Mathematics - Sequence & Series Question with Solution | TestHub
Match List I with List II.
List I | List II |
|---|---|
(A) If the distinct pth, qth, rth and sth terms of a nonconstant A.P. form a nonzero G.P., classify p-q, q-r, r-s. | (P) All equal |
(B) If ln x, ln y, ln z (x,y,z>1) are in G.P., classify 2x+ln(ln x), 3x+ln(ln y), 4x+ln(ln z). | (Q) In A.P. |
(C) If n!, 3n!, (n+1)! form a G.P., classify n!, 5n!, (n+1)!. | (R) In G.P. |
(D) If the arithmetic means of (b-c)²,(c-a)²,(a-b)² and (b+c-2a)²,(c+a-2b)²,(a+b-2c)² are equal, classify a,b,c. | (S) In H.P. |
Options:
Answer:
Solution:
(A) Write the four nonzero G.P. terms as u,uk,uk²,uk³, with k≠1 since the indices are distinct and the A.P. is nonconstant. Dividing successive term differences by the A.P. difference shows that p-q,q-r,r-s are proportional to 1,k,k², hence form a G.P. (B) Since ln x,ln y,ln z are positive G.P. terms, their logarithms form an A.P. Adding the A.P. 2x,3x,4x preserves this. (C) (3n!)²=n!(n+1)! gives n+1=9, so n=8. The new terms are 8!,5·8!,9·8!, an A.P. (D) The second sum of three squares is three times the first. Equality of their arithmetic means therefore forces the first sum to be zero, so a=b=c. Thus A-R; B-Q; C-Q; D-P.
