If in a GP, the sum of the first 18 terms is equal to the sum of first 22 terms and the sum of the first 19 terms is 65, then what will be the sum of first 4 terms? (Note: \( i=\sqrt{-1} \))
Show Hint
- \(-65i\)
- \(65i\)
- \(0\)
- \(-65\)
The Correct Option is C
Solution and Explanation
We are given a Geometric Progression (GP) where \( S_{18} = S_{22} \) and \( S_{19} = 65 \). We need to find \( S_4 \). Step 2: Key Formula or Approach:
Sum of first \( n \) terms of a GP: \( S_n = \sum_{k=1}^{n} T_k \). Relationship between sums: \( S_{22} = S_{18} + T_{19} + T_{20} + T_{21} + T_{22} \). Step 3: Detailed Explanation:
Given \( S_{18} = S_{22} \), we have: \[ S_{22} - S_{18} = 0 \] \[ T_{19} + T_{20} + T_{21} + T_{22} = 0 \] Let the first term be \( a \) and the common ratio be \( r \). Then \( T_n = ar^{n-1} \). \[ ar^{18} + ar^{19} + ar^{20} + ar^{21} = 0 \] Factor out \( ar^{18} \): \[ ar^{18} (1 + r + r^2 + r^3) = 0 \] Since \( S_{19} = 65 \), the GP is not a trivial zero sequence, so \( a \neq 0 \) and \( r \neq 0 \). Thus, we must have: \[ 1 + r + r^2 + r^3 = 0 \] Now, we need to find the sum of the first 4 terms, \( S_4 \): \[ S_4 = a + ar + ar^2 + ar^3 \] \[ S_4 = a(1 + r + r^2 + r^3) \] Substitute the condition found above (\( 1 + r + r^2 + r^3 = 0 \)): \[ S_4 = a(0) = 0 \] Step 4: Final Answer:
The sum of the first 4 terms is 0.
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