Sequences And Series Ques 47
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 resulting sum is the square of an integer
(2000, 4M)
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Solution:
- Let four consecutive terms of the $AP$ are $a-3 d, a-d$, $a+d, a+3 d$, which are integers.
Again, required product
$ P=(a-3 d)(a-d)(a+d)(a+3 d)+(2 d)^{4} $
[by given condition]
$ \begin{aligned} & =\left(a^{2}-9 d^{2}\right)\left(a^{2}-d^{2}\right)+16 d^{4} \\ & =a^{4}-10 a^{2} d^{2}+9 d^{4}+16 d^{4}=\left(a^{2}-5 d^{2}\right)^{2} \end{aligned} $
Now, $a^{2}-5 d^{2}=a^{2}-9 d^{2}+4 d^{2}$
$ \begin{aligned} & =(a-3 d)(a+3 d)+(2 d)^{2} \\ & =I \cdot I+I^{2} \quad \text { [given] }\\ & =I^{2}+I^{2}=I^{2} \\ & =I \end{aligned} $
Therefore, $P=(I)^{2}=$ Integer
BETA
