Let a1,a2,a3..., a10 be in arithmetic progression and h1,h2,h3,..., h10 be in harmonic progression.If a1=h1=2 and a10=h10=3,then a4*h7=?

an = a1 + (n-1)d

a10 = a1 + (10-1)d
3 = 2 + 9d
d = 1/9

a4
= a1 + (n-1)d
= 2 + (4-1)(1/9)
= 2 + 1/3
= 7/3

h1 = 2 = 1 / (1/2)
Hence, we may assume that
hn = 1 / [ 1/2 + (n-1)p ]

h10 = 1 / [ 1/2 + (10-1)p ] = 3
1/2 + (10-1)p = 1/3
9p = - 1/6
p = - 1/54

h7
= 1 / [ 1/2 + (n-1)p ]
= 1 / [ 1/2 + (7-1)(-1/54) ]
= 1 / ( 1/2 - 1/9 )
= 18/7

a4 * h7
= (7/3) * (18/7)
= 6

Ans: 6