- Read the subscript on \(a_4\) as a position in the sequence. - Read \(S_4\) as a total that includes several terms. - Write out exactly which terms contribute to each quantity.

The fourth term is \(a_4=13\). The partial sum is \(S_4=4+7+10+13=34\). The symbol \(a_4\) names one term of the sequence, while \(S_4\) names the cumulative total of the first four terms.

\(a_4=13\); \(S_4=34\). The first is a single term, and the second is a partial sum.

- Compare each term with the term immediately before it by subtraction. - Count the displayed terms after deciding whether the additive change is constant.

1. The consecutive differences are \(17-12=5\), \(22-17=5\), and \(27-22=5\). 2. Because the difference is constant, the sum is an arithmetic series with first term \(a_1=12\), common difference \(d=5\), and \(n=4\) terms.

Yes. The consecutive differences are all \(5\), so it is an arithmetic series with \(a_1=12\), \(d=5\), and \(n=4\).

- Match each given value to the corresponding symbol in the formula. - Substitute all three given values before doing the arithmetic.

1. Substitute \(n=7\), \(a_1=6\), and \(a_7=30\): \(S_7=\frac{7}{2}(6+30)\). 2. Simplify: \(S_7=\frac{7}{2}\cdot36=126\).

\(S_7=\frac{7}{2}(6+30)=126\).

- Look for the repeated change between neighboring terms. - When you move one place inward from each end, how do the two selected numbers change? - How many pairs make up the original sum?

1. The outer pair gives \(2+11=13\), and the inner pair gives \(5+8=13\). 2. Moving inward adds \(3\) to the smaller term and subtracts \(3\) from the larger term, so the pair sum stays the same. 3. Two pairs of \(13\) give the total \(26\).

The pairs are \(2+11=13\) and \(5+8=13\). Moving inward adds \(3\) to one term and subtracts \(3\) from the other, so their sum stays the same. The total is \(26\).

- Determine the number of terms before applying a finite-sum formula. - Use the first term, last term, and term count together. - Check that the last term occurs at an integer index.

The common difference is \(5\). The number of terms satisfies \(83=8+5(n-1)\), so \(n=16\). Therefore, \(S_{16}=\frac{16}{2}(8+83)=8\cdot91=728\).

\(728\)

- Use the form of the arithmetic-series formula that contains \(a_1\), \(d\), and \(n\). - Keep the negative first term when substituting. - Simplify the bracket before multiplying by the outside factor.

Use the arithmetic-series formula with the first term and common difference: \(S_{25}=\frac{25}{2}[2(-6)+(25-1)4]\). Thus, \(S_{25}=\frac{25}{2}(84)=1050\).

\(1050\)

- Model the number of seats in each row as an arithmetic sequence. - Determine the number of seats in the final row before finding the total. - The total number of seats is a sum of all row values, not just the last-row count.

The row counts form an arithmetic sequence with \(a_1=18\), \(d=2\), and \(n=15\). The last row has \(a_{15}=18+14\cdot2=46\) seats. Therefore, \(S_{15}=\frac{15}{2}(18+46)=480\).

\(480\) seats

- What change repeats between the displayed daily counts? - How can you use that change to find the missing fourth day? - Which quantities must be included in the total for all four days?

1. The first three daily counts are \(3,5,7\), with common difference \(2\). 2. Day \(4\) continues the arithmetic pattern with \(9\) bottles. 3. The four-day total is \(3+5+7+9=24\) bottles. One bar gives a daily amount, not the sum of all days.

Day \(4\): \(9\) bottles. Four-day total: \(24\) bottles. A single bar represents one day, not the sum of all four days.

- A single term is the amount added when one partial sum becomes the next. - Compare the terms you recover, not just the partial sums themselves. - Extend the term pattern once before finding the new total.

1. Subtract consecutive sums to find terms: \(a_1=5\), \(a_2=6\), \(a_3=7\), and \(a_4=26-18=8\). 2. The terms increase by \(1\), so \(d=1\) and \(a_5=9\). 3. Add the next term to the previous total: \(S_5=26+9=35\).

\(a_4=8\), \(d=1\), and \(S_5=35\).

- Find the third partial sum before extending it. - How can you get the fifth partial sum without adding the first three terms again? - Does adding two more terms always increase a running total?

1. The first three terms sum to \(S_3=10+6+2=18\). 2. Adding the next two terms gives \(S_5=18+(-2)+(-6)=10\). 3. The fourth and fifth terms are negative and together subtract \(8\) from the earlier total.

\(S_3=18\) and \(S_5=10\). The two added terms are negative and reduce the total by \(8\).

- Is the money already in the fund one of the weekly contributions? - How could you find the total added over five weeks without adding every deposit separately? - Check whether your result includes every source of money described in the problem.

1. The weekly amounts raised form an arithmetic sequence with \(a_1=8\), \(d=3\), and \(n=5\). The fifth amount is \(8+4\cdot3=20\) dollars. 2. The five weekly amounts total \(\frac{5}{2}(8+20)=70\) dollars. 3. Include the \(12\) dollars already in the fund: \(12+70=82\) dollars.

\(\$82\)

- Use the arithmetic-series formula that includes the first term and common difference. - Set up an equation with \(d\) as the only unknown. - Then use the explicit arithmetic-sequence formula to find \(a_{15}\).

1. Use the arithmetic-series formula: \(S_n=\frac n2[2a_1+(n-1)d]\). Substituting the given values gives \(1000=\frac{20}{2}[2\cdot12+19d]=10(24+19d)\). 2. Solve for \(d\): \(100=24+19d\), so \(76=19d\) and \(d=4\). 3. Then \(a_{15}=a_1+14d=12+14\cdot4=68\).

\(d=4\) and \(a_{15}=68\).

- Find the first and last multiples of \(13\) in the interval. - Recognize that the values form an arithmetic sequence. - Determine the number of terms before applying the sum formula.

1. The first multiple of \(13\) in the interval is \(104=8\cdot13\). The last is \(988=76\cdot13\). 2. These numbers form an arithmetic sequence with common difference \(13\). Find the number of terms: \(988=104+(n-1)13\). Thus, \(884=13(n-1)\), so \(n=69\). 3. Apply the arithmetic-series formula: \(S_{69}=\frac{69}{2}(104+988)=\frac{69}{2}\cdot1092=37{,}674\).

The sum is \(37{,}674\).

- Use the spacing between the known indices to find the common difference. - Work backward from the third term to the first term. - Then use the finite arithmetic-series formula.

1. Since \(a_7=a_3+4d\), \(45=17+4d\), so \(d=7\). 2. Since \(a_3=a_1+2d\), \(17=a_1+14\), so \(a_1=3\). 3. Find the twentieth term: \(a_{20}=a_1+19d=3+19\cdot7=136\). 4. Apply the arithmetic-series formula: \(S_{20}=\frac{20}{2}(a_1+a_{20})=10\cdot(3+136)=1390\).

The sum of the first \(20\) terms is \(1390\).

- Identify the first term, last term, and number of terms before substituting into the arithmetic-series formula. - Keep the result symbolic in part a rather than substituting a particular value of \(n\). - In part b, set the derived sum formula equal to the given total and solve for a positive integer.

1. The series has first term \(1\), last term \(n\), and \(n\) terms. Substituting into the arithmetic-series formula gives \(S_n=\frac{n}{2}(1+n)=\frac{n(n+1)}{2}\). 2. Set the derived formula equal to the given sum: \(\frac{n(n+1)}{2}=1275\). 3. This gives \(n^2+n-2550=0\). The discriminant is \(10{,}201=101^2\), so \(n=\frac{-1\pm101}{2}\). 4. The positive solution is \(n=50\).

a) \(S_n=\frac{n}{2}(1+n)=\frac{n(n+1)}{2}\) b) \(n=50\)

- Write the first few terms to identify the arithmetic pattern. - Express the \(n\)th term using the first term and common difference. - For the partial sum, use the first term together with your expression for \(a_n\).

1. The sequence begins \(3, 7, 11, \ldots\). It is arithmetic with first term \(3\) and common difference \(4\), so \(a_n=3+4(n-1)=4n-1\). 2. Apply the arithmetic-series formula: \(S_n=\frac{n}{2}(a_1+a_n)=\frac{n}{2}[3+(4n-1)]=\frac{n}{2}(4n+2)=n(2n+1)=2n^2+n\).

1) \(a_n=4n-1\). 2) \(S_n=2n^2+n\).

- Divide the distance between endpoints by the common difference to count equal steps. - Distinguish between the number of steps and the number of inserted values. - For the total, first determine how many terms run from the first endpoint through the last.

1) Between \(3\) and \(15\), the number of common-difference steps is \(\frac{15-3}{1.5}=8\). Therefore, there are \(8-1=7\) inserted values. The same count applies between \(15\) and \(27\). 2) Beginning at \(3\) and adding \(1.5\), the first six terms are \(3, 4.5, 6, 7.5, 9, 10.5\). 3) The number of terms from \(3\) through \(27\) satisfies \(27=3+(N-1)(1.5)\), so \(N=17\). Thus, \(S_{17}=\frac{17}{2}(3+27)=255\).

1) Insert \(7\) values between each pair. 2) The first six terms are \(3, 4.5, 6, 7.5, 9, 10.5\). 3) The sum is \(255\).

- Use the endpoints and total sum to determine the total number of terms first. - Keep the total number of terms separate from the number of inserted terms. - After finding the term count, use the endpoint spacing to determine the common difference.

1) Use the finite arithmetic-series formula: \(275=\frac{k}{2}(5+45)=25k\). Thus, \(k=11\). Since the two endpoints are not inserted, \(n=k-2=9\). 2) Use the explicit term formula: \(45=5+(11-1)d\). Therefore, \(40=10d\), so \(d=4\). 3) The first five terms are \(5, 9, 13, 17, 21\), and their sum is \(65\). Equivalently, \(S_5=\frac52[2\cdot5+4\cdot4]=65\).

1) \(n=9\) inserted terms. 2) \(d=4\). 3) The sum of the first five terms is \(65\).

- Substitute the first term and common difference into a finite arithmetic-series formula. - Expect an equation in \(n\) with more than one possible positive solution. - Compare the terms added between the two candidate partial sums.

1. Apply the arithmetic-series formula: \(27=\frac{n}{2}[2\cdot12+(n-1)(-3)]\). Simplifying gives \(n^2-9n+18=0\), so \((n-3)(n-6)=0\). Therefore, \(n=3\) or \(n=6\). 2. The sequence begins \(12, 9, 6, 3, 0, -3, \ldots\). The first three terms sum to \(27\). The next three terms sum to \(3+0+(-3)=0\), so adding them does not change the partial sum. Thus, \(S_3=S_6=27\).

1) \(n=3\) and \(n=6\). 2) The additional terms from \(a_4\) through \(a_6\) sum to \(0\), so \(S_3=S_6\).

- Use the relationship among three equally spaced terms of a geometric sequence. - Positivity determines the sign of the middle term. - Substitute the resulting first term into the arithmetic-series formula.

1. In a geometric sequence, terms equally spaced around a middle term satisfy \(b_3^2=b_1b_5\). Therefore, \(b_3^2=144\). Since all terms are positive, \(b_3=12\). Thus, \(a_1=12\). 2. Apply the arithmetic-series formula: \(255=\frac{10}{2}[2\cdot12+9d]\). Thus, \(255=5(24+9d)\), so \(51=24+9d\), and \(d=3\).

The common difference is \(d=3\).

- Set the finite arithmetic-series formula equal to zero. - Determine where the sequence terms change sign. - Use the sign of all later terms to explain what happens after the zero partial sum.

1) Set the arithmetic-series formula equal to zero: \(\frac n2[2\cdot15+(n-1)(-1.5)]=0\). Since \(n>0\), \(30-1.5(n-1)=0\). Thus, \(31.5=1.5n\), so \(n=21\). 2) The sequence is strictly decreasing because \(d<0\). Also, \(a_{11}=15+10\cdot(-1.5)=0\), so every term beginning with \(a_{12}\) is negative. At \(n=21\), the negative terms exactly balance the positive terms. Every additional term is negative, so \(S_n<0\) for all \(n>21\).

1) \(n=21\). 2) Beginning with the twelfth term, all added terms are negative; after \(S_{21}=0\), every further partial sum is negative.

- Interpret the summand values as a finite arithmetic sequence. - Identify its first and last terms in terms of \(n\). - Keep the final result symbolic rather than substituting a particular value of \(n\).

The summand produces the arithmetic sequence \(1,5,9,\ldots,4n-3\). Its sum is \(\frac n2[1+(4n-3)]\) \(=\frac n2(4n-2)=n(2n-1)\).

\(n(2n-1)\)

- Write the same finite sum once from first term to last and once from last term to first. - Compare the sum of the first term in one line with the first term in the reversed line, then do the same for another position. - Count how many equal paired sums appear after adding the two lines.

1. Write the sum in forward and reverse order: \(S_n=a_1+(a_1+d)+\cdots+[a_1+(n-1)d]\), \(S_n=[a_1+(n-1)d]+[a_1+(n-2)d]+\cdots+a_1\). 2. Adding corresponding terms gives the same value each time, \(2a_1+(n-1)d\). 3. There are \(n\) such paired sums, so \(2S_n=n[2a_1+(n-1)d]\). 4. Dividing by \(2\) gives \(S_n=\frac{n}{2}[2a_1+(n-1)d]\).

\(2S_n=n[2a_1+(n-1)d]\), so \(S_n=\frac{n}{2}[2a_1+(n-1)d]\).

- Here the number of terms is the unknown, so keep \(n\) in the finite-sum formula. - Substitute the known first term and common difference before simplifying the equation. - After solving the resulting quadratic, check which solution can represent a number of terms.

Use the arithmetic-series formula: \(440=\frac{n}{2}[2(5)+(n-1)3]\). This simplifies to \(3n^2+7n-880=0\). Factoring gives \((n-16)(3n+55)=0\). The only positive integer solution is \(n=16\).

\(n=16\)

- Read the lower and upper sigma bounds to identify the first and last terms. - Remember that inclusive integer bounds determine the number of terms. - Once you recognize an arithmetic list, use its first term, last term, and term count rather than expanding every term.

1. The first term is \(7-2(4)=-1\), and the last term is \(7-2(24)=-41\). 2. There are \(24-4+1=21\) terms. 3. Therefore, \(S=\frac{21}{2}(-1-41)=-441\).

First term: \(-1\); last term: \(-41\); number of terms: \(21\). Thus \(S=\frac{21}{2}(-1-41)=-441\).

- Read each partial sum from its plotted coordinate; the markers do not contain value labels. - The difference between two consecutive partial sums isolates one sequence term. - After recovering the arithmetic pattern, extend it one more term before updating the partial sum.

1. Reading the graph gives \(S_1=3\), \(S_2=9\), \(S_3=18\), \(S_4=30\), and \(S_5=45\). 2. The first term is \(a_1=S_1=3\). For later terms, subtract consecutive partial sums: \(a_2=9-3=6\), \(a_3=18-9=9\), \(a_4=30-18=12\), and \(a_5=45-30=15\). 3. The recovered terms \(3,6,9,12,15\) have common difference \(d=3\). 4. The sixth term is \(a_6=18\), so \(S_6=S_5+a_6=45+18=63\).

a) \(3,6,9,12,15\) b) \(d=3\) c) \(S_6=63\)

- Count how many terms are included in \(S\). - Use the equal pair-sum property of an arithmetic sequence. - Express the finite sum using its first and last inserted terms.

1. Let \(a_1=x\) and \(a_{n+2}=y\). The inserted terms are \(a_2, a_3, \ldots, a_{n+1}\). 2. In an arithmetic sequence, terms equally distant from the endpoints have the same pair sum. Thus, \(a_2+a_{n+1}=a_1+a_{n+2}=x+y\). 3. The sum of the \(n\) inserted terms is \(S=\frac n2(a_2+a_{n+1})=\frac n2(x+y)\). 4. Since \(x+y\ne0\), solve for \(n\): \(n=\frac{2S}{x+y}\).

\(n=\frac{2S}{x+y}\), provided \(x+y\ne0\).

- Describe exactly which terms are included in \(S_{15}\). - Think of the desired block as part of a larger partial sum with an initial block removed. - Check the number of terms from position \(6\) through position \(20\) after setting up the correct subtraction.

Maya's \(S_{15}\) adds the first fifteen terms, \(a_1\) through \(a_{15}\), so it does not represent \(a_6+\cdots+a_{20}\). The required sum is \(S_{20}-S_5\). \(S_{20}=\frac{20}{2}[2(7)+19(5)]=1090\), and \(S_5=\frac{5}{2}[2(7)+4(5)]=85\). Therefore, \(S_{20}-S_5=1090-85=1005\).

Maya used the wrong block of fifteen terms. The correct sum is \(S_{20}-S_5=1005\).