Question

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.

Answer 1

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

$\frac{(a^2–9d^2)}{(a^2–d^2)}=\frac{2}{3}$

$3(a^2–9d^2)=2(a^2–d^2)$

⇒ $3a^2–27d^2=2a^2–2d^2$

⇒ $3a^2–2a^2=27d^2–2d^2$

⇒ $a^2=25d^2$

⇒ $25=25d^2$

⇒ $d^2=1$

$\therefore d=\pm 1$

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.