Question

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

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

Hint:

For multiple-choice questions on general terms, you can quickly test $n = 1$ in the options:
For Option (A): $3(1) - 4 = -1 \neq -\frac{1}{3}$.
For Option (B): $1 - \frac{4}{3} = -\frac{1}{3}$ (Matches first term).
For Option (C): $\frac{1 - 2}{3} = -\frac{1}{3}$ (Matches first term).
Now test $n = 2$ for Option (B) and (C):
For Option (B): $2 - \frac{4}{3} = \frac{2}{3}$ (Matches second term).
For Option (C): $\frac{2 - 2}{3} = 0 \neq \frac{2}{3}$.
This substitution method is extremely fast and reliable.

Answer 1

The Correct Option is B

Step 1: Look only at the numerators, since every term already has denominator 3.
Write each term with denominator 3: $\frac{-1}{3}, \frac{2}{3}, \frac{5}{3}, \frac{8}{3}, \ldots$. The numerators themselves are $-1, 2, 5, 8, \ldots$, which form their own simple arithmetic progression.

Step 2: Find the first term and common difference of this numerator sequence.
First term of numerators: $a = -1$.
Common difference: $d = 2 - (-1) = 3$.

Step 3: Write the nth-term formula for the numerators alone.
\[ \text{numerator}_n = a + (n-1)d = -1 + (n-1)(3) = -1 + 3n - 3 = 3n - 4 \]

Step 4: Put this numerator back over the denominator 3 to get the nth term of the original sequence.
\[ a_n = \frac{3n - 4}{3} \]

Step 5: Split the fraction to match the answer format.
\[ a_n = \frac{3n}{3} - \frac{4}{3} = n - \frac{4}{3} \]

Final Answer:
The nth term is $n - \frac{4}{3}$, which matches Option (B). \[ \boxed{a_n = n - \dfrac{4}{3}} \]