size restrictions?

1. The original series gives \(\frac{a}{1-r}=6\). 2. Squaring each term produces a geometric series with first term \(a^2\) and ratio \(r^2\), so \(\frac{a^2}{1-r^2}=12\). 3. From the first equation, \(a=6(1-r)\). Substitute to get \(\frac{36(1-r)}{1+r}=12\). 4. Thus \(3(1-r)=1+r\), so \(r=\frac12\), and then \(a=3\).

\(a=3\) and \(r=\frac12\).

- Find the total before excluding any terms. - The excluded indices themselves follow a regular spacing pattern. - Determine the first excluded term and the factor between successive excluded terms.

1. The sum of the full series is \(\frac{1}{1-1/2}=2\). 2. The terms with indices \(5,10,15,\ldots\) form a geometric series with first term \(\frac{1}{16}\) and ratio \(\frac{1}{32}\). 3. Their sum is \(\frac{1/16}{1-1/32}=\frac{2}{31}\). 4. Removing those terms leaves \(2-\frac{2}{31}=\frac{60}{31}\).

The requested sum is \(\frac{60}{31}\).

- Compare two remainders whose indices differ by the same fixed number of terms. - After finding the repeated change, connect an early remainder to the full sum. - Check that the resulting ratio is consistent with convergence.

1. For a geometric series, the remainders satisfy \(E_n=Sr^n\), where \(r\) is the common ratio. 2. Therefore \(\frac{E_5}{E_2}=r^3=\frac{-3/2}{12}=-\frac18\), so \(r=-\frac12\). 3. Since \(E_2=Sr^2\), \(12=S\left(\frac14\right)\), giving \(S=48\). 4. The first term is \(a=S(1-r)=48\left(1+\frac12\right)=72\).

The common ratio is \(-\frac12\), the first term is \(72\), and the series sum is \(48\).

- Separate the motion into horizontal and vertical contributions. - Every second move in one direction is followed by a move in the opposite direction. - Identify the ratio within each coordinate series rather than between consecutive moves.

1. The horizontal displacement is \(8-2+\frac12-\frac18+\cdots\), a geometric series with first term \(8\) and ratio \(-\frac14\). 2. Its sum is \(\frac{8}{1+1/4}=\frac{32}{5}\). 3. The vertical displacement is \(4-1+\frac14-\frac1{16}+\cdots\), with first term \(4\) and ratio \(-\frac14\). 4. Its sum is \(\frac{4}{1+1/4}=\frac{16}{5}\).

The limiting point is \(\left(\frac{32}{5}, \frac{16}{5}\right)\).

- Work first with a finite weighted partial sum rather than manipulating two infinite series immediately. - Compare the finite weighted sum with a shifted version of itself so most coefficients can simplify. - After subtracting, examine what happens to the remaining final term as the partial-sum index grows. - Also check partial sums that stop before a repeated-term block is finished.

1. Let \(T_N=a+2ar+3ar^2+\cdots+(N+1)ar^N\), the sum at the end of block \(N+1\). 2. Then \(rT_N=ar+2ar^2+\cdots+Nar^N+(N+1)ar^{N+1}\), so \((1-r)T_N=a+ar+\cdots+ar^N-(N+1)ar^{N+1}\). 3. As \(|r|<1\), the geometric part tends to \(a/(1-r)\) and \((N+1)ar^{N+1}\to0\). Thus \(T_N\to a/(1-r)^2\). 4. Every partial sum between \(T_N\) and \(T_{N+1}\) differs from \(T_N\) by at most \((N+2)|a||r|^{N+1}\), which tends to \(0\). 5. Hence all partial sums of the new series have the same limit, \(a/(1-r)^2\).

The new series sums to \(\frac{a}{(1-r)^2}\).

- Taking absolute values changes the sign of the ratio but not the first term. - Set up one sum expression for each series. - Eliminate the first term by comparing the two equations.

1. Let the first term be \(a>0\) and the ratio be \(r<0\). The signed sum is \(\frac{a}{1-r}=4\). 2. The series of absolute values is geometric with ratio \(-r\), so its sum is \(\frac{a}{1+r}=12\). 3. Dividing the equations gives \(\frac{1-r}{1+r}=3\), so \(r=-\frac12\). 4. Then \(a=4(1-r)=6\).

The first term is \(6\), and the common ratio is \(-\frac12\).

- What must happen to the individual terms of any convergent series? - Can a sequence of nonzero integers have that limiting behavior? - What does eventual zero behavior imply about the repeated multiplier?

1. Convergence of a geometric series with nonzero first term requires \(|r|<1\), so its terms \(ar^{n-1}\) approach \(0\). 2. An integer sequence that approaches \(0\) must equal \(0\) for all sufficiently large indices. 3. If \(a\ne0\) and \(r\ne0\), then \(ar^{n-1}\ne0\) for every \(n\), which contradicts the preceding conclusion. 4. Therefore \(r=0\). The series then has terms \(a,0,0,\ldots\).

The common ratio must be \(0\).

- Determine how multiplying consecutive geometric terms changes the first term and ratio. - Use the original total to eliminate the first term. - Check every algebraic ratio against the convergence and positivity conditions.

1. Let the first term be \(a>0\). The original sum gives \(a/(1-r)=10\), so \(a=10(1-r)\). 2. The product series has first term \(a^2r\) and ratio \(r^2\), so its sum is \(a^2r/(1-r^2)=\frac{50}{3}\). 3. Substitute \(a=10(1-r)\): \(100r(1-r)/(1+r)=\frac{50}{3}\), or \(6r^2-5r+1=0\). 4. The roots are \(r=\frac12\) and \(r=\frac13\). They give \(a=5\) and \(a=\frac{20}{3}\), respectively.

The possibilities are \((a, r)=\left(5, \frac12\right)\) and \((a, r)=\left(\frac{20}{3}, \frac13\right)\).

- Write the first three partial sums in terms of the original first term and ratio. - Use the defining relation among three consecutive geometric-sequence terms. - Interpret the resulting ratio in the original series.

1. Let the original first term be \(a\ne0\) and its ratio be \(r\). Then \(S_1=a\), \(S_2=a(1+r)\), and \(S_3=a(1+r+r^2)\). 2. Consecutive terms of a geometric sequence satisfy \(S_2^2=S_1S_3\). 3. Cancelling \(a^2\) gives \((1+r)^2=1+r+r^2\), so \(r=0\). 4. The original series is \(a+0+0+\cdots\), and its partial sums are the constant geometric sequence \(a,a,a,\ldots\).

The original common ratio is \(0\), so the series is \(a+0+0+\cdots\) with \(a\ne0\).

- View everything after a chosen term as a geometric tail. - Express that tail in terms of the chosen term and the common ratio. - The same relationship holds at every position, so the chosen term can be canceled.

1. For any term \(a_n\), the following tail has sum \(\frac{a_nr}{1-r}\). 2. The condition gives \(a_n=\frac15\cdot\frac{a_nr}{1-r}\). 3. Cancel \(a_n>0\): \(1=\frac{r}{5(1-r)}\), so \(5(1-r)=r\) and \(r=\frac56\). 4. The total sum is \(\frac{6}{1-5/6}=36\).

The common ratio is \(\frac56\), and the series sum is \(36\).

- Factor out the part of the term that does not change with the summation index. - Identify the expression serving as the common ratio and impose its convergence condition. - Check every algebraic solution against that condition.

1. Rewrite the series as \(x\sum_{n=0}^{\infty}\left(\frac{x^2}{4}\right)^n\). Convergence requires \(|x|<2\). 2. On that interval, the sum is \(\frac{x}{1-x^2/4}\). 3. Solving \(\frac{x}{1-x^2/4}=6\) gives \(3x^2+2x-12=0\). 4. Thus \(x=\frac{-1\pm\sqrt{37}}{3}\). Only \(\frac{-1+\sqrt{37}}{3}\) lies in \((-2,2)\).

\(x=\frac{-1+\sqrt{37}}{3}\).

- Subtract the two partial sums to isolate the next block of two terms. - Compare that block with the first block of two terms. - A squared ratio does not determine the sign of the original ratio.

1. The next pair of terms has sum \(S_4-S_2=\frac52\). 2. In a geometric series, the third and fourth terms together equal \(r^2\) times the first two terms together. Thus \(10r^2=\frac52\), so \(r=\frac12\) or \(r=-\frac12\). 3. From \(a(1+r)=10\), \(a=\frac{20}{3}\) when \(r=\frac12\), and \(a=20\) when \(r=-\frac12\). 4. Both infinite sums equal \(\frac{a}{1-r}=\frac{40}{3}\).

The possibilities are \((a, r)=\left(\frac{20}{3}, \frac12\right)\) and \((a, r)=\left(20, -\frac12\right)\). In both cases, the sum is \(\frac{40}{3}\).

- Express a general term of the original series. - Treat everything after that term as a new geometric tail. - Determine how the product changes when the index increases by one.

1. The \(n\)th original term is \(a_n=4\left(\frac12\right)^{n-1}\). 2. The tail after \(a_n\) begins with \(a_{n+1}\) and has sum \(\frac{4(1/2)^n}{1-1/2}=8\left(\frac12\right)^n\). 3. Their product is \(16\left(\frac14\right)^{n-1}\). 4. The resulting geometric series has sum \(\frac{16}{1-1/4}=\frac{64}{3}\).

The sum of the products is \(\frac{64}{3}\).

- Factor the difference of consecutive terms. - Absolute values affect the repeated multiplier as well as the constant factor. - Simplify separately according to the sign of the ratio.

1. The difference is \(a_{n+1}-a_n=6r^{n-1}(r-1)\). 2. Its magnitude is \(6(1-r)|r|^{n-1}\) because \(r<1\). 3. The series of absolute values is geometric with ratio \(|r|\), so its sum is \(6(1-r)/(1-|r|)\). 4. If \(0\le r<1\), the sum is \(6\). If \(-1<r<0\), it is \(6(1-r)/(1+r)\).

\(6\) for \(0\le r<1\), and \(\frac{6(1-r)}{1+r}\) for \(-1<r<0\).