The sum of first three terms of an AP is 48.

Solution:

Let the first three terms of the AP be (a − d), a and (a + d). Then,

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

$\Rightarrow 3 a=48$

$\Rightarrow a=16$

Now,

$(a-d) \times a=4(a+d)+12$      (Given)

$\Rightarrow(16-d) \times 16=4(16+d)+12$

$\Rightarrow 256-16 d=64+4 d+12$

$\Rightarrow 16 d+4 d=256-76$

$\Rightarrow 20 d=180$

$\Rightarrow d=9$

When a = 16 and d = 9,

$a-d=16-9=7$

$a+d=16+9=25$

Hence, the first three terms of the AP are 7, 16 and 25.