- What happens to the two terms in a pair as you move inward from both ends? - Check whether the resulting pair totals repeat. - Relate the number of pairs to the number of terms in the series.

1. Pair terms from opposite ends: \(4+19=7+16=10+13=23\). 2. There are \(3\) pairs, so the sum is \(3\cdot23=69\). 3. Terms equally far from the ends of any arithmetic series have the same pair-sum. This lets us find the total without adding every term separately.

a) Each pair sums to \(23\). b) The total is \(69\). c) Equal end-pairs let us multiply one pair-sum by the number of pairs.

- Think about the average of the first and last terms of an arithmetic sequence. - Terms equally far from the two ends balance around the same central value. - Relate the total of several values to their average and the number of values.

1. In an arithmetic sequence, terms equally far from the two ends have the same average as the first and last terms. 2. The average value is \(\frac{8+32}{2}=20\). 3. The sum is the number of terms times the average value: \(7\cdot20=140\).

The average value is \(20\), and the sum is \(140\).

- Compare the difference between each pair of neighboring terms. - Check whether the same difference works all the way through, even when the numbers decrease.

1. In A, consecutive terms increase by \(4\), so the common difference is \(4\). 2. In B, the differences are \(2\), \(4\), and \(8\), so it is not arithmetic. 3. In C, consecutive terms decrease by \(6\), so the common difference is \(-6\).

A and C are arithmetic series. Their common differences are \(4\) and \(-6\), respectively. B is not arithmetic.

- Try pairing the first term with the last and then moving inward. - For an odd number of terms, check what is left after all possible end-pairs have been made.

1. Pair the endpoints: \(2+14=16\) and \(5+11=16\). 2. The middle term is \(8\) and does not need a partner. 3. Add the two pair-sums and the middle term: \(16+16+8=40\).

The pairs \(2+14\) and \(5+11\) each total \(16\). Add the unpaired middle term \(8\) once: \(16+16+8=40\).

- Compare the first-plus-last pair with the second-plus-next-to-last pair. - If there is an unpaired middle term, compare its value with the average of the two endpoints. - Think of a sum as the number of terms multiplied by their average value. - Before applying the formula in part c, determine how many terms the displayed arithmetic sequence contains.

1. The first and last terms sum to \(a_1+a_n\). Because the sequence is arithmetic, the second and next-to-last terms have the same sum, and so does every pair equally far from the ends. 2. If \(n\) is even, all terms form such pairs, each with average \(\frac{a_1+a_n}{2}\). If \(n\) is odd, the unpaired middle term lies halfway between the endpoints, so its value is also \(\frac{a_1+a_n}{2}\). Thus the average of all \(n\) terms is \(\frac{a_1+a_n}{2}\). 3. Sum equals number of terms times average term value, so \(S_n=n\cdot\frac{a_1+a_n}{2}=\frac{n}{2}(a_1+a_n)\). 4. For \(12+17+22+\cdots+57\), the common difference is \(5\), so \(n=\frac{57-12}{5}+1=10\). Hence \(S_{10}=\frac{10}{2}(12+57)=345\).

a) Equal end-pairs have average \(\frac{a_1+a_n}{2}\); when \(n\) is odd, the middle term has that same value. b) \(S_n=\frac{n}{2}(a_1+a_n)\). c) \(345\).

- Match terms that are the same number of positions to the left and right of the middle. - Think about how the deviations of such a pair from the middle term are related. - Once the average term value is known, relate the total to the number of terms.

a) In an arithmetic series with an odd number of terms, terms equally far from the middle are equally far below and above the middle term. Their deviations cancel, so the average is the middle term, 24. b) The sum is the number of terms times the average: \(7\cdot24=168\). c) If the middle term is \(m\), every term \(m-r\) on one side is paired with a term \(m+r\) on the other side. Each such pair has average \(m\), and the unpaired middle term is also \(m\). Therefore the average of all the terms is \(m\).

a) \(24\) b) \(168\) c) Symmetric terms around the middle have opposite deviations from the middle term, so their deviations cancel and the middle term is the average.

- Express the last term in terms of the first term, the common difference, and the number of terms. - Substitute that expression for the last term before simplifying the sum formula. - Keep the symbolic derivation separate from the numerical substitution.

1. For an arithmetic sequence, \(a_n=a_1+(n-1)d\). 2. Substitute this into the endpoint form of the sum formula: \(S_n=\frac{n}{2}\left(a_1+a_1+(n-1)d\right)\). 3. Therefore \(S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\). 4. With \(a_1=-4\), \(d=3\), and \(n=20\), \(S_{20}=\frac{20}{2}\left(2(-4)+19\cdot3\right)=490\).

\(S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\), and \(S_{20}=490\).

- Identify the endpoints and determine how many consecutive integers are included. - An inclusive count from one integer endpoint to another needs an adjustment after subtraction. - Apply the endpoint form of the arithmetic-series sum only after expressing the term count in \(p\) and \(q\).

1. The first term is \(p\), the last term is \(q\), and the common difference is \(1\). 2. The number of terms is \(q-p+1\). 3. Using the arithmetic-series formula gives \(S=\frac{q-p+1}{2}(p+q)\). 4. For \(p=37\) and \(q=83\), there are \(83-37+1=47\) terms, so \(S=\frac{47}{2}(37+83)=2820\).

\(S=\frac{(q-p+1)(p+q)}{2}\), and \(37+38+\cdots+83=2820\).

- Count the changes from the first term to the last, including changes past zero. - Once the number of terms is known, use the average of the two endpoints.

1. The common difference is \(-5\). Solving \(28+(n-1)(-5)=-7\) gives \(n-1=7\), so there are \(8\) terms. 2. The sum is \(S_8=\frac{8}{2}(28+(-7))=4\cdot21=84\).

From \(28-5(n-1)=-7\), we get \(n=8\). The sum is \(\frac{8}{2}(28-7)=84\).

- How can you find the seat count in a later row without listing every earlier row? - What feature of the row counts makes adding the first and last counts useful? - Check that the question asks for all rows together, not just the last row.

1. The number of seats per row forms an arithmetic sequence with first term \(8\) and common difference \(3\). 2. The twelfth row has \(8+11\cdot3=41\) seats. 3. Total seats: \(S_{12}=\frac{12}{2}(8+41)=294\).

\(294\) seats.

- Try writing the same finite sum in the opposite order directly underneath the original sum. - Compare the sums of terms that line up in the two rows. - Count how many equal pair-sums appear before isolating \(S_n\).

1. For the given sequence, \(a_n=5+2(n-1)=2n+3\). Write the same sum in the two orders \(S_n=a_1+a_2+\cdots+a_{n-1}+a_n\) and \(S_n=a_n+a_{n-1}+\cdots+a_2+a_1\). 2. In each aligned column, the two terms add to \(a_1+a_n=5+(2n+3)=2n+8\). There are \(n\) columns, so \(2S_n=n(2n+8)\). Therefore \(S_n=n(n+4)=n^2+4n\). 3. For a general arithmetic sequence, the same forward-and-reverse alignment makes every column sum to \(a_1+a_n\). With \(n\) columns, \(2S_n=n(a_1+a_n)\), so \(S_n=\frac{n}{2}(a_1+a_n)\).

a) \(S_n=n^2+4n\). b) Pairing the forward and reversed sums gives \(2S_n=n(a_1+a_n)\), so \(S_n=\frac{n}{2}(a_1+a_n)\).

- Track what is on the left side when the two displayed equations are added together. - Count the equal column sums only after accounting for both copies of the original series. - Apply the repaired formula using the given number of terms and endpoint values.

1. Adding the two displayed equations adds \(S_n\) to another copy of \(S_n\). The left side is therefore \(2S_n\), not \(S_n\). 2. There are \(n\) columns and each column sums to \(a_1+a_n\), so \(2S_n=n(a_1+a_n)\). 3. Dividing by \(2\) gives \(S_n=\frac{n}{2}(a_1+a_n)\). 4. For the given series, \(S_8=\frac{8}{2}(8+50)=232\).

a) The two rows sum to \(2S_n\), not \(S_n\). b) \(S_n=\frac{n}{2}(a_1+a_n)\). c) \(232\)

- Express the last term using the first term, common difference, and an unknown term count. - Substitute that last-term expression into an arithmetic-series sum formula. - After solving the resulting equation, use the meaning of \(n\) to reject any inadmissible solution.

1. If the series has \(n\) terms, then its last term is \(7+5(n-1)\). 2. Using the endpoint form of the sum formula, \(196=\frac{n}{2}\left(7+7+5(n-1)\right)\). 3. Simplifying gives \(5n^2+9n-392=0\). 4. Factoring gives \((5n+49)(n-8)=0\). 5. The number of terms must be positive, so \(n=8\).

\(n=8\).

- Compare two consecutive transformed terms to decide whether their difference is constant. - Express the first and last transformed terms using the original endpoints. - Relate the endpoint sum of the transformed sequence to the endpoint sum that appears in \(S_n\).

1. If the original common difference is \(d\), then \(b_{k+1}-b_k=3(a_{k+1}-a_k)=3d\), which is constant. Thus \(b_k\) is arithmetic. 2. Its endpoints are \(b_1=3a_1+4\) and \(b_n=3a_n+4\). The arithmetic-series formula gives \(T_n=\frac{n}{2}(b_1+b_n)=\frac{n}{2}\left(3(a_1+a_n)+8\right)\). 3. Because \(S_n=\frac{n}{2}(a_1+a_n)\), we obtain \(T_n=3S_n+4n\).

a) \(b_{k+1}-b_k=3(a_{k+1}-a_k)=3d\), so the transformed sequence is arithmetic. b) \(T_n=\frac{n}{2}(3(a_1+a_n)+8)=3S_n+4n\).

- Use the total and number of terms to determine what the two endpoint terms must add to. - Relate the last term to the first term through the common difference and the number of steps between them. - Combine those two endpoint relationships rather than generating the entire sequence.

1. From \(S_n=\frac{n}{2}(a_1+a_n)\), \(66=\frac{12}{2}(a_1+a_{12})\), so \(a_1+a_{12}=11\). 2. Because the common difference is \(-3\), \(a_{12}=a_1+11(-3)=a_1-33\). 3. Substitute into the endpoint-sum equation: \(a_1+(a_1-33)=11\). 4. Thus \(a_1=22\), and \(a_{12}=-11\).

First term: \(22\) Last term: \(-11\)