Mathematics - Sequence & Series Question with Solution | TestHub
Match each item in List I to all applicable entries in List II.
List I | List II |
|---|---|
(A) If three unequal nonzero numbers a,b,c are in A.P. and b-a,c-b,a are in G.P., find (a³+b³+c³)/(3abc). | (P) 4 |
(B) If x is the arithmetic mean and y,z are two geometric means between two positive endpoints, find (y³+z³)/(xyz). | (Q) 1 |
(C) For 50 distinct A.P. terms a₁,...,a₅₀, with a₁²≠a₅₀², suppose a₁²-a₂²+...+a₄₉²-a₅₀²=(5/7)ⁿ(a₁²-a₅₀²), n∈N. Find n. | (R) 2 |
(D) Find tan[Σ(r=1 to infinity) arctan(1/(2r²))]. | (S) 3 |
Options:
Answer:
Solution:
(A) Put d=b-a=c-b≠0. The G.P. d,d,a forces a=d, so the terms are d,2d,3d and the ratio is (1+8+27)/18=2. (B) Write endpoints u,ur³ and geometric means y=ur,z=ur², r>0. Their arithmetic mean is x=u(1+r³)/2, giving y³+z³=2xyz. (C) Pair successive squared terms. Their alternating sum is (25/49)(a₁²-a₅₀²); cancellation is allowed by the stated inequality. Thus (5/7)ⁿ=(5/7)² and n=2. (D) arctan(1/(2r²))=arctan(2r+1)-arctan(2r-1). The positive-angle series telescopes to π/2-π/4=π/4, whose tangent is 1. Therefore A-R; B-R; C-R; D-Q.
