Let the four parts which are in AP are
(a – 3d), (a – d), (a + d), (a + 3d)
According to question,
The sum of these four parts = 20
⇒(a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
⇒ 4a = 20
⇒ a = 5 …(i)
Now, it is also given that
product of the first and fourth : product of the second and third = 2 : 3
i.e. (a – 3d) × (a + 3d) : (a – d) × (a + d) = 2 : 3
⇒ 3(a2 – 9d2) = 2(a2 – d2)
⇒ 3a2 – 27d2 = 2a2 – 2d2
⇒ 3a2 – 2a2 = – 2d2 + 27d2
⇒ (5)2 = – 2d2 + 27d2 [from (i)]
⇒ 25 = 25d2
⇒ 1 = d2
⇒ d = ±1
Case I: if d = 1 and a = 5
a – 3d = 5 – 3(1) = 5 – 3 = 2
a – d = 5 – 1 = 4
a + d = 5 + 1 = 6
a + 3d = 5 + 3(1) = 5 + 3 = 8
Hence, the four parts are
2, 4, 6, 8
Case II: if d = – 1 and a = 5
a – 3d = 5 – 3( – 1) = 5 + 3 = 8
a – d = 5 – ( – 1) = 5 + 1 = 6
a + d = 5 + ( – 1) = 5 – 1 = 4
a + 3d = 5 + 3( – 1) = 5 – 3 = 2
Hence, the four parts are
8, 4, 6, 2