In aptitude (Algebraic Identities), if a + b + c = 13 and a^2 + b^2 + c^2 = 69, what is the value of the expression (ab + bc + ca)?

Introduction / Context:
This problem checks your familiarity with standard algebraic identities involving sums of variables and sums of their squares. Rather than finding individual values of a, b, and c, the question asks for a combined symmetric expression ab + bc + ca. Such questions are frequently used in aptitude and algebra sections to test conceptual understanding of identities and manipulation skills.

Given Data / Assumptions:

• a + b + c = 13.
• a^2 + b^2 + c^2 = 69.
• We must find the value of ab + bc + ca.
• a, b, and c are real numbers, but their exact values are not required.

Concept / Approach:
The key identity relating the square of a sum to sums of squares and pairwise products is: (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). We are given the value of a + b + c and a^2 + b^2 + c^2, so we can substitute them into the identity and solve for ab + bc + ca. This avoids the need to find the individual variables, which may not even be unique.

Step-by-Step Solution:
Step 1: Start from the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). Step 2: Substitute a + b + c = 13, so (a + b + c)^2 = 13^2 = 169. Step 3: Substitute a^2 + b^2 + c^2 = 69 into the identity. Step 4: The equation becomes 169 = 69 + 2(ab + bc + ca). Step 5: Subtract 69 from both sides: 169 - 69 = 2(ab + bc + ca). Step 6: Compute 169 - 69 = 100. Step 7: So 2(ab + bc + ca) = 100. Step 8: Divide by 2: ab + bc + ca = 100 / 2 = 50. Step 9: Hence, the required value is 50.

Verification / Alternative check:
To check consistency, we can attempt constructing one possible triple (a, b, c) that fits the conditions, though it is not necessary. However, the algebraic identity is standard and holds for all real numbers, so once we have substituted correct numeric values, the result is reliable. This kind of cross check reassures us that there is no arithmetic mistake in squaring 13 or subtracting 69 from 169.

Why Other Options Are Wrong:
30, 10, 70, and 40 all fail to satisfy the identity when plugged back into 69 + 2(ab + bc + ca). Only ab + bc + ca = 50 makes a^2 + b^2 + c^2 + 2(ab + bc + ca) equal to 169, which is the square of 13.

Common Pitfalls:
Students sometimes use the wrong identity or forget the factor 2 in front of ab + bc + ca. Another typical mistake is to compute 13^2 incorrectly or to make subtraction errors with 169 and 69. Carefully writing the identity and each arithmetic step significantly reduces these errors in exam settings.

Final Answer:
The value of the expression ab + bc + ca is 50.