Answer :

We know that the sum of 3 numbers in the AP is 18.

Let’s say that those 3 numbers are: a – d, a, a + d

So we can say that, a – d + a + a + d = 18

⇒ 3a = 18

⇒ a= 6

Now, we have that the sum of squares of these 3 numbers is 180.

So we can say that, (a – d) ^{2} + a^{2} + (a + d) ^{2} = 180

⇒ a^{2} – 2ad + d^{2} + a^{2} + a^{2} + 2ad + d^{2} = 180

⇒ 3a^{2} + 2d^{2} = 180

We know that a = 6,

So, 3(6)^{2} + 2d^{2} = 180

⇒ 3(36) + 2d^{2} = 180

⇒ 108 + 2d^{2} = 180

⇒ 2d^{2} = 180 – 108

⇒ 2d^{2} = 72

⇒ d^{2} = 36

⇒ d = 6

Now, our 3 numbers a – d, a, a + d are 6 – 6, 6, 6 + 6 = 0, 6, 12 respectively.

∴ The 3 numbers of the A.P in the increasing order are : 0, 6, 12.

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