Math Problem Statement

Arithmetic Series

I. Find the sum of the following arithmetic sequence:

  1. 3, 8 ½, ..., A20
  2. 19, 13, ..., A10
  3. -17, -12, -7, ..., 33
  4. 9, -1, ..., A8
  5. 4, 11, ..., A16

Solution

To find the sum of an arithmetic sequence, we use the formula for the sum of an arithmetic series:

Sn=n2×(a1+an)S_n = \frac{n}{2} \times (a_1 + a_n)

Where:

  • SnS_n is the sum of the first nn terms.
  • a1a_1 is the first term.
  • ana_n is the nn-th term.
  • nn is the number of terms.

Alternatively, we can also use the general formula for the nn-th term of an arithmetic sequence:

an=a1+(n1)da_n = a_1 + (n-1)d

Where:

  • ana_n is the nn-th term.
  • a1a_1 is the first term.
  • dd is the common difference.
  • nn is the term number.

Let’s solve each sequence one by one.

1. Sequence: 3,812,...,A203, 8 \frac{1}{2}, ..., A_{20}

  • First term: a1=3a_1 = 3
  • Second term: a2=812=8.5a_2 = 8 \frac{1}{2} = 8.5
  • Common difference: d=8.53=5.5d = 8.5 - 3 = 5.5

Now, find the 20th term a20a_{20}: a20=3+(201)×5.5=3+19×5.5=3+104.5=107.5a_{20} = 3 + (20 - 1) \times 5.5 = 3 + 19 \times 5.5 = 3 + 104.5 = 107.5

Now, calculate the sum S20S_{20}: S20=202×(3+107.5)=10×110.5=1105S_{20} = \frac{20}{2} \times (3 + 107.5) = 10 \times 110.5 = 1105

2. Sequence: 19,13,...,A1019, 13, ..., A_{10}

  • First term: a1=19a_1 = 19
  • Second term: a2=13a_2 = 13
  • Common difference: d=1319=6d = 13 - 19 = -6

Find the 10th term a10a_{10}: a10=19+(101)×(6)=1954=35a_{10} = 19 + (10 - 1) \times (-6) = 19 - 54 = -35

Now, calculate the sum S10S_{10}: S10=102×(19+(35))=5×(16)=80S_{10} = \frac{10}{2} \times (19 + (-35)) = 5 \times (-16) = -80

3. Sequence: 17,12,7,...,33-17, -12, -7, ..., 33

  • First term: a1=17a_1 = -17
  • Common difference: d=12(17)=5d = -12 - (-17) = 5
  • Last term: an=33a_n = 33

To find the number of terms nn, use the formula for the nn-th term: 33=17+(n1)×533 = -17 + (n-1) \times 5 33+17=(n1)×533 + 17 = (n-1) \times 5 50=(n1)×5n1=10n=1150 = (n-1) \times 5 \quad \Rightarrow \quad n-1 = 10 \quad \Rightarrow \quad n = 11

Now, calculate the sum S11S_{11}: S11=112×(17+33)=112×16=88S_{11} = \frac{11}{2} \times (-17 + 33) = \frac{11}{2} \times 16 = 88

4. Sequence: 9,1,...,A89, -1, ..., A_8

  • First term: a1=9a_1 = 9
  • Second term: a2=1a_2 = -1
  • Common difference: d=19=10d = -1 - 9 = -10

Find the 8th term a8a_8: a8=9+(81)×(10)=970=61a_8 = 9 + (8 - 1) \times (-10) = 9 - 70 = -61

Now, calculate the sum S8S_8: S8=82×(9+(61))=4×(52)=208S_8 = \frac{8}{2} \times (9 + (-61)) = 4 \times (-52) = -208

5. Sequence: 4,11,...,A164, 11, ..., A_{16}

  • First term: a1=4a_1 = 4
  • Second term: a2=11a_2 = 11
  • Common difference: d=114=7d = 11 - 4 = 7

Find the 16th term a16a_{16}: a16=4+(161)×7=4+105=109a_{16} = 4 + (16 - 1) \times 7 = 4 + 105 = 109

Now, calculate the sum S16S_{16}: S16=162×(4+109)=8×113=904S_{16} = \frac{16}{2} \times (4 + 109) = 8 \times 113 = 904

Final Results:

  1. S20=1105S_{20} = 1105
  2. S10=80S_{10} = -80
  3. S11=88S_{11} = 88
  4. S8=208S_8 = -208
  5. S16=904S_{16} = 904

Would you like any further details on the steps or concepts used?

Here are five related questions:

  1. How do you derive the common difference for an arithmetic sequence?
  2. What is the formula to find the sum of the first nn terms of an arithmetic sequence?
  3. How do you determine the number of terms in a sequence if you know the first and last terms?
  4. What is the relationship between the common difference and the progression of an arithmetic sequence?
  5. How can you calculate the sum of an arithmetic sequence with a negative common difference?

Tip: For arithmetic sequences, once you know the common difference, calculating any term is straightforward using the general formula.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Series
Summation
Sequences and Series

Formulas

Sum of arithmetic series: S_n = (n/2) × (a_1 + a_n)
nth term of an arithmetic sequence: a_n = a_1 + (n-1) × d

Theorems

Arithmetic Sequence Formula
Summation of Arithmetic Series

Suitable Grade Level

Grades 9-11