Arithmetic and geometric sequences Knowledge-Based Study

Math · Algebra · Math · Sequences & Series

Math topic explanations on Arithmetic and geometric sequences.

Study notes

  1. Definition and General Term

    An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. This fixed difference is called the common difference, denoted by dd.

    If the first term of the sequence is a1a_1, then consecutive terms are generated by repeatedly adding dd:

    • a1a_1
    • a2=a1+da_2 = a_1 + d
    • a3=a1+2da_3 = a_1 + 2d
    • an=a1+(n−1)da_n = a_1 + (n - 1)d

    The formula for the nn-th term (the general term) is therefore:

    an=a1+(n−1)da_n = a_1 + (n - 1)d

    where nn is a positive integer representing the term number.

    Example

    Consider the sequence 5, 8, 11, 14, ...

    • First term: a1=5a_1 = 5
    • Common difference: d=8−5=3d = 8 - 5 = 3
    • The 20th term is: a20=5+(20−1)(3)=5+19(3)=5+57=62a_{20} = 5 + (20 - 1)(3) = 5 + 19(3) = 5 + 57 = 62

    Sum of the First n Terms

    An arithmetic series is the sum of the terms of an arithmetic sequence. The sum of the first nn terms, denoted SnS_n, can be determined using either of two equivalent formulas:

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

    Substituting an=a1+(n−1)da_n = a_1 + (n - 1)d into this equation gives:

    Sn=n2(2a1+(n−1)d)S_n = \frac{n}{2}\left(2a_1 + (n - 1)d\right)

    Example

    Find the sum of the first 15 terms of the arithmetic sequence 4, 9, 14, 19, ...

    • Here, a1=4a_1 = 4, d=5d = 5, and n=15n = 15.
    • Using the second formula:

    S15=152(2(4)+(15−1)(5))=152(8+70)=152(78)=15×39=585S_{15} = \frac{15}{2}\left(2(4) + (15 - 1)(5)\right) = \frac{15}{2}(8 + 70) = \frac{15}{2}(78) = 15 \times 39 = 585

    Definition and General Term

    A geometric sequence is a sequence in which each term after the first is found by multiplying the preceding term by a non-zero constant called the common ratio, denoted by rr.

    If the first term is g1g_1, the terms follow the pattern:

    • g1g_1
    • g2=g1⋅rg_2 = g_1 \cdot r
    • g3=g1⋅r2g_3 = g_1 \cdot r^2
    • gn=g1⋅rn−1g_n = g_1 \cdot r^{n-1}

    The general formula for the nn-th term is:

    gn=g1⋅rn−1g_n = g_1 \cdot r^{n-1}

    To find the common ratio between consecutive terms, compute r=gk+1gkr = \frac{g_{k+1}}{g_k} for any valid index kk.

    Example

    Consider the sequence 3, 6, 12, 24, ...

    • First term: g1=3g_1 = 3
    • Common ratio: r=63=2r = \frac{6}{3} = 2
    • The 8th term is: g8=3⋅28−1=3⋅27=3⋅128=384g_8 = 3 \cdot 2^{8 - 1} = 3 \cdot 2^7 = 3 \cdot 128 = 384

    Finite and Infinite Sums

    A geometric series is the sum of terms in a geometric sequence.

    Finite Geometric Series

    The sum of the first nn terms of a geometric sequence with r≠1r \neq 1 is given by:

    Sn=g1(1−rn)1−rS_n = \frac{g_1(1 - r^n)}{1 - r}

    Infinite Geometric Series

    When the absolute value of the common ratio satisfies ∣r∣<1|r| < 1, the terms approach zero as nn increases without bound. The infinite series converges to a finite sum:

    S=g11−rS = \frac{g_1}{1 - r}

    If ∣r∣≥1|r| \ge 1 and g1≠0g_1 \neq 0, the infinite geometric series diverges and has no finite sum.

    Example

    Find the sum of the infinite series 12+6+3+1.5+…12 + 6 + 3 + 1.5 + \dots

    • First term: g1=12g_1 = 12
    • Common ratio: r=612=12r = \frac{6}{12} = \frac{1}{2}
    • Since ∣r∣<1|r| < 1, the sum is:

    S=121−12=120.5=24S = \frac{12}{1 - \frac{1}{2}} = \frac{12}{0.5} = 24