Free math worksheets

- Match the first term with the last, then move one position inward from both ends. - Check whether the resulting pair totals repeat. - Relate the number of pairs to the number of terms in the series.

a) Pair the first and last terms, then the second and next-to-last terms, and so on. Each pair has the same sum: \(4+19=7+16=10+13=23\). b) There are \(3\) pairs, so the total is \(3\cdot23=69\). c) In an arithmetic series, terms equally far from the two ends have the same pair-sum. That lets the total be found from one pair-sum and the number of pairs instead of adding every term separately.

a) Each pair sums to \(23\). b) The total is \(69\). c) End-pairs in an arithmetic series have equal sums, so equal pairing compresses the addition.

- 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.

- Compare the first-plus-last pair with the second-plus-next-to-last pair. - 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.

a) 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 the same is true for every pair equally far from the ends. Therefore each pair has average \(\frac{a_1+a_n}{2}\), which is also the average of all the terms. b) Sum equals number of terms times average term value, so \(S_n=n\cdot\frac{a_1+a_n}{2}=\frac{n}{2}(a_1+a_n)\). c) The common difference is \(5\), so the number of terms is \(n=\frac{57-12}{5}+1=10\). Thus \(S_{10}=\frac{10}{2}(12+57)=345\).

a) Equal end-pairs all have average \(\frac{a_1+a_n}{2}\). b) \(S_n=\frac{n}{2}(a_1+a_n)\). c) \(345\).

- 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 \(37+38+\cdots+83=2820\).

- 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 \(S_n\).

a) Write \(S_n=5+7+\cdots+(2n+1)+(2n+3)\) and \(S_n=(2n+3)+(2n+1)+\cdots+7+5\). Adding corresponding terms gives \(2S_n=n(2n+8)\), because there are \(n\) equal pairs. Therefore, \(S_n=n(n+4)=n^2+4n\). b) For a general arithmetic sequence, write the sum in forward and reverse order. Each paired column has sum \(a_1+a_n\), and there are \(n\) columns. Hence \(2S_n=n(a_1+a_n)\), 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.

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

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 \(S_n\).

a) If the common difference of the original sequence is \(d\), then \(b_{k+1}-b_k=3a_{k+1}+4-(3a_k+4)=3(a_{k+1}-a_k)=3d\). Because this difference is constant, the \(b_k\) form an arithmetic sequence. b) The endpoints are \(b_1=3a_1+4\) and \(b_n=3a_n+4\). Therefore \(T_n=\frac{n}{2}(b_1+b_n)=\frac{n}{2}\left(3(a_1+a_n)+8\right)\). Since \(S_n=\frac{n}{2}(a_1+a_n)\), it follows that \(T_n=3S_n+4n\).

a) The transformed sequence is arithmetic with common difference \(3d\). b) \(T_n=3S_n+4n\).

- 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: \(-11\)