Question

Let u, v, w be real numbers in geometric progression such that u>v>w. suppose u40 = vn = w60. find the value of n

Answer 1

n=48
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Answer 2

Answer:

The value of n in Geometric Progression is 48

Step-by-step explanation:

Step 1:

Given Data:

u, v, w be real number in geometric progression:

Let $u^{40}=y^{n}=w^{60}=K So,\left\{u=K^{1 / 40}, v=K^{1 / n}, w=K^{1 / 60}\right\}$...(1)

Step 2:

Given u, v, w are in Geometric Progression

To calculate the value of n:

⇒v2=uw

Step 3:

So, $\mathrm{K}^{2 / \mathrm{n}}=\mathrm{K}^{1 / 40} \times \mathrm{K}^{1 / 60}$ [using (1)]

$\Rightarrow \mathrm{K}^{2 / \mathfrak{n}}=\mathrm{K}^{1 / 40+1 / 60}$

 [Using $a^{m} \times a^{n}=a^{m+n}$]

Step 4:

$\Rightarrow \mathrm{K}^{2 / \mathrm{n}}=\mathrm{K}^{1 / 24}$

Now comparing powers as bases are same

⇒2/n=1/24

⇒ n=48