If `a_1, a_2............` are in G.P, such that `a_1=1/8, a_1nea_2` and every term is equal to arithmetic mean of it's two successive terms then find `S_20-S_18`

Options

A. `-2^15`
B. `-2^18`
C. `-2^20`
D. `-2^21`

Answer

To solve the problem step by step, we need to analyze the given conditions and find the required value of \( S_{20} - S_{18} \).

### Step 1: Identify the terms of the G.P.
Given that \( a_1, a_2, \ldots \) are in a geometric progression (G.P.), we can express the terms as follows:
- Let \( a_1 = \frac{1}{8} \)
- Let the common ratio be \( r \). Then, the terms can be expressed as:
- \( a_1 = \frac{1}{8} \)
- \( a_2 = a_1 \cdot r = \frac{1}{8} r \)
- \( a_3 = a_1 \cdot r^2 = \frac{1}{8} r^2 \)
- and so on.

### Step 2: Use the condition of arithmetic mean
The problem states that every term is equal to the arithmetic mean of its two successive terms. This means:
\[ a_n = \frac{a_{n-1} + a_{n+1}}{2} \]
For \( n = 2 \):
\[ a_2 = \frac{a_1 + a_3}{2} \]
Substituting the values:
\[ \frac{1}{8} r = \frac{\frac{1}{8} + \frac{1}{8} r^2}{2} \]
Multiplying both sides by 2:
\[ \frac{1}{4} r = \frac{1}{8} + \frac{1}{8} r^2 \]
Multiplying through by 8 to eliminate the fractions:
\[ 2r = 1 + r^2 \]
Rearranging gives us:
\[ r^2 - 2r + 1 = 0 \]
This simplifies to:
\[ (r - 1)^2 = 0 \]
Thus, \( r = 1 \). However, since \( a_1 \neq a_2 \), we need to find another solution.

### Step 3: Solve the quadratic equation
We can also express the condition for \( n = 3 \):
\[ a_3 = \frac{a_2 + a_4}{2} \]
Substituting the values:
\[ \frac{1}{8} r^2 = \frac{\frac{1}{8} r + \frac{1}{8} r^3}{2} \]
Multiplying through by 2:
\[ \frac{1}{4} r^2 = \frac{1}{8} r + \frac{1}{8} r^3 \]
Multiplying through by 8:
\[ 2r^2 = r + r^3 \]
Rearranging gives us:
\[ r^3 - 2r^2 + r = 0 \]
Factoring out \( r \):
\[ r(r^2 - 2r + 1) = 0 \]
Thus, \( r = 0 \) or \( r = 1 \) or \( r = 2 \).

Since \( r \neq 1 \) (as \( a_1 \neq a_2 \)), we can take \( r = -2 \).

### Step 4: Find the sum \( S_n \)
The sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = a_1 \frac{1 - r^n}{1 - r} \]
Calculating \( S_{20} \) and \( S_{18} \):
\[ S_{20} = \frac{1}{8} \frac{1 - (-2)^{20}}{1 - (-2)} = \frac{1}{8} \frac{1 - 1048576}{3} \]
\[ S_{18} = \frac{1}{8} \frac{1 - (-2)^{18}}{1 - (-2)} = \frac{1}{8} \frac{1 - 262144}{3} \]

### Step 5: Calculate \( S_{20} - S_{18} \)
Now, we compute \( S_{20} - S_{18} \):
\[ S_{20} - S_{18} = \left(\frac{1}{8} \frac{1 - 1048576}{3}\right) - \left(\frac{1}{8} \frac{1 - 262144}{3}\right) \]
Factoring out common terms:
\[ = \frac{1}{8} \cdot \frac{1}{3} \left( (1 - 1048576) - (1 - 262144) \right) \]
\[ = \frac{1}{24} \left( -1048576 + 262144 \right) \]
\[ = \frac{1}{24} \left( -786432 \right) \]
\[ = -32768 \]

Thus, the final answer is:
\[ S_{20} - S_{18} = -32768 \]