Free math worksheets

- 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.

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

The series is geometric with common ratio \(r=-\frac{2}{3}\), 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 \(3\).

- 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 \(90\,\text{ft}\).

- Find the first correction from the initial error. - What fraction of one correction becomes the next correction? - Does that ratio allow an infinite geometric sum?

1. The first move is \(\frac23\cdot18=12\,\text{cm}\). 2. Each later move is \(\frac13\) of the preceding move because the remaining error is reduced by a factor of \(\frac13\). 3. The total distance is \(\frac{12}{1-\frac13}=18\,\text{cm}\).

The total distance moved is \(18\,\text{cm}\).

- 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 \(\left(\frac{32}{5}, \frac{16}{5}\right)\).

- Give the weighted series a name, then multiply the entire series by the common ratio. - Align the original and shifted series term by term and subtract. - The difference reduces to an ordinary geometric series.

1. Let \(T=a+2ar+3ar^2+4ar^3+\cdots\). 2. Multiplying by \(r\) gives \(rT=ar+2ar^2+3ar^3+\cdots\). 3. Subtracting the second series from the first leaves \((1-r)T=a+ar+ar^2+\cdots=\frac{a}{1-r}\). 4. Because \(|r|<1\), division by \(1-r\) is valid and \(T=\frac{a}{(1-r)^2}\).

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 \(-1<r<0\).