Question

If m is the A.M. of two distinct real numbers l and n (l , n > 1) and G1, G2 and G3 are three geometric means between and n, then G1 4 + G2 4 + G3 4equals​​​​

4l2mn

4lm2n

4lmn2

4l2m2n​2

Answer 1

m is the Arithmetic mean of l and n
So, m = (l + n)/2 ⇒2m = l + n ----+(1)

Again,l , G₁ , G₂ , G₃ and n are in GP .
So, we can assume common ratio r
G₁ = lr , G₂ = lr² , G₃ = lr³ and n = lr⁴
n/l = r⁴ -------(2)

Now, G₁⁴ + 2G₂⁴ + G₃⁴ = (lr)⁴ + 2(lr²)⁴ + (lr³)⁴ = l⁴r⁴[1 +2 r⁴ + r⁸ ]
= l⁴ × n/l × [1 + r⁴]² [ from equation (2) , r⁴= n/l]
= l³n × [1 + n/l]² = l³n/l²[l + n]² = nl (2m)² = 4m²nl

Answer 2

Step-by-step explanation:

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