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

 Jul 8, 2021
 #1
avatar+238 
0

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.

 Jul 8, 2021
 #2
avatar+36915 
+1

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

 Jul 8, 2021
 #3
avatar+238 
0

ohhhhhhh oops lol

Awesomeguy  Jul 8, 2021

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