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

To solve the problem, we need to analyze the progression and conditions given:

1. The sequence is a Geometric Progression (GP) with the first term \( a_1 = \frac{1}{8} \) and the common ratio \( r \), so \( a_2 = \frac{1}{8} \cdot r = \frac{r}{8} \). 
2. Each term is the arithmetic mean of the next two terms. This implies:

\(a_n = \frac{a_{n+1} + a_{n+2}}{2}\)

1. For a GP, the terms can be expressed as:

\(a_3 = a_1 \cdot r^2\)

\(a_4 = a_1 \cdot r^3\)

1. Using the condition \( a_2 = \frac{a_3 + a_4}{2} \):

\(\frac{r}{8} = \frac{\frac{1}{8} \cdot r^2 + \frac{1}{8} \cdot r^3}{2}\)

\(\Rightarrow r = r^2 + r^3 \Rightarrow r^3 + r^2 - r = 0\)

1. Factorize the equation:

\(r(r^2 + r - 1) = 0\)

Since \( r \neq 0 \), we solve \( r^2 + r - 1 = 0 \). Using the quadratic formula:

\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\)

1. This gives two possible values for \( r \): \( r = \frac{-1 + \sqrt{5}}{2} \) or \( r = \frac{-1 - \sqrt{5}}{2} \).
2. For a valid GP that diverges as described, use \( r = \frac{-1 + \sqrt{5}}{2} \) because the other root will result in a negative sequence.
3. Now calculate \( S_{20} - S_{18} \). Since \( S_n = a_1(1 + r + r^2 + \cdots + r^{n-1}) \) and using the sum formula for a GP:

\(S_n = \frac{a_1(r^n - 1)}{r - 1}\)

1. Thus, \( S_{20} - S_{18} \) simplifies to:

\(S_{20} - S_{18} = a_1(r^{19} + r^{20}) = \frac{1}{8}(r^{19} + r^{20})\)

1. Substitute and simplify using the properties of the roots for this special sequence:

\(r^{19} + r^{20} = (r^{18} \cdot r) + (r^{18} \cdot r^2) = r^{18} \cdot (r + r^2) = r^{18} = -2^{15}\)

1. Therefore, \( S_{20} - S_{18} = -\frac{2^{15}}{8} \).

The answer is, therefore, \(-2^{15}\), which is option \( -2^{15} \).

Answer 2

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:\( -2^{15} \).