Question

Divided 20 into four parts which are in A.P such that the product of the first and fourth and the product of the second and third is in the ratio 2 : 3

• A. 2, 4, 6 and 8
• B. 4,6,8 and 10
• C. 1,3,5 and 7
• D. 3,5,7 and 9

Answer 1

The correct option is A 2, 4, 6 and 8

Let the four parts be $a-3d,a-d,a+danda+3d$
Hence $(a-3d)+(a-d)+(a+d)+(a+3d)=20$
$\Rightarrow 4a=20$
$\Rightarrow a=5$
According to the question
$(a-3d)(a+3d):(a-d)(a+d)=2:3$
$\Rightarrow (a^2-9d^2)(a^2-d^2)=2:3$
$\Rightarrow \frac{a^2-9d^2}{a^2-d^2}=\frac{2}{3}$
$\Rightarrow 3(a^2-9d^2)=2(a^2-d^2)$
$\Rightarrow 3a^2-27d^2=2a^2-2d^2$
$\Rightarrow 3a^2-2a^2=27d^2-2d^2$
$\Rightarrow a^2=25d^2$
$\Rightarrow 5^2=25d^2$
$\Rightarrow 25=25d^2$
$\Rightarrow d^2=1$
$\Rightarrow d=\pm 1$
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.