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What is the 32nd term of the arithmetic sequence if the 2nd and 6th terms are 9 and 11, respectively?​

Sagot :

Arithmetic Sequence Problem
Given:
a2 = 9
a6 = 11

Find value of a1

an = a1 + (n - 1)d
a2 = a1 + (2 - 1)d
a2 = a1 + d
a1 = a2 - d

Substitute the value of a1

a6 = a1 + (6 - 1)d

a6 = a1 + 5d
a6 = (a2 - d) + 5d
a6 = a2 + 4d

11 = 9 + 4d
4d = 2
d = 1/2

Substitute the value of d

a1 = 9 - 1/2 = 17/2 = 8.5

Find 32nd Term
a32 = 8.5 + (32 - 1)1/2
a32 = 8.5 + (31)0.5
a32 = 8.5 +15.5
a32 = 24

The 32nd term is 24