如何做?關於等差數列的一條問題.

Let T(1) = a
T(2) = a + d
.....
T(n) = a + (n-1)d

T(4) = a + 3d
T(8) = a + 7d
T(12) = a + 11d
T(16) = a + 15d

4a + 36d = 224
a + 9d = 56

T(1) + T(2) + ... + T(19)
= [a + a + 18d]*19/2
=19 (a + 9d)
=19*56
=1064