Trigonometry & Inverse Trigonometry
Application problem (GP — misplaced in chapter)
nta_pyq_2023_jan
Grade 11
Question:
If the sum and product of four positive consecutive terms of a G.P. are 126 and 1296, respectively, then the sum of common ratios of all such GPs is
Step-by-Step Solution
Key Concept: Let the four terms be a, ar, ar^2, ar^3. Use sum = 126 and product = 1296 to find r.
Product: a^4 r^6 = 1296, so a^2 r^3 = 36. Sum divided by a: (r^{-3/2}+r^{-1/2}+r^{1/2}+r^{3/2}) = 21. Let A = r^{1/2}+r^{-1/2}, then A^3 - 2A = 21, A = 3, r^2 - 7r + 1 = 0. Sum of both values of r = 7.
Correct Answer: 1