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

The given Arithmetic Progression (A.P.) is \( \frac{13}{3}, \frac{9}{3}, \frac{5}{3}, \dots \) First term, \( a = \frac{13}{3} \). Common difference, \( d = \frac{9}{3} - \frac{13}{3} = -\frac{4}{3} \). The \(n\)-th term of an A.P. is given by \( a_n = a + (n-1)d \). We need to find the 15th term, so \( n = 15 \). \[ a_{15} = \frac{13}{3} + (15-1)\left(-\frac{4}{3}\right) \] \[ a_{15} = \frac{13}{3} + 14\left(-\frac{4}{3}\right) \] \[ a_{15} = \frac{13}{3} - \frac{56}{3} \] \[ a_{15} = \frac{13 - 56}{3} = \frac{-43}{3} \] \[ \boxed{-\frac{43}{3}} \]