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 a₁,...,a₁₀₀ form an H.P., find [Σ(i=1 to 99) aᵢaᵢ₊₁]/(a₁a₁₀₀). | (P) 99 |
(B) Find the largest positive term of the H.P. whose first two terms are 2/5 and 12/13. | (Q) 7 |
(C) Let x be the probability that the first row in a random 3×3 arrangement of 1,...,9 is decreasing. Find 36x. | (R) 6 |
(D) Let x be the probability that a random three-digit integer has exactly three positive divisors. Find 900x. | (S) 12/13 |
| (T) 3 |
Options:
Answer:
Solution:
(A) For reciprocal A.P. difference d≠0, telescope aᵢ-aᵢ₊₁=d aᵢaᵢ₊₁ and use 1/a₁₀₀-1/a₁=99d. The ratio is 99. For d=0 the 99 equal products give the same result. (B) Reciprocal terms start 5/2,13/12 with difference -17/12. Only the first two terms are positive, so the largest H.P. term is 12/13. (C) For each set of three first-row values exactly one of the 3! orders decreases. Thus x=1/6 and 36x=6. (D) Exactly three positive divisors occurs only for prime squares. The seven three-digit prime squares are 121,169,289,361,529,841,961. Since there are 900 three-digit integers, 900x=7. Thus A-P; B-S; C-R; D-Q.
