Free math worksheets

- Compare consecutive terms using subtraction first. - If the differences are not constant, compare consecutive terms using division.

Sequence \(A\) is arithmetic because each term increases by \(4\), so \(d=4\). Sequence \(B\) is geometric because each term is multiplied by \(\frac13\), so \(r=\frac13\).

\(A\): arithmetic with \(d=4\) \(B\): geometric with \(r=\frac13\)

- Find the change from one displayed term to the next. - Apply the same additive change repeatedly.

The difference between consecutive terms is \(-2-(-5)=3\), so \(d=3\). Continue adding \(3\): the next three terms are \(4, 7, 10\).

\(d=3\); next terms: \(4, 7, 10\)

- Compare each term with the term immediately before it using a ratio. - Pay attention to the alternating signs as well as the magnitudes.

Each term is multiplied by \(-3\), so \(r=-3\). Continuing the pattern gives \((-54)(-3)=162\) and \(162(-3)=-486\).

\(r=-3\); next terms: \(162, -486\)

- How many times is the first term multiplied by the common ratio to reach the fifth term? - Write an equation in the form \(x_5=x_1r^n\). - After substituting the known terms, isolate the power of \(r\) before taking a root. - 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 \(-7, -1, 5, 11\).

- Check whether consecutive terms have a constant difference or a constant ratio. - For an arithmetic sequence, relate the requested index to the first term and common difference. - For a geometric sequence, relate the requested index to the first term and common ratio.

a) Sequence \(A\) is arithmetic because consecutive terms have the constant difference \(4\). Sequence \(B\) is geometric because consecutive terms have the constant ratio \(1.5\). b) For sequence \(A\), \(a_{30}=10+29\cdot4=126\). c) For sequence \(B\), \(b_{10}=10(1.5)^9=384.43359375\), so \(b_{10}\approx384.43\).

a) Sequence \(A\) is arithmetic with common difference \(4\). Sequence \(B\) is geometric with common ratio \(1.5\). b) \(126\) c) \(384.43\)

- 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 \(d=3\).

- Identify the constant amount added at each recursive step. - An explicit formula must show how many common-difference steps occur from term \(1\) to term \(n\). - Substitute the requested index only after writing the general rule.

The common difference is \(-4\). Starting from \(a_1=9\), the explicit form is \(a_n=9-4(n-1)\). Therefore, \(a_{12}=9-4(11)=-35\).

\(a_n=9-4(n-1)\); \(a_{12}=-35\)

- Identify the constant factor used in each recursive step. - Count how many multiplicative steps separate the first term from the \(n\)th term. - Keep the exponent tied to the index before evaluating a particular term.

The common ratio is \(0.5\). From term \(1\) to term \(n\), the ratio is applied \(n-1\) times, so \(g_n=160(0.5)^{n-1}\). Then \(g_6=160(0.5)^5=5\).

\(g_n=160(0.5)^{n-1}\); \(g_6=5\)

- Identify the first term by using the smallest allowed index. - Identify the constant additive change encoded by the explicit formula. - A recursive definition needs both an initial value and a rule for later terms.

The explicit form shows first term \(7\) and common difference \(3\). An equivalent recurrence is \(a_1=7\), \(a_n=a_{n-1}+3\) for \(n\ge2\). Also, \(a_5=7+3(4)=19\).

\(a_1=7\), \(a_n=a_{n-1}+3\) for \(n\ge2\); \(a_5=19\)

- Use the explicit formula to identify the value at the first index. - Compare what happens when the exponent increases by \(1\). - A recursive rule should relate each term directly to the preceding term.

The first term is \(5\), and each increase of \(1\) in the index multiplies the value by \(2\). Therefore, \(g_1=5\), \(g_n=2g_{n-1}\) for \(n\ge2\). Also, \(g_4=5\cdot2^3=40\).

\(g_1=5\), \(g_n=2g_{n-1}\) for \(n\ge2\); \(g_4=40\)

- 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) \(1000\,\text{cm}^2\)

- 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 \(6\): \(2\sqrt[7]{5^6}\approx7.95\,\text{m}^2\)

- Test whether consecutive differences or consecutive ratios are constant. - Express the relevant difference or ratio symbolically before deciding what kind of sequence results. - Watch for whether the expression you obtain still depends on \(n\).

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, the consecutive differences and ratios are not constant when \(q\ne1\), so \((b_n)\) is neither arithmetic nor geometric. 3) If \((a_n)\) is arithmetic, \(b_{n+1}-b_n=d(a_{n+1}+a_n)\), which depends on \(n\), so \((b_n)\) is not arithmetic; its ratios are also not constant in general. 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 \(\frac{b_{n+1}}{b_n}=\frac{a_n}{a_n+d}\), so neither is constant. 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 \(\frac1q\).

- Compare consecutive transformed terms using a ratio or a difference that matches the sequence type you want to prove. - Use exponent and logarithm identities only after writing the consecutive-term relationship. - For the product sequence, use the defining recurrence of each geometric sequence rather than dividing by terms that might be zero.

a) \(\frac{b_{n+1}}{b_n}=\frac{2^{a_{n+1}}}{2^{a_n}}=2^{a_{n+1}-a_n}=2^d\). Since this ratio is constant, \((b_n)\) is geometric with ratio \(2^d\). b) \(c_{n+1}-c_n=\log_2(g_{n+1})-\log_2(g_n)=\log_2\!\left(\frac{g_{n+1}}{g_n}\right)=\log_2(q)\). Since this difference is constant, \((c_n)\) is arithmetic with difference \(\log_2(q)\). c) Because \(u_{n+1}=q_uu_n\) and \(v_{n+1}=q_vv_n\), \(w_{n+1}=u_{n+1}v_{n+1}=(q_uu_n)(q_vv_n)=q_uq_vw_n\). Therefore, \((w_n)\) is geometric with common 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 \(q = 10^d\).

- Read the plotted y-values in order of increasing integer index. - Compare consecutive values first by differences and then by ratios. - Use the first plotted term together with the constant change to write the explicit rule.

a) Panel a) has values \(2, 5, 8, 11, 14\). The consecutive differences are \(3\), so it is arithmetic with \(a_n=2+3(n-1)\). b) Panel b) has values \(2, 4, 8, 16, 32\). Each term is twice the preceding term, so it is geometric with \(a_n=2\cdot2^{n-1}\).

a) Panel a): arithmetic, \(d=3\), \(a_n=2+3(n-1)\) b) Panel b): geometric, \(r=2\), \(a_n=2\cdot2^{n-1}\)

- Read the y-values at consecutive integer indices. - Compare additive change and multiplicative change separately. - Think about which kind of function has a constant rate of change and which has a constant growth factor.

a) Sequence \(L\) has values \(5, 8, 11, 14, 17, 20\), so its consecutive differences are constant at \(3\). It is arithmetic. Sequence \(E\) has values \(5, 7.5, 11.25, 16.875, 25.3125, 37.96875\), so its consecutive ratio is \(1.5\). It is geometric. b) \(L_n=5+3(n-1)\) and \(E_n=5(1.5)^{n-1}\). c) An arithmetic sequence changes by a constant amount per index step, matching a linear function on the integer inputs. A geometric sequence changes by a constant factor per index step, matching an exponential function on the integer inputs.

a) \(L\) is arithmetic with \(d=3\); \(E\) is geometric with \(r=1.5\). b) \(L_n=5+3(n-1)\); \(E_n=5(1.5)^{n-1}\) c) Constant differences give linear behavior; constant ratios give exponential behavior.

- Test each proposed formula at the first allowed index. - An explicit sequence formula must reproduce the stated first term before anything else. - Apply the same indexing idea to the exponent in a geometric formula.

a) Jordan's formula is correct. When \(n=1\), \(6+4(n-1)=6\), the required first term. Alex's formula gives \(10\) at \(n=1\), so it shifts every term one index too far. b) With first term \(6\), common ratio \(4\), and indexing from \(1\), the formula is \(g_n=6\cdot4^{n-1}\).

a) Jordan is correct; Alex's formula is shifted by one index. b) \(g_n=6\cdot4^{n-1}\)

- 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 \(3, 6, 12\).

- 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 so the sum condition becomes simple. - For three consecutive geometric terms, compare the middle term with the product of the outer terms. - In part b, use the first term and common difference from the increasing arithmetic sequence.

a) Write the arithmetic terms as \(a-d, a, a+d\). Their sum is \(3a=21\), so \(a=7\). The terms are \(7-d, 7, 7+d\). 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 \(d=-11\). Therefore, the two arithmetic triples are \((3, 7, 11)\) and \((18, 7, -4)\). b) For the sequence with positive common difference, \(a_1=3\) and \(d=4\). Thus, \(a_n=3+4(n-1)=4n-1\), and \(a_{15}=4(15)-1=59\).

a) The sequences begin \((3, 7, 11)\) and \((18, 7, -4)\). b) \(a_n=4n-1\), and \(a_{15}=59\).