The sum of the first 10 terms of an arithmetic series is 390, and the third term of the series is 19. What is the value of the first term of this series?

Introduction / Context:
This question combines the formula for the sum of an arithmetic series with the formula for the nth term. It tests whether you can handle multiple relationships at once: the total sum over a number of terms and the value of a particular term. Such questions help build algebraic manipulation skills and are frequently used in aptitude tests involving sequences.

Given Data / Assumptions:

• The series is arithmetic.
• The sum of the first 10 terms, S10, is 390.
• The third term, T3, is 19.
• We are required to find the first term a.
• Let d be the common difference of the series.

Concept / Approach:
For an arithmetic progression, the nth term is Tn = a + (n - 1) * d, and the sum of the first n terms is Sn = n / 2 * (2a + (n - 1) * d). We are given T3 and S10, so we can write one equation using T3 and another using S10. Solving these two equations simultaneously will give us a and d. Once a is determined, we have answered the question. The important skill is substituting correctly into the formulas and keeping track of the algebraic manipulations.

Step-by-Step Solution:
Step 1: Let a be the first term and d be the common difference.Step 2: The third term is T3 = a + (3 - 1) * d = a + 2d. We are given T3 = 19, so a + 2d = 19.Step 3: The sum of the first 10 terms is S10 = 10 / 2 * (2a + (10 - 1) * d) = 5 * (2a + 9d).Step 4: We are given S10 = 390, so 5 * (2a + 9d) = 390.Step 5: Divide both sides by 5 to get 2a + 9d = 78.Step 6: Now we have a system of equations: (1) a + 2d = 19 and (2) 2a + 9d = 78.Step 7: From equation (1), express a in terms of d: a = 19 - 2d.Step 8: Substitute a = 19 - 2d into equation (2): 2(19 - 2d) + 9d = 78.Step 9: Simplify: 38 - 4d + 9d = 78, so 38 + 5d = 78.Step 10: Subtract 38 from both sides: 5d = 40, so d = 8.Step 11: Substitute d = 8 back into a = 19 - 2d: a = 19 - 16 = 3.

Verification / Alternative check:
We can quickly check by constructing the series. With a = 3 and d = 8, the first few terms are: 3, 11, 19, 27, 35, 43, 51, 59, 67, 75. The third term is 19, as required. Now sum these 10 terms: (3 + 11 + 19 + 27 + 35) + (43 + 51 + 59 + 67 + 75). The first group sums to 95, the second group also sums to 295, giving a total of 390. This confirms the correctness of a = 3.

Why Other Options Are Wrong:
Option 5: If a = 5, we would not be able to get T3 = 19 and S10 = 390 with a single common difference.
Option 7: Leads to inconsistent values for the sum when combined with T3 = 19.
Option 8: Would make T3 too large relative to 19 for any reasonable d.
Option 1: Would force d to be unrealistically large to reach T3 = 19, and the sum would not match 390.

Common Pitfalls:
A common error is misusing the sum formula or writing S10 as 10 * (a + d) instead of 10 / 2 * (2a + 9d). Another pitfall is trying to guess the first term and common difference without forming equations, which is inefficient and error-prone. Careful substitution and stepwise algebra are the best way to handle such problems.

Final Answer:
The first term of the arithmetic series is 3.