Show that (x2+xy+y2),(z2+xz+x2) and (y2+yz+z2) are consecutive terms of an A.P., if x, y and z are in A.P.
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Step-by-Step Solution
Step 1: Understand the condition for A.P.
For three terms to be in an Arithmetic Progression (A.P.), the middle term must be the average of the other two terms. This means that the difference between the second and first term is equal to the difference between the third and second term. So, if x, y, z are in A.P., then y−x=z−y, which simplifies to 2y=x+z.
Step 2: Define the terms to be checked
We are given three expressions and need to show that they form an A.P. Let's label these expressions as A, B, and C for clarity. We need to prove that 2B=A+C.
Step 3: Calculate A+C
First, we calculate the sum of the first and third terms, A+C. We combine like terms to simplify the expression.
Step 4: Substitute 2y=x+z into A+C
From the condition that x, y, z are in A.P., we know x+z=2y. We substitute this into the expression for A+C to simplify it further. This substitution helps us relate the expression to the middle term B.
Step 5: Calculate 2B
Next, we calculate two times the middle term, 2B. This is the expression we need to show is equal to A+C.
Step 6: Substitute x+z=2y into 2B
We substitute x+z=2y into the expression for 2B. We can rewrite 2xz as part of (x+z)2. This allows us to express 2B in terms of x, z, and y.
Step 7: Compare A+C and 2B
We have A+C=x2+xy+3y2+yz+z2 and 2B=x2+4y2+z2. These expressions do not immediately look equal. Let's re-evaluate the substitution in A+C more carefully.
Step 8: Re-evaluate A+C using x+z=2y
Let's re-substitute x+z=2y into the expression for A+C. We group the terms involving y and x+z to make the substitution clearer. This leads to A+C=x2+4y2+z2.
Step 9: Conclusion
Since we have shown that A+C=x2+4y2+z2 and 2B=x2+4y2+z2, it follows that A+C=2B. This proves that the given terms are in an A.P.