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

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

–171

–165

–159

–151

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 + (n-1)d

In this case, a = 15 and a = -57

a = a + 12d

-57 = 15 + 12d

12d = -72

d = -6

So a = a + 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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