Question

\(n^{\text{th}}\) term of the A.P. : \(-\frac{1}{3}, \frac{4}{3}, 3, \ldots\) is

• A. \(\frac{5n - 9}{3}\)
• B. \(\frac{5n - 6}{3}\)
• C. \(\frac{3n - 4}{3}\)
• D. \(\frac{3n + 2}{3}\)

Hint:

An easy way to double check your answer is to substitute \(n = 1\) and \(n = 2\) into the final expression: - For \(n = 1\): \(\frac{5(1) - 6}{3} = -\frac{1}{3}\) (matches the first term).
- For \(n = 2\): \(\frac{5(2) - 6}{3} = \frac{4}{3}\) (matches the second term).
This confirms the result is correct!

Answer 1

The Correct Option is B

Step 1: Identify the first term and common difference.
The A.P. is \(-\frac{1}{3}, \frac{4}{3}, 3, \ldots\)
First term: \(a = -\frac{1}{3}\). Common difference: \(d = \frac{4}{3} - \left(-\frac{1}{3}\right) = \frac{5}{3}\).
Step 2: Recall the formula for the nth term.
\[ T_n = a + (n - 1)d \]
Step 3: Substitute values.
\[ T_n = -\frac{1}{3} + (n-1) \cdot \frac{5}{3} \]
Step 4: Expand and simplify.
\[ T_n = \frac{-1 + 5(n-1)}{3} = \frac{-1 + 5n - 5}{3} = \frac{5n - 6}{3} \]
Step 5: Verify with known terms.
For \(n = 1\): \(T_1 = \frac{5-6}{3} = \frac{-1}{3}\). Correct! For \(n = 2\): \(T_2 = \frac{10-6}{3} = \frac{4}{3}\). Correct!
Step 6: Select the correct option.
The nth term is \(\frac{5n-6}{3}\), which matches option 2.
\[ \boxed{\dfrac{5n-6}{3}} \]