wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If a1,a2,a3,....,an1,an are in A.P., then show that 1a1an+1a2an1+1a3an2+.....1ana1=2(a1+an)[1a1+1a2+....1an].

Open in App
Solution

a1,a2,a3.....an1,ana1+an=a2+an1=a3+an2=K(letsay)

1a1an=a1+an(a1+an)a1an=1a1+an(1a1+1an)1a1an=1K(1a1+1an)

So, sum of

1a1an+1a2an1+...1ana1=1K(1a1+1an)+1K(1a2+1an1)+....1K(1a1+1an)=2K(1a1+1an+1a2+1an1+...)=2a1+an(1a1+1a2+....1an)


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Arithmetic Progression
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon