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

15th term of the A.P $\frac{13}{3}, \frac{9}{3}, \frac{5}{3}, .............$ is
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Answers 1

The given arithmetic progression (A.P.) is:

$\frac{13}{3}, \frac{9}{3}, \frac{5}{3}, \dots$

First term ($a$):
$a = \frac{13}{3}$

Common difference ($d$):
$d = \frac{9}{3} - \frac{13}{3} = \frac{-4}{3}$

The general formula for the $n$th term of an A.P. is:

$a_n = a + (n-1) d$

Substituting $n = 15$:

$a_{15} = \frac{13}{3} + (15-1) \times \frac{-4}{3}$

$= \frac{13}{3} + 14 \times \frac{-4}{3}$

$= \frac{13}{3} - \frac{56}{3}$

$= \frac{13 - 56}{3}$

$= \frac{-43}{3}$
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