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Find the ninth term if the a sub 1 is 10 and the common diff. Is negative ½. With solution thanks guys.

Sagot :

Okay, I am assuming that [tex] a_{1} [/tex] is equal to [tex] t_{1} [/tex]

Formula:
               [tex] a_{n} = a_{1} + (n -1) d[/tex]
Substitute:
                n for 9
               [tex] a_{1} [/tex] for 10
                d for [tex] -\frac{1}{2} [/tex]
Solution:
               [tex] a_{9} [/tex] = 10 + ( 9 -1)[tex]- \frac{1}{2} [/tex]
                 [tex] a_{9} [/tex] = 10 + 8 ([tex]- \frac{1}{2} [/tex])
                 [tex] a_{9} [/tex] = 10 + -4
                  [tex] a_{9} [/tex] = 10 - 4
                   [tex] a_{9} [/tex] = 6
Check:
  First 9 terms:           10, 9.5, 9, 8.5, 8, 7.5, 7, 6.5, 6

Answer:
               The 9th term is 6.

[tex]a_n=a_1+(n-1)d \\ a_9=10+(8)(-1/2)=10-4=6[/tex]