Question

The sum of three terms of an A.P is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

Answer 1

Let the three terms of AP are a,a+d,a+2d.
Now,
As per the question ,
a+a+d+a+2d = 3a+3d = 21
i.e. a+d = 7 or, d= 7-a.

Now,
$a(a+2d)-(a+d)=6a[a+2(7-a\left)\right]-7=6a[a+14-2a]=13a[14-a]=1314a-a^2=13a^2-14a+13=0$
By applying quadratic formula, you will get
a= 13, 1.
d will be 7-a i.e. -6 ,6.
Now, there will be two AP
13,7, 1 and 1,7,13.