If a1, a2, ...a10 be in AP, h1, h2...h10 be in HP. a1 = h1 = 2, a10 = h10 = 3, then a4h7 =

(1) 2

(2) 3

(3) 5

(4) 6

Solution:

Given a1, a2, …a10 be in AP.

a1 = h1 = 2, a10 = h10 = 3

a10 = a1+9d

3 = 2+9d

9d = 1

d = 1/9

a4 = a1+3d

= 2+3(1/9)

= 7/3

h1, h2…h10 be in HP

So 1/h1, 1/h2…1/h10 are in AP.

1/h10 = 1/h1+9D

1/3 = ½ + 9D

D = -1/54

1/h7 = 1/h1+6D

= ½ + 6(-1/54)

= ½ – (1/9)

= 7/18

h7 = 18/7

a4h7 = (7/3)(18/7)

= 18/3

= 6

Hence option (4) is the answer.

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