Question

What is the 32nd term of the arithmetic sequence where a1 = 15 and a13= −57?
−171

−165

−159

−151

Answer 1

@jdoe0001 i know you can figure this out xD

Answer 2

Do you know this ? @Peaches15

Answer 3

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

Answer 4

@LexiLuvv2431 do I get a medal

Answer 5

yess thank youu