Problem 6. (Arithmetic Sequence)

If the numbers a, b and c are terms of an arithmetic sequence, then show that the numbers

A = a² + ab + b²
B = a² + ac + c²
C = b² + bc + c²

are also terms of an arithmetic sequence.

Solution

If a, b, and c are consecutive terms of an arithmetic sequence, then we have

b = a + h
c = a + 2h

and so

A = a² + a(a + h) + (a + h)² = 3 a² + 3ah + h²
B = a² + a(a + 2h) + (a + 2h)² = 3 a² + 6ah + 4h²
C = (a + h)² + (a + h)(a + 2h) + (a + 2h)² = 3 a² + 2ah + 7h²

Hence, by direct computation

B - A = 3 h² + 3ah
C - B = 3 h² + 3ah

Hence, the differences are the same.

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Last modified on Wednesday, March 17, 1999