The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?
If the first term is x, second is x + y, and so on,
The fifth term would be \(x + 4y = 9\) and 32nd would be \(x + 31y = -84\). Subtract and solve for x and y,
\(x + 4y = 9 \\ x+ 31y = -84 \\ 27y = -93 \\ y = -\frac{31}{9} \\ x + 4(-\frac{31}{9})=9 \\ x - 13 \frac{7}{9} = 9 \\ x = 22 \frac{7}{9}\)
Equation for 23nd term would be \(x + 22y\).
\(22 \frac{7}{9} + 4(-\frac{31}{9}) \\ 22 \frac{7}{9} + - 13\frac{7}{9} = 9\)
So the 23rd term is 9.
from 9 to -84 is 27 more of the common diference , d , of the sequence
(-84 - 9 )/27 = -93/27 = -31 / 9
from 5 th tem to 23 is addition of 18 d's
9 + 18 ( -31/9) = -53