# The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.

Let the numbers be (a - d), a, (a + d).

Now, sum of the numbers = 48

(a - d) + a + (a + d) = 48

3a = 48

a = 16

Now, we are given that,

Product of first and second terms exceeds 4 times the third term by 12.

(a - d) × a = 4(a + d) + 12

a2 - ad = 4a + 4d + 12

On putting the value of a in the above equation, we get,

256 - 16d = 64 + 4d + 12

20 d = 180

d = 9

The numbers are a - d, a, a + d

i.e. the numbers are 7, 16, 25.

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  Quiz on Arithmetic Progression Quiz32 mins  Become a Master of A.P. in 45 Minutes!!47 mins  Arithmetic Progression Tricks and QUIZ37 mins  Quiz | Group of Questions on General Term of an A.P49 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 