Test: Relation Between AM And GM - JEE MCQ

Geometric mean of two numbers a and b is √ab
As two numbers are 14 and 56
Geometric mean is √14×56
= ± √2×7×2×2×2×7
= ±(2×2×7)
= ±28

nth G.M. between a and b is
G_n = arⁿ
Where common ratio is r = (b/a)^(1/(n+1))
​So, to insert 3 geometric means between 1/4 and 64
 r = (b/a)^(1/(n+1))
r = (64/(¼)^(1/(3+1))
r = (256)^1/4
r  = (4)^(4)1/4
r = 4
G_n = 1 * (4)ⁿ
G₀ = 1 * (4)⁰ = 1
G₁ = 1 * (4)¹ = 4
G₂ = 1 * (4)² = 16
The terms are 1, 4, 16

nth G.M. between a and b is
G_n = arⁿ
Where common ratio is r = (b/a)^(1/(n+1))
​So, to insert 3 geometric means between 1 and 256
 r = (b/a)^(1/(n+1))
r = (256/(1)^(1/(3+1))
r = (256)^1/4
r  = (4)^(4)1/4
r = 4
G_n = 1 * (4)ⁿ
G₁ = 1 * (4)¹ = 4
G₂ = 1 * (4)² = 16
G₃ = 1 * (4)³ = 64
The terms are  4, 16, 64

A=a+b/2. ,G=√ab
A-G=(a+b/2)-√ab
=(a+b-2√ab)/2
=(√a-√b)²/2 is greater than or equal to zero
A-G is greater than or equal to zero so
So A is greater than equal to zero

The given series is a geometric series in which a=2,r=3,l=4374.
Therefore,
Required sum = (lr−a)/(r−1)
​= (4374×3−2)/(3−1)
​= 6560

a, b, c are in AP
Let d be the common difference.
b = a + d,
c = a + 2d 

There are in GP with common ratio: kd
 

Let the numbers be x, y 

Then arithmatic mean = (x+y)/2 =34 

→x+y =68 

Also geometric mean =√(xy)=16 

oy xy=16^2=256 

Hence 

x(68−x)=256 

or x^2−68x+256=0 

(x−64)(x−4)=0 

Hence x=64 or x=4 

and y=4 or 64 

Larger number is 64

b² = ac
a = 3/2, c = 27/2
⇒ b² = 3/2 * 27/2
⇒ b = (√81/4)
⇒ b = 9/2

a(rⁿ-1)/(r-1) = 5460
=> 4(4^n-1)/4-1 = 5460
=> 4ⁿ = 4096
=> 2²ⁿ = 2¹²
=> n = 6