# determine arithmetic and geometric series

• Determine whether each of the following series is an arithmetic or a geometric series

Find the value of the common difference/ratio.

Find the sum of the terms.

1.) -2/5+(-3/20)+1/10+7/20+.....+8/5

2.) 0.5+(-0.25)+0.125+(-0.0625)+...+0.0078125

3.) 75+63+51+...+(-69)

4.) 2+4+8+16+...+16384

### Comments for determine arithmetic and geometric series

 Oct 27, 2012 arithmetic and geometric series by: Staff Answer Part I Final Answer for Problem 1 : -2/5 + (-3/20) + 1/10 + 7/20 + ..... + 8/5 Arithmetic Series. n, number of terms = 9 d, the common difference = ¼ Sum of the 9 terms = 5 ⅖, or 5.4 Final Answer for Problem 2 : 0.5 + (-0.25) + 0.125 + (-0.0625) + ... + 0.0078125 Geometric series. n, number of terms = 7 r, the common ratio = -0.5. Since -1< r < 1, this is a “convergent series”. Sum of the 7 terms = 0.3359375 Final Answer for Problem 3 : 75 + 63 + 51 + ... + (-69) Arithmetic Series. n, number of terms = 13 d, the common difference = -12 Sum of the 13 terms = 39 Final Answer for Problem 4 : 2 + 4 + 8 + 16 + ... + 16384 Geometric series. n, number of terms = 14 r, the common ratio = 2. Since r ≥ 1, this is a “divergent series”. Sum of the 14 terms = 32766 -------------------------------------------------------------------------------------------------------- What an arithmetic sequence (arithmetic progression) looks like . You can calculate the nth term in an arithmetic sequence using the following formula: -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part II an = a₁ + (n - 1) * (d) [An arithmetic sequence has the same form as a linear function: y = mx + b. However, the domain (the set of all input values; in this case n - 1) for an arithmetic sequence is limited to positive integers.] an = a with a subscript of n (this is the nth term in the series) a₁ = a with a subscript of 1 (this is the 1st term in the series) n = number of terms d = difference between consecutive terms (the common difference) d, the common difference, can be calculated as follows: d = an - an-1 (n must be greater than 1) An “arithmetic series” is the “sum of an arithmetic sequence”. It is “n” multiplied by the average (arithmetic mean) of the first and last terms. [A sequence is an ordered list of numbers such as 1, 2, 3, 4. The sum of the terms of a sequence is called a series (for example, 1+2+3+4 is a series).] Sn = (n/2)*(a1 + an) or Sn = (n/2)*[2a1 + (n – 1)d] An arithmetic progression is a “divergent series”. Therefore, the sum of an infinite arithmetic progression cannot be calculated. What an geometric sequence (geometric progression) looks like . A geometric sequence has the (general) form: an = a1 * (r)(n - 1) an = a with a subscript of n is the nth term in the sequence a1 = a with a subscript of 1 is the 1st term in the sequence n = number of terms r = the common ratio r, the common ratio, can be calculated as follows: r = an/an-1 (n must be greater than 1) -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part III A “geometric series” is the “sum of an geometric sequence”. [A sequence is an ordered list of numbers such as 1, 2, 4, 8. The sum of the terms of a sequence is called a series (for example, 1+2+4+8 is a series).] Sn = a1(1 – rn) / (1 – r) S∞ = a1 / (1 – r) This formula is only valid when -1< r < 1 (“convergent series”). 1.) -2/5 + (-3/20) + 1/10 + 7/20 + ..... + 8/5 . This is an arithmetic series because the difference between each of the terms is a constant, d: d, the common difference = ¼       (-3/20) - (-2/5) = ¼       (1/10) - (-3/20) = ¼       (7/20) - (1/10) = ¼ To find the sum of the terms, you must first determine how many terms are in the sequence: -2/5, -3/20, 1/10, 7/20, ....., 8/5 Use the following formula. Substitute 8/5 for an. Substitute -2/5 for a1. Substitute ¼ for d. Solve for “n”. -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part IV an = a1 + (n - 1) * (d)       8/5 = -2/5 + (n - 1) * (1/4)       8/5 + 2/5 = -2/5 + (n - 1) * (1/4) + 2/5       10/5 = -2/5 + (n - 1) * (1/4) + 2/5       2 = -2/5 + (n - 1) * (1/4) + 2/5       2 = -2/5 + 2/5 + (n - 1) * (1/4)       2 = 0 + (n - 1) * (1/4)       2 = (n - 1) * (1/4)       2 * 4 = (n - 1) * (1/4) * 4       8 = (n - 1) * (1/4) * 4       8 = (n - 1) * (4/4)       8 = (n - 1) * 1       8 = (n - 1)       8 = n - 1       8 + 1 = n - 1 + 1       9 = n - 1 + 1       9 = n + 0       9 = n n = 9, there are 9 terms in the series To find the sum of the series, use the following formula. Substitute 9 for “n”. Substitute ¼ for “d”. Substitute -2/5 for a1. Sn = (n/2)*[2a1 + (n – 1)d]       S9 = (9/2) * [2 * (-2/5) + (9 – 1) * ¼]       S9 = (9/2) * [(-4/5) + 8 * ¼]       S9 = (9/2) * [(-4/5) + 2]       S9 = (9/2) * (2 - 4/5)       S9 = (9/2) * (10/5 - 4/5)       S9 = (9/2) * (6/5)       S9 = (9 * 6)/(2 * 5)       S9 = (54)/(10) S9 = 5 ⅖, or 5.4 Final Answer for Problem 1: Arithmetic Series. n, number of terms = 9 d, the common difference = ¼ Sum of the 9 terms = 5 ⅖, or 5.4 -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part V 2.) 0.5 + (-0.25) + 0.125 + (-0.0625) + ... + 0.0078125 . This is a geometric series because the ratio between successive the terms is a constant, r: r, the common ratio = -0.5       (-0.25) / 0.5 = -0.5       0.125 / (-0.25) = -0.5       (-0.0625) / 0.125 = -0.5 Since -1< r < 1, this is a “convergent series”. To find the sum of the terms, you must first determine how many terms are in the sequence: 0.5, (-0.25), 0.125, (-0.0625), ..., 0.0078125 Use the following formula. Substitute 0.0078125 for an. Substitute 0.5 for a1. Substitute -0.5 for r. Solve for “n”. -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part VI an = a1 * (r)(n - 1)       0.0078125 = 0.5 * (-0.5)(n - 1)       0.0078125 / 0.5 = 0.5 * (-0.5)(n - 1) / 0.5       0.015625 = 0.5 * (-0.5)(n - 1) / 0.5       0.015625 = (-0.5)(n - 1) * (0.5 / 0.5)       0.015625 = (-0.5)(n - 1) * (1)       0.015625 = (-0.5)(n - 1)       (-0.5)(n - 1) = 0.015625 Since the last term in the series is a positive number, you know that n - 1 must be an even number. To solve for n, change the -0.5 to 0.5.       (0.5)(n - 1) = 0.015625       log(0.5(n - 1)) = log(0.015625)       (n – 1) * log(0.5) = log(0.015625)       (n – 1) * log(0.5) / log(0.5) = log(0.015625) / log(0.5)       (n – 1) * 1 = log(0.015625) / log(0.5)       (n – 1) = log0.5(0.015625) / log0.5(0.5)       (n – 1) = log0.5(0.015625) / 1       (n – 1) = log0.5(0.015625)       n - 1 = 6       n - 1 + 1 = 6 + 1       n + 0 = 6 + 1       n = 6 + 1 n = 7, there are 7 terms in the series To find the sum of the series, use the following formula. Substitute 7 for “n”. Substitute -0.5 for “r”. Substitute 0.5 for a1. Sn = a1(1 – rn) / (1 - r)       S7 = 0.5 *(1 - (-0.5)7) / (1 - (-0.5))       S7 = 0.5 *(1 - (-0.5)7) / (1 + 0.5)       S7 = 0.5 *(1 - (-0.5)7) / (1.5)       S7 = 0.5 * (1 - (-0.0078125))/ (1.5)       S7 = 0.5 * (1.0078125)/ (1.5)       S7 = (1.0078125)/ (3) S7 = 0.3359375 -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part VII Final Answer for Problem 2: Geometric series. n, number of terms = 7 r, the common ratio = -0.5. Since -1< r < 1, this is a “convergent series”. Sum of the 7 terms = 0.3359375 3.) 75 + 63 + 51 + ... + (-69) . This is an arithmetic series because the difference between each of the terms is a constant, d: d, the common difference = -12 (63) - (75) = -12 (51) - (63) = -12 To find the sum of the terms, you must first determine how many terms are in the sequence: 75, 63, 51, ..., -69 Use the following formula. Substitute -69 for an. Substitute 75 for a1. Substitute -12 for d. Solve for “n”. an = a1 + (n - 1) * (d)       -69 = 75 + (n - 1) * (-12)       -69 - 75 = 75 + (n - 1) * (-12) - 75       -144 = 75 + (n - 1) * (-12) - 75       -144 = 75 - 75 + (n - 1) * (-12)       -144 = 0 + (n - 1) * (-12)       -144 = (n - 1) * (-12)       -144 / (-12) = (n - 1) * (-12) / (-12)       12 = (n - 1) * (-12) / (-12)       12 = (n - 1) * 1       12 = (n - 1)       n - 1 = 12       n - 1 + 1 = 12 + 1       n + 0 = 12 + 1       n = 12 + 1 n = 13, there are 13 terms in the series To find the sum of the series, use the following formula. Substitute 13 for “n”. Substitute -12 for “d”. Substitute 75 for a1. -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part VIII Sn = (n/2)*[2a1 + (n – 1)d]       S13 = (13/2)*[2*75 + (13 – 1)*(-12)]       S13 = (13/2)*[150 + (12)*(-12)]       S13 = (13/2)*(150 - 144)       S13 = (13/2)*(6)       S13 = (13)*(3) S13 = 39 Final Answer for Problem 3: Arithmetic Series. n, number of terms = 13 d, the common difference = -12 Sum of the 13 terms = 39 4.) 2 + 4 + 8 + 16 +... + 16384 . This is a geometric series because the ratio between successive the terms is a constant, r: r, the common ratio = 2       (4) / 2 = 2       8 / 4 = 2       16 / 8 = 2 Since r ≥ 1, this is a “divergent series”. To find the sum of the terms, you must first determine how many terms are in the sequence: 2, 4, 8, 16, ..., 16384 Use the following formula. Substitute 16384 for an. Substitute 2 for a1. Substitute 2 for r. Solve for “n”. -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part IX an = a1 * (r)(n - 1)       16384 = 2 * (2)(n - 1)       16384 / 2 = 2 * (2)(n - 1) / 2       8192 = 2 * (2)(n - 1) / 2       8192 = (2)(n - 1) * (2/ 2)       8192 = (2)(n - 1) * (1)       8192 = (2)(n - 1)       (2)(n - 1) = 8192       (2)(n - 1) = 213       (n - 1) = 13       n - 1 = 13       n - 1 + 1 = 13 + 1       n + 0 = 13 + 1       n = 13 + 1 n = 14, there are 14 terms in the series Now, you have a problem. This geometric series is a divergent series. You can’t be sure if the total of the 14 terms will be accurate using the formula. You will need to compute the total more than one way. To find the sum of the series using the following formula: Substitute 14 for “n”. Substitute 2 for “r”. Substitute 2 for a1. Sn = a1(1 – rn) / (1 - r)       S7 = 2 *(1 - (2)14) / (1 - 2)       S7 = 2 *(1 - 16384) / (1 - 2)       S7 = 2 *(-16383) / (-1)       S7 = 32766 -------------------------------------------------------

 Oct 27, 2012 arithmetic and geometric series by: Staff ------------------------------------------------------- Part X Is this number an accurate number? In this particular case it is. But the formula does not always work for a divergent series. You can verify the sum by calculating the value of each term, and then adding the terms using Excel. n      2 * (2)(n - 1)   1         2   2         4   3         8   4         16   5         32   6         64   7         128   8         256   9         512 10         1024 11         2048 12         4096 13         8192 14         16384       sum 32766 Both totals agree. Final Answer for Problem 4: Geometric series. n, number of terms = 14 r, the common ratio = 2. Since r ≥ 1, this is a “divergent series”. Sum of the 14 terms = 32766 Thanks for writing. Staff www.solving-math-problems.com