Question

the 8th term of AP 12,8,4,0....is​

Answer 1

Step-by-step explanation:

$a = 12 \\ d = 8 - 12= - 4 \\ tn = a + (n - 1)d \\ t8 = 12 + (8 - 1) ( - 4) \\ t8 = 12 - (7)(4) \\ t8 = - 16$

Answer 2

Given:

AP: $12,8,4,0,------$

To find: the 8th term of the given AP.

Solution:

Observe from the given AP that,

$\[\begin{array}{l}a = 12\\d = 8 - 12\end{array}\]$

  $\[ = - 4\]$

$\[n = 8\]$

Find the 8th term of the AP.

$\[\begin{array}{l}{t_n} = a + \left( {n - 1} \right)d\\ \Rightarrow {t_8} = 12 + \left( {8 - 1} \right)\left( { - 4} \right)\\ \Rightarrow {t_8} = 12 + \left( 7 \right)\left( { - 4} \right)\\ \Rightarrow {t_8} = 12 - 28\\ \Rightarrow {t_8} = - 16\end{array}\]$

Hence, the 8th term of the AP is $-16.$