Divide 20 into four parts which are in A
P such that the ratio between the products of the extremes to the product of the means is 2:3. Find the A.P.
Answers
Answer:
Let the four parts be a – 3d, a – d, a + d and a +3d
Hence (a – 3d) + (a – d) + (a + d) + (a +3d) = 20
⇒ 4a = 20
∴ a = 5
It is also given that (a – 3d)(a + 3d) : (a – d)(a + d) = 2 : 3
(a2–d2)(a2–9d2)=32
3(a2–9d2)=2(a2–d2)
⇒ 3a2–27d2=2a2–2d2
⇒ 3a2–2a2=27d2–2d2
⇒ a2=25d2
⇒ 25=25d2
⇒ d2=1
∴d=±1
Case (i): If d=1
Hence (a–3d)=(5–3)=2
(a–d)=(5–1)=4
(a+d)=(5+1)=6
(a+3d)=(5+3)=8
Hence the four numbers are 2, 4, 6 and 8.
Case (ii): If d=–1
Hence (a–3d)=(5+3)=8
(a–d)=(5+1)=6
(a+d)=(5–1)=4
Step-by-step explanation:
Let the four parts be a – 3d, a – d, a + d and a +3d
Hence (a – 3d) + (a – d) + (a + d) + (a +3d) = 20
⇒ 4a = 20
∴ a = 5
It is also given that (a – 3d)(a + 3d) : (a – d)(a + d) = 2 : 3
(a
2
–d
2
)
(a
2
–9d
2
)
=
3
2
3(a
2
–9d
2
)=2(a
2
–d
2
)
⇒ 3a
2
–27d
2
=2a
2
–2d
2
⇒ 3a
2
–2a
2
=27d
2
–2d
2
⇒ a
2
=25d
2
⇒ 25=25d
2
⇒ d
2
=1
∴d=±1
Case (i): If d=1
Hence (a–3d)=(5–3)=2
(a–d)=(5–1)=4
(a+d)=(5+1)=6
(a+3d)=(5+3)=8
Hence the four numbers are 2, 4, 6 and 8.
Case (ii): If d=–1
Hence (a–3d)=(5+3)=8
(a–d)=(5+1)=6
(a+d)=(5–1)=4
(a+3d)=(5–3)=2
Hence the four numbers are 8, 6, 4 and 2.