# Question #88190

Feb 6, 2017

$13 + \left(n - 1\right) \cdot \left(- \frac{5}{3}\right)$

#### Explanation:

I think this is an arithmetic progression. Therefore,
${T}_{n} = a + \left(n - 1\right) d$, where $a$ = first term, $n$ = number of term and $d$=common different.

$13 + 27 \cdot \left(- \frac{5}{3}\right) = 13 + \left(28 - 1\right) \cdot \left(- \frac{5}{3}\right) = {T}_{28}$

Therefore we can say that,
$a = 13 , d = \left(- \frac{5}{3}\right)$

so, ${T}_{n} = 13 + \left(n - 1\right) \cdot \left(- \frac{5}{3}\right)$