I. The geometric mean of 2,4,16 and 32 is a
II. The strength of 7 colleges in a city are 385, 1748, 1343, 1935, 786, 2874 , 2108. Then the median strength is b.
II. The algebric sum of the deviations of 20 observations measured from 30 is 2. The mean of these observations is c.

Options

A. `a lt b lt c`
B. `b lt c lt a`
C. `c lt a lt b`
D. `a lt c lt b`

Answer

Let's solve the given problems step by step.

### Part I: Finding the Geometric Mean (a)

The geometric mean (GM) of a set of numbers is calculated using the formula:

\[ GM = (x_1 \times x_2 \times \ldots \times x_n)^{\frac{1}{n}} \]

For the numbers 2, 4, 16, and 32:

1. **Identify the numbers**: \( x_1 = 2, x_2 = 4, x_3 = 16, x_4 = 32 \)
2. **Calculate the product**:
\[ Product = 2 \times 4 \times 16 \times 32 \]
- \( 2 \times 4 = 8 \)
- \( 8 \times 16 = 128 \)
- \( 128 \times 32 = 4096 \)

3. **Count the numbers**: \( n = 4 \)

4. **Calculate the geometric mean**:
\[ GM = (4096)^{\frac{1}{4}} = 8 \]

Thus, the geometric mean \( a = 8 \).

### Part II: Finding the Median Strength (b)

To find the median of the given strengths of 7 colleges: 385, 1748, 1343, 1935, 786, 2874, 2108.

1. **Arrange the data in ascending order**:
- Ordered data: 385, 786, 1343, 1748, 1935, 2108, 2874

2. **Find the median**:
- Since there are 7 observations (an odd number), the median is the middle value.
- Median position: \( \frac{n + 1}{2} = \frac{7 + 1}{2} = 4 \)
- The 4th value in the ordered list is 1748.

Thus, the median strength \( b = 1748 \).

### Part III: Finding the Mean of Observations (c)

We know that the algebraic sum of the deviations of 20 observations from 30 is 2. This means:

\[ \sum (x_i - 30) = 2 \]

1. **Rearranging the equation**:
\[ \sum x_i - 20 \times 30 = 2 \]
\[ \sum x_i - 600 = 2 \]
\[ \sum x_i = 602 \]

2. **Calculate the mean**:
\[ Mean = \frac{\sum x_i}{n} = \frac{602}{20} = 30.1 \]

Thus, the mean of these observations \( c = 30.1 \).

### Final Answers:
- \( a = 8 \)
- \( b = 1748 \)
- \( c = 30.1 \)

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