Free Math Worksheets

- How many times is the first term multiplied by the common ratio to reach the fifth term? - Write an equation in the form \(x_5=x_1r^n\). - What positive number has fourth power \(16\)? - How does the increasing order restrict the common ratio?

1. For a geometric sequence, \(x_5=x_1r^4\). 2. Substitute the known values: \(8=0.5r^4\), so \(r^4=16\). 3. Since the sequence has positive terms and is increasing, \(r=2\). 4. The missing terms are \(x_2=0.5\cdot2=1\), \(x_3=1\cdot2=2\), and \(x_4=2\cdot2=4\).

\(r=2\), \(x_2=1\), \(x_3=2\), and \(x_4=4\)

- Count how many common-difference steps lie between the fourth and eleventh terms. - Use the two known terms to find the common difference. - Then use an explicit arithmetic-sequence formula.

1. The common difference is \(d=\frac{a_{11}-a_4}{11-4}=\frac{-11-17}{7}=-4\). 2. Use \(a_4\) to find the first term: \(a_1=a_4-3d=17-3\cdot(-4)=29\). 3. Apply the explicit formula: \(a_{50}=a_1+49d=29+49\cdot(-4)=-167\).

\(a_{50}=-167\).

- Count the total number of terms after inserting the new values. - Use the explicit arithmetic-sequence formula with the first and last terms. - Once the common difference is known, generate successive terms.

1. The endpoints are \(a_1=-7\) and \(a_6=23\). 2. Use the explicit formula: \(23=-7+5d\). Thus, \(30=5d\), so \(d=6\). 3. Starting from \(-7\) and adding \(6\), the first four terms are \(-7, -1, 5, 11\).

The common difference is \(d=6\). The first four terms are \(-7, -1, 5, 11\).

- 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 \(d=3\).

- How can you find the common difference when the first and fourth terms are known? - How can you find the common ratio when three equal multiplicative steps connect the first and fourth terms? - Find the two middle terms before adding them.

1. For arithmetic sequence A, let the common difference be \(d\). Since \(a_4=a_1+3d\), \(64=1+3d\), so \(d=21\). The middle terms are \(a_2=22\) and \(a_3=43\), and their sum is \(22+43=65\). 2. For geometric sequence B, let the common ratio be \(r\). Since \(b_4=b_1r^3\), \(64=1\cdot r^3\), so \(r=4\). The middle terms are \(b_2=4\) and \(b_3=16\), and their sum is \(4+16=20\). 3. Since \(65>20\), the sum of the middle terms is greater for the arithmetic sequence.

Sequence A: \(65\) Sequence B: \(20\) The arithmetic sequence has the greater sum of its two middle terms.

- What equation represents multiplying by the same factor four times? - How can you evaluate a fourth root using two square roots? - For part b, can you use an exact value for \(k^2\) instead of the rounded value of \(k\)?

1. After four reductions, \(A_4=A_0k^4\), so \(400=2500k^4\). 2. Therefore, \(k^4=\frac{400}{2500}=0.16\), and \(k=(0.16)^{1/4}\approx0.6325\). 3. For the second reduction, \(A_2=A_0k^2\). Since \(k^2=\sqrt{0.16}=0.4\), \(A_2=2500\cdot0.4=1000\,\text{cm}^2\).

a) \(k\approx0.6325\) b) \(1000\,\text{cm}^2\)

- Check whether consecutive terms have a constant difference or a constant ratio. - Use the arithmetic-sequence formulas for the last term and the sum of the first \(n\) terms. - Use the explicit formula for a geometric sequence to find a specified term.

1. Sequence \(A\) is arithmetic because each term increases by the constant difference \(d=4\). Sequence \(B\) is geometric because each term is multiplied by the constant ratio \(r=\frac{15}{10}=1.5\). 2. For sequence \(A\), \(a_1=10\), \(d=4\), and \(n=30\). The \(30\)th term is \(a_{30}=10+29\cdot4=126\). Therefore, \(S_{30}=\frac{30}{2}\cdot(10+126)=2040\). 3. For sequence \(B\), \(b_{10}=10\cdot(1.5)^9=384.43359375\), so \(b_{10}\approx384.43\).

a) Sequence \(A\) is arithmetic with common difference \(4\). Sequence \(B\) is geometric with common ratio \(1.5\). b) \(2040\) c) \(384.43\)

- Exponential growth sampled at equal time intervals forms a geometric sequence. - Count the number of daily steps from the initial value to day \(7\). - Solve for the common ratio. - Use the relationship between roots and rational exponents.

1. Let \(b_0=2\) and \(b_7=10\). For common ratio \(q\), \(10=2q^7\), so \(q^7=5\) and \(q=\sqrt[7]{5}\). 2. The sequence is \(b_n=2\left(\sqrt[7]{5}\right)^n=2\sqrt[7]{5^n}\). 3. The six values are: Day \(1\): \(2\sqrt[7]{5}\approx2.52\,\text{m}^2\) Day \(2\): \(2\sqrt[7]{5^2}\approx3.17\,\text{m}^2\) Day \(3\): \(2\sqrt[7]{5^3}\approx3.99\,\text{m}^2\) Day \(4\): \(2\sqrt[7]{5^4}\approx5.02\,\text{m}^2\) Day \(5\): \(2\sqrt[7]{5^5}\approx6.31\,\text{m}^2\) Day \(6\): \(2\sqrt[7]{5^6}\approx7.95\,\text{m}^2\).

Day \(1\): \(2\sqrt[7]{5}\approx2.52\,\text{m}^2\) Day \(2\): \(2\sqrt[7]{5^2}\approx3.17\,\text{m}^2\) Day \(3\): \(2\sqrt[7]{5^3}\approx3.99\,\text{m}^2\) Day \(4\): \(2\sqrt[7]{5^4}\approx5.02\,\text{m}^2\) Day \(5\): \(2\sqrt[7]{5^5}\approx6.31\,\text{m}^2\) Day \(6\): \(2\sqrt[7]{5^6}\approx7.95\,\text{m}^2\)

- Represent the geometric sequence with a first term and common ratio. - For three arithmetic terms, the middle term is the average of the outer terms. - Combine the sum condition with the arithmetic condition. - Use the fact that the original geometric sequence is increasing.

1. Write the geometric sequence as \(a, aq, aq^2\), where \(q>1\). 2. The sum condition gives \(a(1+q+q^2)=21\). 3. The modified terms are \(a+2\), \(aq+2\), and \(aq^2-1\). For them to be arithmetic, \(2(aq+2)=(a+2)+(aq^2-1)\). Simplifying gives \(a(q-1)^2=3\). 4. Substitute \(a=\frac{21}{1+q+q^2}\): \(\frac{21(q-1)^2}{1+q+q^2}=3\). This simplifies to \(2q^2-5q+2=0\), so \(q=2\) or \(q=\frac12\). 5. Since the original sequence is increasing, \(q=2\). Then \(a=3\), so the original numbers are \(3, 6, 12\).

The original numbers are \(3, 6, 12\).

- Test whether consecutive differences or consecutive ratios are constant. - Express the result symbolically before drawing a conclusion. - A quantity that still depends on \(n\) is generally not constant.

1. If \((a_n)\) is arithmetic, then \(b_{n+1}-b_n=5(a_{n+1}-a_n)=5d\), so \((b_n)\) is arithmetic. If \((a_n)\) is geometric, then \(\frac{b_{n+1}}{b_n}=\frac{5a_{n+1}}{5a_n}=q\), so \((b_n)\) is geometric. 2. If \((a_n)\) is arithmetic, adding \(5\) preserves the difference \(d\), so \((b_n)\) is arithmetic. If \((a_n)\) is geometric, neither \(\frac{a_{n+1}+5}{a_n+5}\) nor \((a_{n+1}+5)-(a_n+5)\) is generally constant. Thus, \((b_n)\) is neither arithmetic nor geometric. 3. If \((a_n)\) is arithmetic, \(b_{n+1}-b_n=d(a_{n+1}+a_n)\), which generally depends on \(n\), and the ratio also generally depends on \(n\). Thus, \((b_n)\) is neither. If \((a_n)\) is geometric, \(\frac{b_{n+1}}{b_n}=q^2\), so \((b_n)\) is geometric. 4. If \((a_n)\) is arithmetic, \(b_{n+1}-b_n=-\frac{d}{a_n(a_n+d)}\) and \(\frac{b_{n+1}}{b_n}=\frac{a_n}{a_n+d}\), neither of which is generally constant. Thus, \((b_n)\) is neither. If \((a_n)\) is geometric, \(\frac{b_{n+1}}{b_n}=\frac1q\), so \((b_n)\) is geometric.

1) Arithmetic remains arithmetic; geometric remains geometric. 2) Arithmetic remains arithmetic; geometric becomes neither. 3) Arithmetic becomes neither; geometric remains geometric with ratio \(q^2\). 4) Arithmetic becomes neither; geometric remains geometric with ratio \(\frac1q\).

- Use the quotient rule for powers with the same base. - Use the logarithm identity for a difference of logarithms. - To prove a sequence is geometric, show that consecutive ratios are constant. - To prove a sequence is arithmetic, show that consecutive differences are constant.

a) Compute the ratio: \(\frac{b_{n+1}}{b_n}=\frac{2^{a_{n+1}}}{2^{a_n}}=2^{a_{n+1}-a_n}=2^d\). Since this ratio is constant, \((b_n)\) is geometric with ratio \(2^d\). b) Compute the difference: \(c_{n+1}-c_n=\log_2(g_{n+1})-\log_2(g_n)\) \(=\log_2\left(\frac{g_{n+1}}{g_n}\right)=\log_2(q)\). Since this difference is constant, \((c_n)\) is arithmetic with difference \(\log_2(q)\). c) Compute the ratio: \(\frac{w_{n+1}}{w_n}=\frac{u_{n+1}v_{n+1}}{u_nv_n}=\frac{u_{n+1}}{u_n}\frac{v_{n+1}}{v_n}=q_uq_v\). Therefore, \((w_n)\) is geometric with ratio \(q_uq_v\).

a) The common ratio is \(2^d\). b) The common difference is \(\log_2(q)\). c) The product sequence is geometric with common ratio \(q_uq_v\).

- Write the explicit formula for an arithmetic sequence. - Use exponent properties to simplify a quotient of powers with the same base. - A sequence is geometric when the ratio of consecutive terms is constant. - Relate the exponent difference to the common difference \(d\).

1. The explicit formula for the arithmetic sequence is \(a_n = a_1 + (n-1)d\). 2. Therefore, \(x_n = 10^{a_1+(n-1)d}\), and \(x_1 = 10^{a_1}\). 3. Compute the ratio of consecutive terms: \(\frac{x_{n+1}}{x_n} = \frac{10^{a_{n+1}}}{10^{a_n}} = 10^{a_{n+1}-a_n}\). 4. Since \(a_{n+1}-a_n = d\), the ratio is \(q = 10^d\). 5. The ratio is constant, so \(x_n\) is geometric with first term \(x_1 = 10^{a_1}\) and common ratio \(q = 10^d\).

Yes. \(x_n\) is geometric with \(x_1 = 10^{a_1}\) and \(q = 10^d\).

- Center the arithmetic sequence at its middle term. - For three geometric terms, use the relationship between the middle term and its neighbors. - Do not discard a decreasing sequence unless the problem requires an increasing one.

1. Write the arithmetic terms as \(a-d, a, a+d\). Their sum is \(3a=15\), so \(a=5\). The terms are \(5-d, 5, 5+d\). 2. The modified terms are \(4-d, 4, 6+d\). For three consecutive geometric terms, the square of the middle term equals the product of the outer terms: \(4^2=(4-d)(6+d)\). 3. Simplifying gives \(d^2+2d-8=0\), so \(d=2\) or \(d=-4\). 4. For \(d=2\), the original triple is \((3, 5, 7)\), and the modified triple is \((2, 4, 8)\). For \(d=-4\), the original triple is \((9, 5, 1)\), and the modified triple is \((8, 4, 2)\). Both modified triples are geometric.

The possible original triples are \((3, 5, 7)\) and \((9, 5, 1)\).

- Express the geometric condition using the middle term squared. - Express the arithmetic condition using the middle term as an average. - Use the pair sums to reduce the number of variables. - Apply the positive-integer restriction after solving the quadratic.

1. The sequence conditions give \(b^2=ac\) and \(2c=b+d\). The sum conditions are \(b+c=18\) and \(a+d=21\). 2. Write \(b=18-c\). Then \(a=\frac{(18-c)^2}{c}\) and \(d=2c-b=3c-18\). 3. Substitute into \(a+d=21\): \(\frac{(18-c)^2}{c}+3c-18=21\). Multiplying by \(c\) and simplifying gives \(4c^2-75c+324=0\). 4. The solutions are \(c=12\) and \(c=6.75\). Since all four numbers must be integers, \(c=12\). Then \(b=6\), \(a=3\), and \(d=18\).

The four integers are \(3, 6, 12, 18\).

- Represent the geometric terms using a first term and common ratio. - Set consecutive differences of the modified terms equal. - Check exceptional cases before dividing equations. - Solve first for the common ratio, then for the first term.

1. Write the geometric sequence as \(a, aq, aq^2, aq^3\). The modified terms are \(a-2, aq-3, aq^2-9, aq^3-25\). 2. Equality of the first two arithmetic differences gives \(a(q-1)-1=aq(q-1)-6\), so \(a(q-1)^2=5\). 3. Equality of the next two differences gives \(aq(q-1)-6=aq^2(q-1)-16\), so \(aq(q-1)^2=10\). 4. The cases \(a=0\) and \(q=1\) do not satisfy the required arithmetic condition. Dividing the second equation by the first gives \(q=2\). 5. Then \(a(q-1)^2=5\) gives \(a=5\). Therefore, the original sequence is \(5, 10, 20, 40\).

\(5, 10, 20, 40\).

- Center the arithmetic sequence at its middle term. - Use the middle-term property of three consecutive geometric terms. - For part b), apply the finite arithmetic-series formula.

1. Write the arithmetic terms as \(a-d, a, a+d\). Their sum is \(3a=21\), so \(a=7\). The terms are \(7-d, 7, 7+d\). 2. The modified terms are \(9-d, 10, 16+d\). For three consecutive geometric terms, \(10^2=(9-d)(16+d)\). This simplifies to \(d^2+7d-44=0\), so \(d=4\) or \(d=-11\). 3. The two arithmetic triples are therefore \((3, 7, 11)\) and \((18, 7, -4)\). 4. For the sequence with \(d=4\) and \(a_1=3\), set the arithmetic-series sum equal to \(465\): \(\frac n2[2\cdot3+(n-1)4]=465\). This gives \(2n^2+n-465=0\), whose positive solution is \(n=15\).

a) The sequences begin \((3, 7, 11)\) and \((18, 7, -4)\). b) \(15\) terms must be added.