Question

If each term of a geometric progression \( a_1, a_2, a_3, \dots \) with \( a_1 = \frac{1}{8} \) and \( a_2 \neq a_1 \), is the arithmetic mean of the next two terms and \( S_n = a_1 + a_2 + \dots + a_n \), then \( S_{20} - S_{18} \) is equal to

• A. \( 2^{15} \)
• B. \( -2^{18} \)
• C. \( 2^{18} \)
• D. \( -2^{15} \)

Answer 1

The Correct Option is D

Let the r-th term of the geometric progression (GP) be \( ar^{r-1} \).

Step 1. Given condition: Since each term is the arithmetic mean of the next two terms, we have:  
  \(2a_r = a_{r+1} + a_{r+2}\)
  Substituting the terms, this becomes:  
  \(2ar^{r-1} = ar^r + ar^{r+1}\)
  Dividing by \( ar^{r-1} \) (assuming \( a \neq 0 \)), we get:  
  \(2 = r + r^2\)

Step 2. Solve for \( r \): Rearranging, we have:  
  [\(r^2 + r - 2 = 0\)
   Factoring, we get:  
   \((r - 1)(r + 2) = 0\)
  Thus, \( r = 1 \) or \( r = -2 \). Since \( a_2 \neq a_1 \), \( r \neq 1 \), so \( r = -2 \).

Step 3. Calculate \( S_{20} - S_{18} \): Since the sum of the first \( n \) terms of a GP is given by  
\(S_n = \frac{a(r^n - 1)}{r - 1},\) 
  we find \( S_{20} \) and \( S_{18} \):  
  \(S_{20} = \frac{1}{-2}((-2)^{20} - 1), \quad S_{18} = \frac{1}{-2}((-2)^{18} - 1).\)

 Therefore,
\(S_{20} - S_{18} = \frac{1}{-2}((-2)^{20} - (-2)^{18})\)
Simplifying, we get:  
\(S_{20} - S_{18} = -2^{15}.\)
The Correct Answer is:\( -2^{15} \).