CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If a1,a2,a3,an are in AP, where ai>0 for all i, then the value of 1a1+a2+1a2+a3++1an2+an is



 


A
1a1+an
loader
B
1a1an
loader
C
na1+an
 
loader
D
(n1)a1+an
loader

Solution

The correct option is D (n1)a1+an
1a1+a2+1a2+a3++1an1+an=1a1+a2×a1a2a1a2+1a2+a3×a2a3a2a3++1a1+a2×an1anan1an=a1a2a1a2+a2a3a2a3++an1anan1an=a1a2d+=a2a3d++an1and
[d= common difference]
=a2a1d+a3a2d++anan1d=1d[a2a1+a3a2++anan1]=1d[a1+an]  (Rest all terms will cancelled)
=1d[a1+an]=1d[ana1]=1dana1×an+a1an+a1=1dana1an+a1=1da1+(n1)da1an+a1=1d(n1)dan+a1=n1a1+an

All India Test Series

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image