How to Find an AP from Two Given Terms

How can two terms determine an AP?

If the pth term is P and the qth term is Q, with p and q different, the common difference is d = (Q − P)/(q − p). Then a = P − (p − 1)d. Subtracting the two term equations removes the first term and leaves the number of equal gaps.

Example: fifth term 18, twelfth term 46

Write a + 4d = 18 and a + 11d = 46. Subtract the first equation from the second: 7d = 28. Therefore d = 4. Substitute into a + 4d = 18 to get a + 16 = 18, hence a = 2.

The AP is 2, 6, 10, 14, 18, … . Its general term is 2 + 4(n − 1), or 4n − 2. Check the twelfth term separately: 4 × 12 − 2 = 46. You can now find any requested term without listing the whole progression; the twentieth term is 78.

Why not divide by the larger position?

Between the fifth and twelfth positions there are seven gaps, not twelve. Dividing the value difference 28 by twelve would underestimate d. Draw two labelled positions if the subtraction is confusing. Keep the order consistent: reversing both differences is fine, but reversing only one changes the sign incorrectly.

Decreasing and fractional cases

The method also works when d is negative or fractional. If the third term is 14 and seventh term is 2, then d = (2 − 14)/(7 − 3) = −3 and a = 20. The result is a valid decreasing AP. Different values assigned to the same position would instead be inconsistent data.

Practice

1. Second term 9 and sixth term 25: d = 4, a = 5. Checks: 5 + 4 = 9 and 5 + 20 = 25.
2. Fourth term 8 and eighth term 0: d = −2, a = 14. Checks: 14 − 6 = 8 and 14 − 14 = 0.

Two distinct positions determine a unique arithmetic progression when its terms are real numbers.

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