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# HELP ASAP!!

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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
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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
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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
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ohhhhhhh oops lol

Awesomeguy  Jul 8, 2021