MATH EXPONENTIAL HELP?

A concert hall has 32 rows of seats. There are 26 seats in the first row, 30 seats in the second row, 34 seats in the third row and 38 seats in the fourth row. If the price per concert ticket is $40 and all the seats are sold for the show, what will be the total sales for a one-night show?

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  • 3 months ago
    Favourite answer

    The number of seats in a row is an arithmetic sequence with the first term = 26 and the common difference = 4.  We can use this generic equation for the n'th term of an arithmetic sequence to find an expression for each row of seats, given the row number:

    a(n) = a + b(n - 1)

    where:

    a(n) is the number of seats in row "n"

    a is the number of seats in the first row (26)

    b is the common difference (4)

    Substitute and simplify to get:

    a(n) = 26 + 4(n - 1)

    a(n) = 26 + 4n - 4

    a(n) = 4n + 22

    the sum of the first "n" terms of an arithmetic sequence can be found with:

    S(n) = [ a + a(n)] n / 2

    We have values and expressions for "a" and a(n).  Substitute and simplify:

    S(n) = (26 + 4n + 22) n / 2

    S(n) = (4n + 48) n / 2

    S(n) = 2(2n + 24) n / 2

    S(n) = (2n + 24) n

    S(n) = 2n² + 24n

    There are 32 rows of seats, so solving for S(32) will tell us how many seats are in the hall:

    S(32) = 2(32)² + 24(32)

    S(32) = 2(1024) + 24(32)

    S(32) = 2048 + 768

    S(32) = 2816

    And finally, at $40 per seat, you bring in:

    2816 * 40 = $112,640 each night if all seats are sold.

  • sepia
    Lv 7
    3 months ago

    A concert hall has 32 rows of seats. 

     There are 26 seats in the first row, 30 seats in the second row, 

     34 seats in the third row and 38 seats in the fourth row. 

     26, 30, 34, 38, ...

     a_n = 2 (2 n + 11) 

     a32 = 150

     26 + 30 + 34 + 38 + .... + 150 = 2816

     (sum_(n=1)^32(4 n + 22) = 2816)

     If the price per concert ticket is $40 and all the seats are sold for the show,

     the total sales will be $112,640 for a one-night show.

  • 3 months ago

    The crucial step is finding the

    number of seats. We have an

    arithmetic sequence, first term (a) = 26,

    common difference (d) = 4, number of

    terms (n) = 32.

    The sum is (n/2)(2a + [ n-1 ]d)

    = 16(52 + 31*4) = 2816 seats.

    At $40 a seat . . . I leave the math

    to you.

  • No exponents needed at all

    r[1] = 26 = 22 + 4 = 22 + 4 * 1

    r[2] = 30 = 26 + 4 = 22 + 8 = 22 + 4 * 2

    r[3] = 30 = 22 + 12 = 22 + 4 * 3

    r[n] = 22 + 4 * n

    r[32] = 22 + 4 * 32 = 22 + 128 = 150

    S[n] = (n/2) * (r[1] + r[n])

    S[32] = (32/2) * (26 + 150)

    S[32] = 16 * 176

    S[32] = 16 * (175 + 1)

    S[32] = 16 * 175 + 16

    S[32] = 4 * 4 * 700 + 16

    S[32] = 4 * 2800 + 16

    S[32] = 11200 + 16

    S[32] = 11216

    That's a pretty big concert hall.

    40 * 11216 =>

    448640

    Not bad for a single night.  Must be one hell of a show.

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