Free Math Worksheets

- 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 pattern. - Determine the first term and common difference. - Substitute the explicit term into the arithmetic-series formula.

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 steps. - Distinguish between the number of steps and the number of inserted values. - Use the arithmetic-series formula for the total sum.

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 \(27\) is determined by \(27=3+(N-1)(1.5)\). Thus, \(N=17\). The sum is \(S_{17}=\frac{17}{2}(3+27)=255\).

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 \(255\).

- 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 the partial sums.

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.

- Identify each list as an arithmetic sequence. - Use the finite arithmetic-series formula. - Express the last odd number in terms of \(n\). - Translate the condition in part 3 into an equation.

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 find the total number of terms. - Distinguish the total number of terms from the number inserted. - Then use the first and last terms to find 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 given values into the finite arithmetic-series formula. - Expect a quadratic equation in \(n\). - Examine the terms added between the two solutions.

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 \(S_3=S_6\).