Answer:
The four parts are 9, 15, 21, 27
Given problem:
Divide 72 in four parts in AP, such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 27:35. Find the four parts.
Step-by-step explanation:
Given number 72
let the (a-3d), (a-d), (a+d), (a+3d) are be the 4 parts which are in AP
[with common difference 2d ]
here 72 is divided as (a-3d), (a-d), (a+d), (a+3d)
then sum of 4 terms will be equals to 72
a - 3d + a - d + a + d + a + 3d = 72
4a = 72
a = 18
product of extremes (1st and 4th terms) = (a-3d)(a+3d) = a²-9d²
product of means (2nd and 3rd terms) = (a-d) (a+d) = a²-d²
given that the ratio of the product of their extremes to product of their means = 27:35
(a²-9d²) : (a²-d²) = 27 : 35
[tex]\frac{ a^{2} -9d^{2} }{a^{2} - d^{2} } = \frac{27}{35}[/tex]
[tex]\frac{ 324 -9d^{2} }{324 - d^{2} } = \frac{27}{35}[/tex] [ ∵ a² = 18² = 324 ]
35(324 - 9d²) = 27 (324 -d²)
11340 - 315d² = 8748 - 27d²
288 d² = 2592
d² = 9
d = 3
the required numbers (a-3d) = 18 - 9 = 9
(a-d) = 18-3 = 15
(a+d) = 18+3 = 21
(a+3d) = 18+9 = 27
