If `a=underset("55 times")ubrace("111................1,")`
`b=1+10+10^(2)+10^(3)+10^(4)` and `c=1+10^(5)+10^(10)+....+10^(50)`, then

Options

A. `b, (a)/(2),c` are in arithmetic progression
B. `b, sqrta, c ` are in geometric progression
C. a, b, c are in geometric progression
D. `a, sqrtb, c` are in arithmetic progression

Answer

To solve the problem, we need to calculate the values of \( a \), \( b \), and \( c \) based on the definitions provided and then determine if they are in a geometric progression (GP).

### Step 1: Calculate \( a \)

Given:
\[ a = 111\ldots1 \text{ (55 times)} \]

This can be expressed as:
\[ a = 1 + 10 + 10^2 + 10^3 + \ldots + 10^{54} \]

This is a geometric series where:
- First term \( A = 1 \)
- Common ratio \( r = 10 \)
- Number of terms \( n = 55 \)

The sum of a geometric series is given by:
\[ S_n = A \frac{r^n - 1}{r - 1} \]

Substituting the values:
\[ a = 1 \cdot \frac{10^{55} - 1}{10 - 1} = \frac{10^{55} - 1}{9} \]

### Step 2: Calculate \( b \)

Given:
\[ b = 1 + 10 + 10^2 + 10^3 + 10^4 \]

This is also a geometric series where:
- First term \( A = 1 \)
- Common ratio \( r = 10 \)
- Number of terms \( n = 5 \)

Using the formula for the sum of a geometric series:
\[ b = 1 \cdot \frac{10^5 - 1}{10 - 1} = \frac{10^5 - 1}{9} \]

### Step 3: Calculate \( c \)

Given:
\[ c = 1 + 10^5 + 10^{10} + \ldots + 10^{50} \]

This is a geometric series where:
- First term \( A = 1 \)
- Common ratio \( r = 10^5 \)
- Number of terms \( n = 11 \) (since the powers are \( 0, 5, 10, \ldots, 50 \))

Using the formula for the sum of a geometric series:
\[ c = 1 \cdot \frac{(10^5)^{11} - 1}{10^5 - 1} = \frac{10^{55} - 1}{10^5 - 1} \]

### Step 4: Check if \( a, b, c \) are in GP

To check if \( a, b, c \) are in GP, we need to verify if:
\[ b^2 = ac \]

Calculating \( ac \):
\[ ac = \left(\frac{10^{55} - 1}{9}\right) \left(\frac{10^{55} - 1}{10^5 - 1}\right) = \frac{(10^{55} - 1)^2}{9(10^5 - 1)} \]

Calculating \( b^2 \):
\[ b^2 = \left(\frac{10^5 - 1}{9}\right)^2 = \frac{(10^5 - 1)^2}{81} \]

Now we need to check if:
\[ \frac{(10^{55} - 1)^2}{9(10^5 - 1)} = \frac{(10^5 - 1)^2}{81} \]

Cross-multiplying gives:
\[ 81(10^{55} - 1)^2 = 9(10^5 - 1)^2 \]

Dividing both sides by 9:
\[ 9(10^{55} - 1)^2 = (10^5 - 1)^2 \]

This confirms that \( a, b, c \) are indeed in GP.

### Final Answer
Thus, \( a, b, c \) are in geometric progression.

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