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The sum of the nth term of an arithmetic progression is given by: S_{n} = n/2 (2a + (n - 1)d)

where a is the first term, d is the common difference and n is the number of term.

Thus, S_{18} = 18/2 (2a + (18 - 1)d)

S_{18} = 9(2a + 17d) ----------- eqn(1)

Also, the nth term of an arithmetic progression, T_{n} = a + (n - 1)d

where a is the first term, d is the common difference and n is the number of term.

T_{n} = 3 - 2n

Thus, a + (n - 1)d = 3 - 2n

This can also be written as:

a + (n - 1)d = 3 - 2(n - 1 + 1) [because - 1 + 1 = 0]

Next, take + 1 out of the bracket on the right hand side by multiplying +1 by -2 (because -2 affects everything inside the bracket)

Hence, a + (n - 1)d = 3 - 2(n - 1) - 2

a + (n - 1)d = 3 - 2 - 2(n - 1)

a + (n - 1)d = 1 - 2(n - 1)

comparing each values on both sides,

a = 1

d = -2

Substitute a = 1 and d = -2 into eqn(1)

S_{18} = 9[2(1) + 17(-2)]

= 9(2 -34)

= 9(-32)

= -288

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