# How do you find the first five terms given a_1=-6 and a_(n+1)=a_n+3?

Nov 2, 2016

First five terms are $\left\{- 6 , - 3 , 0 , 3 , 6\right\}$

#### Explanation:

As ${a}_{n + 1} = {a}_{n} + 3$ and ${a}_{1} = - 6$

${a}_{2} = {a}_{1 + 1} = {a}_{1} + 3 = - 6 + 3 = - 3$

${a}_{3} = {a}_{2 + 1} = {a}_{2} + 3 = - 3 + 3 = 0$

${a}_{4} = {a}_{3 + 1} = {a}_{3} + 3 = 0 + 3 = 3$

${a}_{5} = {a}_{4 + 1} = {a}_{4} + 3 = 3 + 3 = 6$

Hence, first five terms are $\left\{- 6 , - 3 , 0 , 3 , 6\right\}$

Nov 2, 2016

The first five terms of the arithmetic progression are $- 6 , - 3 , 0 , + 3 , + 6.$

#### Explanation:

${a}_{n + 1} = {a}_{n} + 3.$ [GIVEN].
$\therefore {a}_{n + 1} - {a}_{n} = 3.$

$\therefore$When, $n = 1 ,$
${a}_{2} - {a}_{1} = 3.$
$\therefore {a}_{2} + 6 = 3.$
$\therefore {a}_{2} = - 3.$

$\therefore$Common Difference $\left(d\right) = {a}_{2} - {a}_{1} = \left(- 3\right) - \left(- 6\right) = + 3.$

$\therefore {a}_{1} = - 6 , {a}_{2} = - 3 , {a}_{3} = 0 , {a}_{4} = + 3 , {a}_{5} = + 6.$

Therefore, the first five terms of the arithmetic progression are $- 6 , - 3 , 0 , + 3 , + 6.$ (answer).