One of the basic question from arithmetic progression chapter in which the product of first and second terms exceeds 4 times the third term by 12 and the sum of first three terms is 48. We have to the AP.
Book – RS Aggarwal, Class 10, chapter 5B, question no 13.
Let a−d,a,a+d are the three terms of AP
So,
Sum=a−d+a+a+d=48
3a=48
or a=16
And,
a(a−d)=4(a+d)+12
16(16−d)=4(16+d)+12
256−16d=64+4d+12=4d+76
256−76=4d+16d
180=20d
d=9
Which implies:
Numbers are: (7,16,25).