The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

The n'th term is given by an = a1 + (n-1)d

Here:   

9 = a1 + 4*d

-84 = a1 + 31*d

These allow you to find a1 and d, so you can now find: 

a23 = a1 + 22*d

We can solve these two equations to find a₁ [the first term]  and d [ the common difference between terms]

9 = a₁ + d(5 -1)  →           9 =  a₁ + 4d      (1)

-84 = a₁ + d(32 -1)   →   -84 = a₁ + 31d    (2)

Multiply (1)  by -1 and add it to (2)....this gives

-93  = 27d      divide both sides by  27

-93/27   = d  = -31/9

Using (1), we can find a₁ thusly

9 = a₁ + 4( -31/9)

81/9 = a₁ - 124/9       [ add 124/9 to both sides ]

205/9 = a₁ 

So........the 23rd term, a₂₃,  is given by

a₂₃  = 205/9 + (-31`/9)(23 -1)

a₂₃ = 205/9  + (-31/9)(22)   = -53

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

The difference between 32nd term and 5th term is:

32 - 5 =27

The difference between the values of the two terms is:

-84 - 9 =-93, therefore their common difference is:

-93 / 27 =- 31/9, therefore the 1st term is:

[4 x 31/9] + 9 =22 7/9 =205/9, therefore:

23rd term=an = a1 + (n-1)d =205/9 + (23 -1)*-31/9=-53