Q. 124.6( 30 Votes )

# Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.

Let 32 be divided into parts as (a - 3d), (a - d), (a + d) and (a + 3d).

Now (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

4a = 32

a = 8

Now, we are given that product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.

i.e. [(a - 3d) × (a + 3 d)] : [(a - d) × (a + d)] = 7 : 15

=

15[(a - 3d) × (a + 3 d)] = 7[(a - d) × (a + d)]

15[a2 - 9d2] = 7 [a2 - d2]

15a2 - 135d2 = 7a2 - 7d2

8a2 - 128d2 = 0

8a2 = 128d2

Putting the value of a, we get,

512 = 128 d2

d2 = 4

d =  ±2

If d = 2, then the numbers are 2, 6, 10, 14.

If d = - 2, then the numbers are 14, 10, 6, 2.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Champ Quiz | Arithmetic Progression34 mins
Champ Quiz | Arithmetic Progression30 mins
Lets Check Your Knowledge in A.P.49 mins
Arithmetic progression: Previous Year NTSE Questions34 mins
Arithmetic Progression Tricks and QUIZ37 mins
Quiz | Group of Questions on General Term of an A.P49 mins
Quiz on Arithmetic Progression Quiz32 mins
Become a Master of A.P. in 45 Minutes!!47 mins
Get to Know About Geometric Progression41 mins
NCERT | Solving Questions on Introduction of A.P42 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses