Let 
, 
, . . . , 
be an arithmetic sequence. If 
and 
, then find .
Subtracting the second sum from the first, we have
(a2 - a1) + (a4 - a3) + (a6 - a5) + (a8 - a7) + (a10 - a9) = -2 or
d + d + d + d + d = -2 or
5d = -2 → d = -2/5
Also, the sum of the arithmetic series is given by
S = (10/2)(a1 + a10) = 5(a1 + a10) = 32 divide both sides by 5
So
a1 + a10 = 32/5 → a10 = 32/5 - a1 ( 1)
And the last term is given by
a10 = a1 + d(9) (2)
Therefore, setting (1) = (2), we have
32/5 - a1 = a1 + d(9) ... solve for a1
(32/5 -2a1) = d(9)
32/5 -2a1 = (-2/5)(9) = -18/5
32/5 + 18/5 = 2a1
2a1 = 50/5
a1 = 5
And a10= 32/5 - a1 = 32/5 - 5 = 1.4
Proof
a1 + a3 + a5 + a7 + a9 = 5 + 4.2 + 3.4 + 2.6 + 1.8 = 17
And since each of the other 5 terms is (2/5) less than it's preceding term.....
[17 - 5(2/5)] = [17 - 2] = 15 = the sum of the terms with even subscripts
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