A number 20 is divided into 4 parts that are in AP such that the product of first and fourth is to the product of second and third is 2:3. Find the numbers.
Answers
Answered by
0
(a1)(a4) : (a2)(a3) solve it yourself please
a^2+3ad : a^2+3ad+2d^2 = 2:3
a^2+3ad / a^2+3ad+2d^2 = 2 / 3 (please solve this equation)
Therefore, (a+d) (a-4d) = 0
So , a=-d or a=4d
a1 + a2 + a3 + a4 = 20
a+d+a+2d+a+3d=20
4a+6d=20
2a+3d=10 ———————(1)
Putting value of a=-d in (1)
D=10 (please solve it)
And similarly d=10/11
Now find a by putting value of d in (1)
And then find a1 ,a2, a3, a4
HOPE IT HELPS
a^2+3ad : a^2+3ad+2d^2 = 2:3
a^2+3ad / a^2+3ad+2d^2 = 2 / 3 (please solve this equation)
Therefore, (a+d) (a-4d) = 0
So , a=-d or a=4d
a1 + a2 + a3 + a4 = 20
a+d+a+2d+a+3d=20
4a+6d=20
2a+3d=10 ———————(1)
Putting value of a=-d in (1)
D=10 (please solve it)
And similarly d=10/11
Now find a by putting value of d in (1)
And then find a1 ,a2, a3, a4
HOPE IT HELPS
Similar questions