Sequences and Series: Algebra 2 Math Worksheet Answer Key – PDF Download

Problem 1: Arithmetic or geometric

1. Differences: \(9-5=4\), \(13-9=4\), \(17-13=4\). Arithmetic, \(d=4\).
2. Differences \(3,\,6,\,12\) are not equal; ratios \(\frac{6}{3}=\frac{12}{6}=\frac{24}{12}=2\). Geometric, \(r=2\).
3. Differences are all \(-3\). Arithmetic, \(d=-3\).
4. Ratios: \(\frac{27}{81}=\frac{9}{27}=\frac{3}{9}=\frac{1}{3}\). Geometric, \(r=\frac{1}{3}\).

Problem 2: Reading a graph

1. Each step is \(-6\): 40, 34, 28, 22, 16. Arithmetic, \(d=-6\).
2. \(a_{n}=40+(n-1)(-6)=46-6n\). \(a_{n}=46-6n\)
3. \(a_{12}=46-6(12)=46-72=\)\(-26\).

Problem 3: Explicit arithmetic rule

\(a_{n}=7+(n-1)(5)=5n+2\). \(a_{n}=5n+2\)

\(a_{20}=5(20)+2=\)\(102\).

Problem 4: Explicit geometric rule

\(a_{n}=3\cdot 2^{n-1}\)

\(a_{10}=3\cdot 2^{9}=3\cdot 512=\)\(1536\).

Problem 5: Terms from a recursive rule

1. Multiply by \(-2\) each time: \(4,\,-8,\,16,\,-32,\,64\).
2. Subtract 6 each time: \(50,\,44,\,38,\,32,\,26\).

Problem 6: Writing a recursive rule

1. Differences are all 7: \(a_{1}=12,\quad a_{n}=a_{n-1}+7\).
2. Ratios are all \(\frac{1}{2}\): \(a_{1}=80,\quad a_{n}=\frac{1}{2}a_{n-1}\).

Problem 7: How many terms

Solve \(8+(n-1)(5)=163\): \(5(n-1)=155\), so \(n-1=31\). \(n=32\) terms.

Problem 8: Rule from two terms

From \(a_{2}\) to \(a_{7}\) there are 5 steps: \(5d=39-9=30\), so \(d=6\).

\(a_{1}=9-6=\)\(3\). Then \(a_{n}=3+6(n-1)=\)\(6n-3\). Check: \(a_{7}=6(7)-3=39\).

Problem 9: Geometric from two terms

From \(a_{2}\) to \(a_{5}\) there are 3 steps: \(r^{3}=\frac{324}{12}=27\), so \(r=3\).

\(a_{1}=\frac{12}{3}=\)\(4\). Then \(a_{8}=4\cdot 3^{7}=4\cdot 2187=\)\(8748\).

Problem 10: Evaluating sigma notation

Substitute \(k=1,\,2,\,3,\,4,\,5,\,6\): \(3+5+7+9+11+13=\)\(48\).

Check with the formula: \(\frac{6(3+13)}{2}=48\).

Problem 11: Geometric sigma

Terms: \(2,\,6,\,18,\,54\). Sum: \(80\).

Check: \(\frac{2(1-3^{4})}{1-3}=\frac{2(-80)}{-2}=80\).

Problem 12: Writing sigma notation

1. Arithmetic with \(a_{1}=6\), \(d=4\): the rule is \(4k+2\). \(\sum_{k=1}^{5}(4k+2)\).
2. Geometric with \(a_{1}=3\), \(r=2\): the rule is \(3\cdot 2^{k-1}\). \(\sum_{k=1}^{6}3\cdot 2^{k-1}\).

Problem 13: Arithmetic series sum

Number of terms: \(4+5(n-1)=99\) gives \(n=20\).

\(S_{20}=\frac{20(4+99)}{2}=10\cdot 103=\)\(1030\).

Problem 14: Geometric series sum

\(S_{8}=\frac{5(1-2^{8})}{1-2}=\frac{5(-255)}{-1}=\)\(1275\).

Problem 15: Ratio less than 1

\(S_{6}=\frac{81(1-(\frac{1}{3})^{6})}{1-\frac{1}{3}}=\frac{81\cdot\frac{728}{729}}{\frac{2}{3}}=\frac{\frac{728}{9}}{\frac{2}{3}}=\)\(\frac{364}{3}\), or about 121.33.

Check by adding: \(81+27+9+3+1+\frac{1}{3}=121\frac{1}{3}\).

Problem 16: Finding the number of terms

\(S_{n}=\frac{n}{2}(2a_{1}+(n-1)d)=\frac{n}{2}(4+3n-3)=\frac{n(3n+1)}{2}\).

Set equal to 155: \(3n^{2}+n=310\), so \(3n^{2}+n-310=0\), which factors as \((3n+31)(n-10)=0\).

\(n=-\frac{31}{3}\) is impossible, so \(n=10\) terms. Check: \(a_{10}=29\) and \(\frac{10(2+29)}{2}=155\).

Problem 17: Weekly savings

1. \(a_{12}=5+11(3)=38\). $38.
2. \(S_{12}=\frac{12(5+38)}{2}=6\cdot 43=258\). $258.

Problem 18: Theater seating

1. \(a_{18}=14+17(2)=48\). 48 seats.
2. \(S_{18}=\frac{18(14+48)}{2}=9\cdot 62=558\). 558 seats.

Problem 19: Bouncing ball

1. Bounce 1 is \(10(0.6)=6\) feet, with ratio 0.6: \(h_{n}=6(0.6)^{n-1}\).
2. \(h_{5}=6(0.6)^{4}=6(0.1296)=\)0.7776 ft.
3. \(S_{5}=\frac{6(1-0.6^{5})}{1-0.6}=\frac{6(0.92224)}{0.4}=\)13.8336 ft.

Problem 20: Chain email

1. Geometric with \(a_{1}=3\), \(r=3\): \(a_{7}=3\cdot 3^{6}=2187\). 2,187 people.
2. \(S_{7}=\frac{3(1-3^{7})}{1-3}=\frac{3(-2186)}{-2}=3279\). 3,279 people.

Problem 21: Raises over ten years

Salaries form a geometric sequence with \(a_{1}=42000\) and \(r=1.03\).

1. \(a_{10}=42000(1.03)^{9}\approx \$54,800.47\). $54,800.47.
2. \(S_{10}=\frac{42000(1.03^{10}-1)}{0.03}\approx \$481,482.93\). $481,482.93 earned in 10 years.

Problem 22: Savings account

The last deposit earns no interest and the first earns interest for 11 months. The values form a geometric series with \(a_{1}=200\), \(r=1.005\), \(n=12\).

\(S_{12}=\frac{200(1.005^{12}-1)}{0.005}\approx \$2,467.11\). $2,467.11.

Problem 23: Car loan

The values form a geometric series with \(a_{1}=\frac{300}{1.005}\), \(r=\frac{1}{1.005}\), \(n=36\).

1. \(P=300\cdot\frac{1-1.005^{-36}}{0.005}\approx \$9,861.30\). $9,861.30.
2. Total paid: \(300\cdot 36=\$10{,}800\). Interest: \(10800-9861.30\approx \$938.70\). $938.70.

Problem 24: Mortgage payment

Payment \(k\) is worth \(M(1.005)^{-k}\) today. The sum of 180 of these is \(M\cdot\frac{1-1.005^{-180}}{0.005}\), and it equals 150,000.

1. \(M=\frac{150000(0.005)}{1-1.005^{-180}}\approx \$1,265.79\). $1,265.79 per month.
2. Total paid: \(1265.79\cdot 180=\$227,842.20\). Interest: $77,842.20.

See every chapter on the Algebra 2 math page.