Question

Divide 20 into 4 parts which are in A.P. such that the product of

the 1st term and 4th term and the product of the 2nd term and 3rd

term is in the ratio 2:3.​

Answer 1

Solutìon :

The four numbers are 2, 4, 6 and 8.(if d= 1)

and the four numbers are 8, 6, 4 and 2.(if d= -1)

Explanatìon :

Let the Four parts be a – 3d, a – d, a + d and a +3d

Then ,

$\sf(a-3d)+(a-d)+(a+d)+(a+3d)=20$

$\sf20=4a$

$\sf\:a=5..(1)$

According to the Question :

The product of the 1st term and 4th term and the product of the 2nd term and 3rd term is in the ratio 2:3.

$\sf\implies\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}= \dfrac{2}{3}$

$\sf\implies\dfrac{a^2-9d^2}{a^2-d^2}= \dfrac{2}{3}$

$\sf\implies3(a^2-9d^2)=2(a^2-d^2)$

$\sf\implies3a^2-27d^2=2a^2-2d^2$

$\sf\implies3a^2-2a^2=27d^2-2d^2$

$\sf\implies\:a^2=25d^2$

$\sf\implies\:d=1\:or\:-1$

Case -1: If d = 1

then, (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 -2:If d = –1

then, (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.