CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let a1, a2, a3, a4 and a5 be such that a1, a2 and a3 are in A.P., a2, a3 and a4 are in G.P., and a3, a4 and a5 are in H.P. Then loge a1, logea3 and loge a5 are in 


  1. A.P.

  2. G.P.

  3. none of these 

  4. H.P.


Solution

The correct option is A

A.P.


We have,

a1,a2,a3 are in A.P. 2a2=a1+a3

a2,a3,a4 are in G.P. a23=a2a4

a3,a4,a5 are in H.P. a4=2a3a5a3+a5

Putting a2=a1+a32 and a4=2a3a5a3+a5 in (2)we get

a23=a1+a32×2a3a5a3+a5

a23=a1a5

Hence, a1,a3, and a5 are in G.P. So loge a1 loge a3 and loge a5 are in A.P.

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More...



footer-image