What is the 32nd term of the arithmetic sequence where a1 = 15 and a13= –57?

NetherCraft 0

What is the 32nd term of the arithmetic sequence where a1 = 15 and a13= –57?

–171

–165

–159

–151

6 Answers

  • Answer is -171

    Okay for a general term, T(n) the formula for an a.p. is

    T(n) = a + (n-1)d where a = first term, n =nth term, d = common difference

    Since a1 = 15 I assume as in the first term T(1) = 15

    thus sub it into the general formula,

    T(1) = a + (1-1)d = 15

    a = 15 since 0 x d gives 0

    Okay now that you know a = 15,

    sub it into the 13th term T(13) = -57

    thus -57 = a + (13 – 1)d

    since you found out that a = 15 earlier

    you get -57 = 15 + 12d —-> d = -6

    so substitute that into the general formula to get the 32nd term

    T(32) = a + (n-1)d since a = 15 (found from the first part), n = 32 (32nd term) and d = -6 (calculated above)

    T(32) = 15 + 31(-6) = -171

    Source(s): H2 Math (:
  • The formula is a[n] = a[1] + (n-1)d

    In this case, a[1] = 15 and a[13] = -57

    a[13] = a[1] + 12d

    -57 = 15 + 12d

    12d = -72

    d = -6

    So a[32] = a[1] + 31d = 15 – 186 = -171

  • Diff. between 1st and thirteenth time period is -fifty seven – 15 = -seventy 2. So each and every time period is 6 decrease than previous. Going lower back yet another 19 words subtracts yet another 6 x 19 = 114 Subtract 114 from -fifty seven provides -171

  • a13=a1+12d

    => -57=15+12d

    => 12d=-72

    => d=-6

    a32=a1+31d

    =-57-31*6

    =-243

  • 12 steps for – 72 so each step -72/12=-6

    nth term is -6n +21

    32nd term -6×32 +21 = -171

  • a2=(15+d)

    a13=(15+12d)

    -57=15+12d

    -72=12d

    -72/12=d

    -6=d

    a32=15+(32-1)*-6

    a32=-171

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