# How do you find the formula for a_n for the arithmetic sequence a_3=94, a_6=85?

May 5, 2018

$n$ th term is ${a}_{n} = 100 - 3 \left(n - 1\right)$

#### Explanation:

$3$ rd term a_3= a_1+2 d =94; (1); a_1 and d  are first term

and common difference of A.P sequence.

$6$ th term a_6= a_1+5 d =85 ; (2); . Subtracting equation (1)

from equation (2) we get, $3 d = - 9 \therefore d = - \frac{9}{3} = - 3$ Putting

$d = - 3$ in equation (1) we get, ${a}_{1} + 2 \cdot \left(- 3\right) = 94$ or

${a}_{1} = 94 + 6 = 100 \therefore$ first term ${a}_{1} = 100$ Therefore,

$n$ th term is ${a}_{n} = {a}_{1} + \left(n - 1\right) d \mathmr{and} {a}_{n} = 100 - 3 \left(n - 1\right)$ [Ans]

May 5, 2018

${a}_{n} = 103 - 3 n$

#### Explanation:

Since this is an arithmetic sequence:
for some constant $k$
$\textcolor{w h i t e}{\text{XXX}} {a}_{6} = \textcolor{b l u e}{{a}_{5}} + k$
$\textcolor{w h i t e}{\text{XXX"a_6)=color(blue)(} \left(\textcolor{\lim e}{{a}_{4}} + k\right)} + k$
color(white)("XXX"a_6)=color(blue)(""(color(lime)(""(a_3+k))+k))+k#
$\textcolor{w h i t e}{\text{XXX} {a}_{6}} = {a}_{3} + 3 k$

since ${a}_{6} = 85$ and ${a}_{3} = 64$
$\textcolor{w h i t e}{\text{XXX}} 3 k = 85 - 94 = - 9$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow k = - 3$

Similarly
$\textcolor{w h i t e}{\text{XXX}} {a}_{1} = {a}_{0} + k$
$\textcolor{w h i t e}{\text{XXX}} {a}_{2} = {a}_{0} + 2 k$
$\textcolor{w h i t e}{\text{XXX}} {a}_{3} = {a}_{0} + 3 k$
and
since ${a}_{3} = 94$ and $k = - 3$
$\textcolor{w h i t e}{\text{XXX}} 94 = {a}_{0} + 3 \cdot \left(- 3\right) = {a}_{0} - 9$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow {a}_{0} = 103$

The general arithmetic formula is
$\textcolor{w h i t e}{\text{XXX}} {a}_{n} = {a}_{0} + k \cdot n$
So, for this specific case,
$\textcolor{w h i t e}{\text{XXX}} {a}_{n} = 103 + \left(- 3\right) n$