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The value of the nth term of an Arithmetic Progression is given by a + (n - 1)d, where a is the first term and d is the common difference.

t5 = a + (5 - 1)d = a + 4d = -1/2

t9 = a + (9 - 1)d = a + 8d = -1/128

Subtracting the first equation in bold from the second equation gives :-

4d = -1/128 - (-1/2) = -1/128 -(-64/128) = 63/128 therefore d = 63/(128x4) = 63/512

Substituting for d in the first equation a + (4x63)/512 = -1/2 : a = -1/2 - 252/512 = -508/512.

t3 = -508/512 + (3 - 1)63/512 = -508/512 + 126/512 = - 382/512 = -191/256

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Q: How do you find the third term of an arithmetic sequence if t to the fifth equals negative one half and t to the ninth equals negative 1 over 128?
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