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A arithmetic sequence follows the form An=A1+(n-1)d

If the third term is equal to 5, then 5=A1+2d.

If the sum of the first 10 terms is 125, then 125=Sum(A1 +[n-1]d) from n=1 to n=10.

Therefore 125=10A1+45d (45=1+2+3+4+5+6+7+8+9)

Solve linear equations:

10A1+45d=125

+ (A1+2d= 5)(-10)

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10A1+45d=125

-10A1-20d=-50

_____________

25d=75

Therefore d=3 and substituting into the first equation A1=-1

First term is -1 and the difference is 3.

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