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Find the number of triplets of integers in arithmetic progression, the sum of whose squares is 1994.

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Does not exit

answer is D.

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Detailed Solution

Suppose that the triplets =

(x - a), x, (x + a)

Sum of square of these integers is 1994. Thus,

(x - a)^{2} + x^{2} + (x + a)^{2} = 1994 x^{2} - 2 xa + a^{2} + x^{2} + x^{2} + 2 xv + a^{2} = 1994 3 x^{2} + 2 a^{2} = 1994 2 a^{2} = 1994 - 3 x^{2} a^{2} = 997 - \frac{3 x^{2}}{2} = integer

Here,

x^{2}

should be integer which can cancle 3/2 term. Thus,

x = 2 n

(even term)

x^{2} = 4 n^{2}

Thus,

a^{2} = 997 - \frac{3 (4 n^{2})}{2}

a^{2} = 997 - 6 n^{2}

997 - 6 n^{2} \geq 0

(for

a^{2} \geq 0

)

997 \geq 6 n^{2}

n^{2} \leq 166.16

n \leq 12.89

So, n can be 0, 1, 2…..12. Thus, at n=0,

a^{2} = 997 - 6 {(0)}^{2} = 997

= not perfect square
At n= 1,

a^{2} = 997 - 6 {(1)}^{2} = 991

= not perfect square
At n= 2,

a^{2} = 997 - 6 {(2)}^{2} = 973

= not perfect square
Now, for any of the value of n (0,1,2,…12) no value of

a^{2}

is an perfect square to make a integer.

So, triplet does not exist for the given conditions.
Correct option is 4.

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