In a sequence of 21 terms the first 11 terms are in A.P. with common difference 2. and the last 11 terms are in G.P. with common ratio 2. If the middle tem of the A.P. is equal to the middle term of the G.P., then the middle term of the entire sequence is

Options

A. `-(10)/(31)`
B. ` (10)/(31)`
C. `-(32)/(31)`
D. `(32)/(31)`

Answer

To solve the problem step by step, we will analyze the given information and derive the required values.

### Step 1: Understand the Sequence
We have a sequence of 21 terms where:
- The first 11 terms are in Arithmetic Progression (A.P.) with a common difference of 2.
- The last 11 terms are in Geometric Progression (G.P.) with a common ratio of 2.

### Step 2: Find the Middle Term of the A.P.
In an A.P. with \( n \) terms, the middle term is the \( \frac{n+1}{2} \)th term. For the first 11 terms:
- The middle term is the 6th term.

Using the formula for the \( n \)th term of an A.P.:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

For the 6th term:
\[ a_6 = a + (6-1) \cdot 2 = a + 10 \]

### Step 3: Find the Middle Term of the G.P.
In a G.P. with \( n \) terms, the middle term is also the \( \frac{n+1}{2} \)th term. For the last 11 terms:
- The middle term is the 6th term.

Using the formula for the \( n \)th term of a G.P.:
\[ a_n = p \cdot r^{n-1} \]
where \( p \) is the first term and \( r \) is the common ratio.

For the 6th term:
\[ g_6 = p \cdot 2^{6-1} = p \cdot 32 \]

### Step 4: Set the Middle Terms Equal
According to the problem, the middle term of the A.P. is equal to the middle term of the G.P.:
\[ a + 10 = 32p \tag{1} \]

### Step 5: Relate the 11th Term of A.P. to the 1st Term of G.P.
The 11th term of the A.P. is:
\[ a_{11} = a + (11-1) \cdot 2 = a + 20 \]
This term is equal to the first term of the G.P.:
\[ p = a + 20 \tag{2} \]

### Step 6: Solve the Equations
Now we have two equations:
1. \( a + 10 = 32p \)
2. \( p = a + 20 \)

Substituting equation (2) into equation (1):
\[ a + 10 = 32(a + 20) \]
Expanding this:
\[ a + 10 = 32a + 640 \]
Rearranging gives:
\[ 10 - 640 = 32a - a \]
\[ -630 = 31a \]
Thus:
\[ a = -\frac{630}{31} \]

Now substitute \( a \) back into equation (2) to find \( p \):
\[ p = -\frac{630}{31} + 20 \]
Converting 20 to a fraction with a denominator of 31:
\[ p = -\frac{630}{31} + \frac{620}{31} = -\frac{10}{31} \]

### Step 7: Find the Middle Term of the Entire Sequence
The middle term of the entire sequence is the 11th term of the A.P. or the 1st term of the G.P.:
\[ \text{Middle term} = p = -\frac{10}{31} \]

### Final Answer
The middle term of the entire sequence is:
\[ \boxed{-\frac{10}{31}} \]