The given question can be solved by using the formula of sum of n terms of an A.P and sum of squares of n terms of an A.P.

• A.P: An Arithmetic Progression (A.P) is a sequence of numbers such that the difference of any two successive numbers is constant.

• Sum of n terms of an A.P: The sum of n terms of an A.P is given by S = n/2 [2a + (n-1)d], where a is the first term, d is the common difference and n is the number of terms.

• Sum of squares of n terms of an A.P: The sum of squares of n terms of an A.P is given by S = n/6 [a2 + (n-1)d2 + 2(2a + (n-1)d].

From the given information,

Sum of three numbers = 21
Sum of squares of three numbers = 179

To solve this equation, the following equation can be formed.

• Sum of three numbers = n/2 [2a + (n-1)d]
• Sum of squares of three numbers = n/6 [a2 + (n-1)d2 + 2(2a + (n-1)d]

Substituting the given values in the above equation,

21 = 3/2 [2a + (3-1)d]
179 = 3/6 [a2 + (3-1)d2 + 2(2a + (3-1)d]

Solving for a and d,

a = 3, d = 4

Therefore, the required three numbers in A.P are 3, 7, 11.

Hence, option D is the correct answer.

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