Question

13. 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.​

Answer 1

Answer:

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).