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Grade 11Algebra

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

Profile image of Simran Bhatia
12 Years agoGrade 11
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1 Answer

Profile image of Aditi Chauhan
12 Years ago
Hello Student,
Please find the answer to your question
Let a – 3d, a – d, a + d and a + 3d be any four consecutive terms of an A. P. with common difference 2d ∵ Terms of A. P. are integers, 2d is also an integer.
Hence p = (2d)4 + (a – 3d) (a – d) (a + d) (a + 3d)
= 16d4 + (a2 – 9d2 ) (a2 – d2) = a2 – 5d2)2
Now, a2 – 5d2 = a2 – 9d2 + 4d2
= (a – 3 d) (a + 3 d) + (2d)2 = some integer
Thus p = square of an integer.

Thanks
Aditi Chauhan
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