If a,b,c,d,e be five numbers such that a,b,c are in A.P., b,c,d are in G.P., and c,d,e are in H.P., then
e=(b−a)2a
We have,
a,b,c are in A.P
So,
2b=a+c ............(1)
Since, b,c,d are in G.P
So,
c2=bd ............(2)
Since, c,d,e are in H.P
So,
1e=2d−1c ..........(3)
From equations (1),(2) and (3), we get
1e=2bc2−12b−a
1e=2b(2b−a)2−12b−a
1e=2b−2b+a(2b−a)2
e=(2b−a)2a
Hence, this is the answer.