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The sum of the squares of three distinct real numbers, which are in G. P., is S 2. If their sum is aS, show that a 2 ∈ (1/3, 1 ) ∪ (1, 3).

The sum of the squares of three distinct real numbers, which are in G. P., is S2. If their sum is aS, show that
a2 ∈ (1/3, 1 ) ∪ (1, 3).

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
Hello Student,
Please find the answer to your question
Let the three distinct real numbers be α/r, α, αr.
Since sum of squares of three numbers be S2
∴ α2/r2 + α2 + α2 r2 = S2
Or α 2(1 + r2 + r4)/r2 = S2 ……… (1)
Sum of numbers is aS
∴ α (1 + r + r2)/r = aS ……… (2)
Dividing eq. (1) by the square (2), we get
α2 (1 + r2 + r4)/r2 * r2/ α2 ( 1 + r + r2)2 = S2/a2S2
(1 + 2r2 + r4) – r2 / (1 + r + r2) 2 = 1/a2, ( 1 + r + r2) (1 – r + r2 ) /(1 + r + r2)2 = 1/a2
1 – r + r2/1 + r + r2 = 1/a2 ⇒ a2r2 – a2r + a2 = 1 + r + r2
⇒ (a2 – 1) r2 – (a2 + 1) r + ( a2 – 1) = 0
⇒ r2 + ( 1 + a2/1 – a2)r + 1 = 0 …… (3)
For real valuew of r, D ≥ 0
⇒ (1 + a2/1 – a2)2 – 4 ≥ 0
⇒ 1 + 2a2 + a2 – 4 + 8a2 – 42 ≥ 0
⇒ 3a4 – 10a2 + 3 ≤ 0 ⇒ (3a2 – 1) (a2 – 3) ≤ 0
⇒ (a2 – 1/3) ( a2 – 3) ≤ 0
Clearly the above inequality holds for
1/3 ≤ a2 ≤ 3
But from eq. (3), a ≠ 1
∴ a2 ∈(1/3, 1) ∪ (1 , 3).

Thanks
Aditi Chauhan
askIITians Faculty

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