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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
This Question has 1 answers.
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 $n = 15$:
$a_{15} = \frac{13}{3} + (15-1) \times \frac{-4}{3}$
$= \frac{13}{3} + 14 \times \frac{-4}{3}$
$= \frac{13}{3} - \frac{56}{3}$
$= \frac{13 - 56}{3}$
$= \frac{-43}{3}$
$\frac{13}{3}, \frac{9}{3}, \frac{5}{3}, \dots$
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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