7.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.
Answers
Answer:
Let the four parts be (a−3d),(a−d),(a+d) and (a+3d).
Then, Sum of the numbers =32
⟹(a−3d)+(a−d)+(a+d)+(a+3d)=32⟹4a=32⟹a=8
It is given that
(a−d)(a+d)
(a−3d)(a+3d)
=
15
7
⟹
a
2
−d
2
a
2
−9d
2
=
15
7
⟹
64−d
2
64−9d
2
=
15
7
⟹128d
2
=512⟹d
2
=4⟹d=±2
Thus, the four parts are a−3d,a−d,a+d and 3d, i.e. 2,6,10,1
Step-by-step explanation:
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
[ ∵ 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