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

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