Free math worksheets

- 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.

- 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 \(1390\).

- 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 \(4\). Therefore, \(a_n=3+4(n-1)=4n-1\). 2) Apply the arithmetic-series formula: \(S_n=\frac n2(a_1+a_n)=\frac n2[3+(4n-1)]\) \(=\frac n2(4n+2)=n(2n+1)=2n^2+n\).

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 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.

- 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

- Identify each list as an arithmetic sequence before choosing a sum formula. - In part 2, express the last odd number in terms of \(n\). - In part 3, translate the total into an equation for \(n\).

1) The first \(40\) positive multiples of \(3\) form an arithmetic sequence with first term \(3\) and fortieth term \(120\). Therefore, \(S_{40}=\frac{40}{2}(3+120)=20\cdot123=2460\). 2) The first \(n\) positive odd integers are \(1, 3, 5, \ldots, 2n-1\). Their sum is \(S_n=\frac n2[1+(2n-1)]=\frac n2(2n)=n^2\). 3) Solve \(\frac{n(n+1)}{2}=1275\). This gives \(n^2+n-2550=0\), so \(n=\frac{-1\pm\sqrt{10{,}201}}{2}=\frac{-1\pm101}{2}\). The positive solution is \(n=50\).

1) \(2460\). 2) \(\frac n2[1+(2n-1)]=n^2\). 3) \(n=50\).

- 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 \(x+y\ne0\).

- 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: \(27=\frac n2[2\cdot12+(n-1)(-3)]\). 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 \(S_3=S_6\).

- 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.

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

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

- 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.

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

\(-441\)

- A partial sum \(S_n\) contains all terms through position \(n\). - 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.

a) The first term is \(a_1=S_1=3\). For later terms, subtract consecutive partial sums: \(a_2=8-3=5\), \(a_3=15-8=7\), \(a_4=24-15=9\), and \(a_5=35-24=11\). Thus the first five terms are \(3,5,7,9,11\). b) Consecutive terms differ by \(2\), so the common difference is \(d=2\). c) The sixth term is \(a_6=13\), so \(S_6=35+13=48\).

a) \(3,5,7,9,11\) b) \(d=2\) c) \(S_6=48\)

- 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 \(S_{20}-S_5=1005\).