Question

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: \( i=\sqrt{-1} \))

• A. \(-65i\)
• B. \(65i\)
• C. \(0\)
• D. \(-65\)

Hint:

If \( S_n = S_{n+k} \) in a sequence, the sum of the terms from \( n+1 \) to \( n+k \) is zero. For a GP, this often implies the common ratio involves complex roots of unity or \(-1\).

Answer 1

The Correct Option is C

Strategy:

We are given a Geometric Progression (GP) where \( S_{18} = S_{22} \) and \( S_{19} = 65 \). We need to find \( S_4 \).
Sum of first \( n \) terms of a GP: \( S_n = \sum_{k=1}^{n} T_k \). Relationship between sums: \( S_{22} = S_{18} + T_{19} + T_{20} + T_{21} + T_{22} \).
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 \( r \neq 0 \). Therefore, we must have: \[ 1 + r + r^2 + r^3 = 0 \] Now, it is necessary 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 (\( 1 + r + r^2 + r^3 = 0 \)): \[ S_4 = a(0) = 0 \]
The sum of the first 4 terms is 0.