Question

The n\(^{th}\) term of the A.P. \(-\frac{1}{3}\), \(\frac{2}{3}\), \(\frac{5}{3}\), \(\frac{8}{3}\), ... is :

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

Hint:

An easy way to verify your answer is to substitute \( n = 1 \) and \( n = 2 \) into the options:
- For option (B), substituting \( n = 1 \) gives \( 1 - \frac{4}{3} = -\frac{1}{3} \), which matches the first term.
- Substituting \( n = 2 \) gives \( 2 - \frac{4}{3} = \frac{2}{3} \), which matches the second term.
This substitution test can be completed very quickly and confirms the correct answer without any risk of algebraic errors.

Answer 1

The Correct Option is B

Step 1: Understand the Question:
This question is from Arithmetic Progressions.
We are given the AP $-\frac{1}{3}, \frac{2}{3}, \frac{5}{3}, \frac{8}{3}, \ldots$ and need its $n^{th}$ term.
Instead of working with the fractions all the way through, we will pull out the common denominator first and study just the numerators.

Step 2: Write every term over the same denominator 3.
The terms are already all written over 3:
\[ -\frac{1}{3}, \frac{2}{3}, \frac{5}{3}, \frac{8}{3}, \ldots \]
So the numerators, taken on their own, form the sequence:
\[ -1, 2, 5, 8, \ldots \]

Step 3: Recognise that the numerators form their own simple AP.
Check the gap between consecutive numerators:
\[ 2 - (-1) = 3, \quad 5 - 2 = 3, \quad 8 - 5 = 3 \]
So the numerator sequence is an AP with first term $-1$ and common difference $3$.
Its $n^{th}$ term is
\[ (\text{numerator})_n = -1 + (n-1)(3) = -1 + 3n - 3 = 3n - 4 \]

Step 4: Divide back by 3 to get the $n^{th}$ term of the original AP.
Since every term of the original AP is (numerator)/3,
\[ a_n = \frac{3n - 4}{3} = \frac{3n}{3} - \frac{4}{3} = n - \frac{4}{3} \]

Final Answer:
The $n^{th}$ term of the AP is $n - \frac{4}{3}$, which matches option (B).
\[ \boxed{a_n = n - \frac{4}{3}} \]