algebra
posted by Anonymous .
the third term of an arithmetic sequence is 14 and the ninth term is 1. Find the first four terms of the sequence

ninth 1, third 14
it goes now by 15 in six steps; step 15/6 or 2.5
third through tenth terms
14, 11.5, 9, 6.5, 4, 1.5, 1, 3.5,...
check that 
third term is 14 and ninth term is 1
14  2.5 = 11.5 > 4th term
11.5  2.5 = 9 > 5th term
9  2.5 = 6.5 > 6th term
6.5  2.5 = 4 > 7th term
4  2.5 = 1.5 > 8th term
1.5  2.5 = 1 > 9th term
The terms are decreasing by 2.5 > (2.5)
To go from one term to the next, subtract 2.5.
The common difference is 2.5
d = 2.5
Arithmetic Sequence Formula:
Tn = Tn + d(n1)
a = 1st term
n = nth term
d = common difference
We don't know the first term yet!
Tn = T1 + d(n1)
Substitute 3 for n, and 2.5 for d
T3 = T1 + 2.5(31)
Now substitute 14 for T3
14 = T1 + 2.5(2)
14 = T1 5
14 + 5 = T1 5 +5
19 = T1
T1 = 19
T2 = 16.5
T3 = 14
T4 = 11.5
Now to find a term:
Tn = T1 + d(n1)
T2 = 19 + 2.5(21) T3 = 19 + 2.5(31)
T2 = 19 + 2.5(1) T3 = 19 + 2.5(2)
T2 = 19 + 2.5 T3 = 19 + 5
T2 = 16.5 T3 = 14
__________________________
You can also use this formula:
Tn = dn + 21.5
T1 = 2.5(1) + 21.5
T1 = 19
T4 = 2.5(4) + 21.5
T4 = 11.5
T9 = 2.5(9) + 21.5
T9 = 1
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