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What is the arithmetic mean of first 16 natural numbers with weights being the number itself?
- a)17/2
- b)33/2
- c)11
- d)187/2
Correct answer is option 'C'. Can you explain this answer?
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Solution:
Mean with Weights:
Arithmetic Mean with weights is defined as the sum of the product of the elements and their respective weights divided by the sum of the weights.
Mean with weights = $\frac{\sum_{i=1}^{n} w_ix_i}{\sum_{i=1}^{n} w_i}$
Given:
Here, the weights are the numbers itself, i.e., the first natural number has a weight of 1, the second natural number has a weight of 2, and so on. We need to find the arithmetic mean of the first 16 natural numbers with weights being the number itself.
Therefore, x1 = 1, x2 = 2, ..., x16 = 16 and w1 = 1, w2 = 2, ..., w16 = 16.
Calculation:
We can calculate the mean with weights using the formula mentioned above.
Mean with weights = $\frac{\sum_{i=1}^{16} w_ix_i}{\sum_{i=1}^{16} w_i}$
= $\frac{1\times1 + 2\times2 + ... + 16\times16}{1 + 2 + ... + 16}$
= $\frac{1^2 + 2^2 + ... + 16^2}{\frac{16\times17}{2}}$
= $\frac{\frac{16\times17\times33}{6}}{\frac{16\times17}{2}}$
= $\frac{33}{2}$
Therefore, the arithmetic mean of the first 16 natural numbers with weights being the number itself is 33/2.
Hence, option (c) is the correct answer.
Mean with Weights:
Arithmetic Mean with weights is defined as the sum of the product of the elements and their respective weights divided by the sum of the weights.
Mean with weights = $\frac{\sum_{i=1}^{n} w_ix_i}{\sum_{i=1}^{n} w_i}$
Given:
Here, the weights are the numbers itself, i.e., the first natural number has a weight of 1, the second natural number has a weight of 2, and so on. We need to find the arithmetic mean of the first 16 natural numbers with weights being the number itself.
Therefore, x1 = 1, x2 = 2, ..., x16 = 16 and w1 = 1, w2 = 2, ..., w16 = 16.
Calculation:
We can calculate the mean with weights using the formula mentioned above.
Mean with weights = $\frac{\sum_{i=1}^{16} w_ix_i}{\sum_{i=1}^{16} w_i}$
= $\frac{1\times1 + 2\times2 + ... + 16\times16}{1 + 2 + ... + 16}$
= $\frac{1^2 + 2^2 + ... + 16^2}{\frac{16\times17}{2}}$
= $\frac{\frac{16\times17\times33}{6}}{\frac{16\times17}{2}}$
= $\frac{33}{2}$
Therefore, the arithmetic mean of the first 16 natural numbers with weights being the number itself is 33/2.
Hence, option (c) is the correct answer.
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What is the arithmetic mean of first 16 natural numbers with weights being the number itself?a)17/2b)33/2c)11d)187/2Correct answer is option 'C'. Can you explain this answer?
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