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The nth term of an arithmetic progression is given by: T_{n} = a + (n - 1)d ------ eqn(1)

where a is the first term, d is the common difference, n is the number of term and T_{n} is the nth term.

for the 9th term,

T_{9} = a + (9 - 1)d

= a + 8d

Also, T_{9} = 499

Thus, a + 8d = 499 ---------- eqn(2)

for the 499th term,

T_{499} = a + (499 - 1)d

= a + 498d

Also, T_{499} = 9

Thus, a + 498d = 9 ---------- eqn(3)

eqn(2) and eqn(3) are simultaneous equations and can be simplified by elimination method.

Hence, subtract eqn(3) from eqn(2)

(a - a) + (8 - 498)d = (499 - 9)

-490d = 490

divide both sides by -490

d = -1

substitute d = -1 into eqn(2)

a + 8(-1) = 499

a - 8 = 499

Add 8 to both sides

a = 499 + 8

a = 507

Also, from eqn(1), to get the term which is equal to zero, T_{n} = 0.

That is, 0 = a + (n - 1)d

0 = 507 + (n - 1)(-1)

0 = 507 - n + 1

0 = 508 - n

n = 508

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