Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

The sum of the squares of three distinct real numbers, which are in G. P., is S2. If their sum is aS, show thata2 ∈ (1/3, 1 ) ∪ (1, 3).

6 years ago
Hello Student,
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