Four numbers are in A.P. and their sum is 72 and the largest of them is twice the smallest. Then the numbers are ………

We have n = 4

⇒ S₄ = 72

Let the 4 nos. Be, a, a + d, a + 2d, a + 3d

Now we have that a + a + d + a + 2d + a + 3d = 72

⇒ 4a + 6d = 72

⇒ 2a + 3d = 36 …….(1)

Now We know that the largest of them is twice the smallest.

So we can say that,

⇒ 2a = a + 3d

⇒ a = 3d

we now put the value of a in (1),

to get :

⇒ 2a + a = 36

⇒ 3a = 36

⇒ a = 12

we know that a is the first term in this AP, so the possible option is (b)

but to check whether it is the correct we will have to check the relation between the first and the last term of the AP of (b) option.

We can see that, in (b), first term is 12

And the last term is 24 so, this is the correct AP.

∴ the correct option is (b)