Let `a_(1),a_(2),a_(3),a_(4)anda_(5)` be such that `a_(1),a_(2)anda_(3)` are in A.P., `a_(2),a_(3)anda_(4)` are in G.P., and `a_(3),a_(4)anda_(5)` are in H.P. Then, `a_(1),a_(3)anda_(5)` are in

Options

A. G.P.
B. A.P.
C. H.P.
D. none of these

Answer

A

Explanation

We have `a_(1),a_(2),a_(3)` are in A.P. `rArr2a_(2)=a_(1)+a_(3)` . . . .(i)
`a_(2),a_(3),a_(4)` are in G.P. `rArr a_(3)^(2)=a_(2)a_(4)` . . .(ii)
`a_(3),a_(4),a_(5)` are in G.P. `rArra_(4)=(2a_(3)a_(5))/(a_(3)+a_(5))` . . .(iii)
Putting `a_(2)=(a_(1)+a_(3))/(2)anda_(4)=(2a_(3)a_(5))/(a_(3)+a_(5))` in (ii), we get
`a_(3)^(2)=(a_(1)+a_(3))/(2)xx(2a_(3)a_(5))/(a_(3)+a_(5))`
`rArr" "a_(3)^(2)=a_(1)a_(5)rArra_(1),a_(3),a_(5)` are in G.P.