Question

15th term of the A.P. \( \frac{13}{3}, \frac{9}{3}, \frac{5}{3}, \dots \) is

• A. 23
• B. \( -\frac{53}{3} \)
• C. \( -11 \)
• D. \( -\frac{43}{3} \)

Answer 1

The Correct Option is D

Given:
An arithmetic progression (A.P.):
\[ \frac{13}{3}, \frac{9}{3}, \frac{5}{3}, \dots \] We need to find the 15th term of this A.P.

Step 1: Identify the first term \(a\) and common difference \(d\)
First term:
\[ a = \frac{13}{3} \] Second term:
\[ a_2 = \frac{9}{3} = 3 \] Calculate common difference \(d\):
\[ d = a_2 - a = 3 - \frac{13}{3} = \frac{9}{3} - \frac{13}{3} = -\frac{4}{3} \]

Step 2: Use formula for \(n\)th term of an A.P.
The \(n\)th term \(a_n\) is given by:
\[ a_n = a + (n - 1)d \] Substitute \(n = 15\), \(a = \frac{13}{3}\), and \(d = -\frac{4}{3}\):
\[ a_{15} = \frac{13}{3} + (15 - 1) \times \left(-\frac{4}{3}\right) = \frac{13}{3} + 14 \times \left(-\frac{4}{3}\right) \] \[ = \frac{13}{3} - \frac{56}{3} = \frac{13 - 56}{3} = -\frac{43}{3} \]

Final Answer:
\[ \boxed{-\frac{43}{3}} \]