Question

If \[a,b,c\] are non-zero real numbers such that \[3({a^2} + {b^2} + {c^2}) = 2(a + b + c + ab + bc + ca)\], then \[a,b,c\] are in
1) \[AP\]
2) \[GP\]
3) \[AP,GP,HP\]
4) \[AP\] as well as \[GP\]

Answer 1

Hint: This question involves the basic concepts of rearrangement and factorization of expressions. The concepts of progressions are also applied. The Formulas involved in this question are:
Arithmetic progression \[2y = x + z\], for the variables \[x,y,z\]
Geometric progression \[{y^2} = xz\], for the variables \[x,y,z\]
Harmonic progression \[y = \dfrac{{2xz}}{{x + z}}\]for the variables \[x,y,z\]

Complete step-by-step answer:
Now, we are given the equation,
\[ \Rightarrow 3({a^2} + {b^2} + {c^2}) = 2(a + b + c + ab + bc + ca)\]
Now, let’s open the brackets on both sides of the equation,
\[ \Rightarrow 3{a^2} + 3{b^2} + 3{c^2} = 2a + 2b + 2c + 2ab + 2bc + 2ca\]
By bringing everything on the left side of the equation to the right side, we get,
\[ \Rightarrow 3{a^2} + 3{b^2} + 3{c^2} - 2a - 2b - 2c - 2ab - 2bc - 2ca = 0\]
Now, grouping terms as shown below
\[ \Rightarrow ({a^2} - 2a) + ({b^2} - 2b) + ({c^2} - 2c) + ({a^2} - 2ab + {b^2}) + ({b^2} - 2bc + {c^2}) + ({c^2} - 2ac + {a^2}) + 3 = 0\]
Now, this is the key step of the question, where we break 3 and write it as 1+1+1,
\[ \Rightarrow ({a^2} - 2a) + ({b^2} - 2b) + ({c^2} - 2c) + ({a^2} - 2ab + {b^2}) + ({b^2} - 2bc + {c^2}) + ({c^2} - 2ac + {a^2}) + 1 + 1 + 1 = 0\]
This allows us to form perfect squares of the initial three brackets
\[ \Rightarrow ({a^2} - 2a + 1) + ({b^2} - 2b + 1) + ({c^2} - 2c + 1) + ({a^2} - 2ab + {b^2}) + ({b^2} - 2bc + {c^2}) + ({c^2} - 2ac + {a^2}) = 0\]
\[ \Rightarrow {(a - 1)^2} + {(b - 1)^2} + {(c - 1)^2} + {(a - b)^2} + {(b - c)^2} + {(c - a)^2} = 0\]
Now, after simplifying the equation to this extent, we observe that all the individual terms are positive as they are squares of non-zero real numbers.
So, the only positive case for the L.H.S = R.H.S is by equating the individual terms to zero
\[ \Rightarrow {(a - 1)^2} = 0,{(b - 1)^2} = 0,{(c - 1)^2} = 0\]
\[ \Rightarrow {(a - b)^2} = 0,{(b - c)^2} = 0,{(c - a)^2} = 0\]
After equating the terms to zero, we get,
\[ \Rightarrow a = 1,b = 1,c = 1\]
\[ \Rightarrow a = b,b = c,c = a\]
Now, checking the condition for arithmetic progression,
\[ \Rightarrow 2b = a + c\]
\[ \Rightarrow 2 = 1 + 1\]
\[ \Rightarrow 2 = 2\]
As, the L.H.S = R.H.S, are in arithmetic progression
Now, checking the condition for geometric progression,
\[ \Rightarrow {b^2} = ac\]
\[ \Rightarrow {1^2} = 1 \times 1\]
\[ \Rightarrow 1 = 1\]
As, the L.H.S = R.H.S, are in geometric progression
Now, checking the condition for harmonic progression,
\[ \Rightarrow b = \dfrac{{2ac}}{{a + c}}\]
\[ \Rightarrow 1 = \dfrac{{2 \times 1 \times 1}}{{1 + 1}}\]
\[ \Rightarrow 1 = \dfrac{2}{2}\]
\[ \Rightarrow 1 = 1\]
As, the L.H.S = R.H.S, are in harmonic progression
 They are in arithmetic progression, geometric progression, harmonic progression
Thus, option (3) is the correct answer.
So, the correct answer is “Option 3”.

Note: This question involves the concepts of analyzing algebraic and bringing meaning out of them. Be careful while applying conditions on the individual expressions of the equation. Keep in mind the restrictions on the original variables. Do not commit calculation mistakes, and be sure of the final answer.