Question:medium

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

Updated On: Jan 13, 2026
  • 23
  • \( -\frac{53}{3} \)
  • \( -11 \)
  • \( -\frac{43}{3} \)
Show Solution

The Correct Option is D

Solution and Explanation

Given:
Arithmetic progression (A.P.):
\[ \frac{13}{3}, \frac{9}{3}, \frac{5}{3}, \dots \] Find the 15th term of this A.P.

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

Step 2: Use the formula for the \(n\)th term
The \(n\)th term \(a_n\):
\[ a_n = a + (n - 1)d \] 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:
\[ \boxed{-\frac{43}{3}} \]
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