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The product of n positive numbers is unity, then their sum is :
- a)a positive integer
- b)divisible by n
- c)equal to n+1/n
- d)never less than n
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The product of n positive numbers is unity, then their sum is :a)a pos...
Let a1, a2, a3…an be the positive numbers.
Product => a1a2..an = 1
We know that AM ≥ GM
(a1+a2+…an)/n ≥ (a1a2..an)1/n
(a1+a2+…an)/n ≥ 1
(a1+a2+…an) ≥ n
So the sum is never less than n.
Most Upvoted Answer
The product of n positive numbers is unity, then their sum is :a)a pos...
Let a1, a2, a3…an be the positive numbers.
Product => a1a2..an = 1
We know that AM ≥ GM
(a1+a2+…an)/n ≥ (a1a2..an)1/n
(a1+a2+…an)/n ≥ 1
(a1+a2+…an) ≥ n
So the sum is never less than n.
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Community Answer
The product of n positive numbers is unity, then their sum is :a)a pos...
Explanation:
Given: The product of n positive numbers is unity.
Let's assume we have n positive numbers: a1, a2, ..., an
Product of n positive numbers:
The product of these n positive numbers is given as:
a1 * a2 * ... * an = 1
This means that the product of these numbers is equal to 1.
Sum of n positive numbers:
Now, we need to find the sum of these n positive numbers.
Let's assume the sum of these numbers is S.
Hence, we can write:
a1 + a2 + ... + an = S
Proof:
We need to prove that the sum of n positive numbers is never less than n.
To prove this, let's assume that the sum of these n positive numbers is less than n.
Hence, S < />
Now, let's consider the product of these n positive numbers:
(a1 * a2 * ... * an) ≤ (a1 + a2 + ... + an)
Since the product of these numbers is equal to 1, we can write:
1 ≤ (a1 + a2 + ... + an)
This implies that the sum of these n positive numbers is always greater than or equal to 1.
But we had assumed that the sum of these numbers is less than n, which contradicts our assumption.
Hence, our assumption is false.
Therefore, the sum of n positive numbers is never less than n.
Conclusion:
Hence, the correct answer is option D: The sum of n positive numbers is never less than n.
Given: The product of n positive numbers is unity.
Let's assume we have n positive numbers: a1, a2, ..., an
Product of n positive numbers:
The product of these n positive numbers is given as:
a1 * a2 * ... * an = 1
This means that the product of these numbers is equal to 1.
Sum of n positive numbers:
Now, we need to find the sum of these n positive numbers.
Let's assume the sum of these numbers is S.
Hence, we can write:
a1 + a2 + ... + an = S
Proof:
We need to prove that the sum of n positive numbers is never less than n.
To prove this, let's assume that the sum of these n positive numbers is less than n.
Hence, S < />
Now, let's consider the product of these n positive numbers:
(a1 * a2 * ... * an) ≤ (a1 + a2 + ... + an)
Since the product of these numbers is equal to 1, we can write:
1 ≤ (a1 + a2 + ... + an)
This implies that the sum of these n positive numbers is always greater than or equal to 1.
But we had assumed that the sum of these numbers is less than n, which contradicts our assumption.
Hence, our assumption is false.
Therefore, the sum of n positive numbers is never less than n.
Conclusion:
Hence, the correct answer is option D: The sum of n positive numbers is never less than n.
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The product of n positive numbers is unity, then their sum is :a)a positive integerb)divisible by nc)equal to n+1/nd)never less than nCorrect answer is option 'D'. Can you explain this answer?
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