nth term of the A.P.: – 1/3, 4/3, 3 ,… is

n^th term of the A.P.: – 1/3, 4/3, 3 ,…in
(A) (5n – 9)/3
(B) (5n – 6)/3
(C) (3n – 4)/3
(D) (3n + 2)/3

Solution : Correct answer is : (B) (5n – 6)/3

Given A.P.:$$-\frac{1}{3}, \; \frac{4}{3}, \; 3, \dots$$Identify first term (a)

$$a = -\frac{1}{3}$$Find common difference (d)

$$d = \frac{4}{3} – \left(-\frac{1}{3}\right)$$$$d = \frac{4}{3} + \frac{1}{3} = \frac{5}{3}$$

Check with next terms:$$3 – \frac{4}{3} = \frac{9}{3} – \frac{4}{3} = \frac{5}{3}$$So,$$d = \frac{5}{3}$$Use nth term formula of A.P.

$$a_n = a + (n-1)d$$an​=a+(n−1)d

Substitute values:$$a_n = -\frac{1}{3} + (n-1)\frac{5}{3}$$Simplify

$$a_n = -\frac{1}{3} + \frac{5(n-1)}{3}$$an​=−31​+35(n−1)​

Take common denominator 3:$$a_n = \frac{-1 + 5(n-1)}{3}$$an​=3−1+5(n−1)​

Now expand:$$= \frac{-1 + 5n – 5}{3}$$$$= \frac{5n – 6}{3}$$

Final Answer:

$$\boxed{\frac{5n – 6}{3}}$$