# How do you write the first five terms of the sequence defined recursively a_1=6, a_(k+1)=a_k+2, then how do you write the nth term of the sequence as a function of n?

Oct 24, 2017

First five terms are $\left\{6 , 8 , 10 , 12 , 14\right\}$ and ${a}_{n} = 2 n + 4$

#### Explanation:

As ${a}_{k + 1} = {a}_{k} + 2$ and${a}_{1} = 6$

${a}_{2} = {a}_{1} + 2 = 6 + 2 = 8$

${a}_{3} = {a}_{2} + 2 = 8 + 2 = 10$

${a}_{4} = {a}_{3} + 2 = 10 + 2 = 12$

and ${a}_{5} = {a}_{4} + 2 = 12 + 2 = 14$

and hence first five terms are $\left\{6 , 8 , 10 , 12 , 14\right\}$

As ${a}_{k + 1} = {a}_{k} + 2$, each term is $2$ more than previous term

it is an arithmetic sequence with first term as ${a}_{1}$ and common difference $d$ and hence ${n}^{t h}$ term is

${a}_{n} = {a}_{1} + \left(n - 1\right) d$ and hence ${n}^{t h}$ term of the sequence is

${a}_{n} = 6 + \left(n - 1\right) \times 2 = 6 + 2 n - 2 = 2 n + 4$