Divide 20 into four parts which are in A.P. and such that the product

1.	Divide 20 into four parts which are in A.P. and such that the product of the first and fourth is to the product of the second and third in the ratio 2 : 3. Find the product of the first and fourth term of the A.P. (a) 12 (b) 16 (c) 20 (d) 25
Answer» (b) 16 Let the required four parts be (a – 3d), (a – d), (a + d), (a + 3d) Given, a – 3d + a – d + a + d + a + 3d = 20 ⇒ 4a = 20 ⇒ a = 5 Also, \(\frac{(a-3d)(a+3d)}{(a-d)(a+d)}\) = \(\frac{2}{3}\) ⇒ 3(a² – 9d²) = 2(a² – d²) ⇒ 3(25 – 9d²) = 2(25 – d²) ⇒ 75 – 27d²= 50 – 2d² ⇒ 25 = 25d² ⇒ d = ± 1 Taking d = 1, the numbers are 5 – 3, 5 – 1, 5 + 1, 5 + 3, i.e., 2, 4, 6, 8 Taking d = –1, the numbers are 5 + 3, 5 + 1, 5 – 1, 5 – 3, i.e., 8, 6, 4, 2 ∴ Required product in either case = 2 × 8 = 16.

Divide 20 into four parts which are in A.P. and such that the product of the first and fourth is to the product of the second and third in the ratio 2 : 3. Find the product of the first and fourth term of the A.P. (a) 12 (b) 16 (c) 20 (d) 25

Answer»

(b) 16

Let the required four parts be (a – 3d), (a – d), (a + d), (a + 3d)

Given, a – 3d + a – d + a + d + a + 3d = 20

⇒ 4a = 20 ⇒ a = 5

Also, \(\frac{(a-3d)(a+3d)}{(a-d)(a+d)}\) = \(\frac{2}{3}\)

⇒ 3(a² – 9d²) = 2(a² – d²)

⇒ 3(25 – 9d²) = 2(25 – d²)

⇒ 75 – 27d²= 50 – 2d² ⇒ 25 = 25d²

⇒ d = ± 1

Taking d = 1, the numbers are 5 – 3, 5 – 1, 5 + 1, 5 + 3, i.e., 2, 4, 6, 8

Taking d = –1, the numbers are 5 + 3, 5 + 1, 5 – 1, 5 – 3, i.e., 8, 6, 4, 2

∴ Required product in either case = 2 × 8 = 16.