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

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Answer 1 · 2017-10-24T16:25:17

First five terms are ${6, 8, 10, 12, 14}$ ${6,8,10,12,14}$ and $a_{n} = 2 n + 4$ $a_n=2n+4$

As $a_{k + 1} = a_{k} + 2$ $a_(k+1)=a_k+2$ and $a_{1} = 6$ $a_1=6$

$a_{2} = a_{1} + 2 = 6 + 2 = 8$ $a_2=a_1+2=6+2=8$

$a_{3} = a_{2} + 2 = 8 + 2 = 10$ $a_3=a_2+2=8+2=10$

$a_{4} = a_{3} + 2 = 10 + 2 = 12$ $a_4=a_3+2=10+2=12$

and $a_{5} = a_{4} + 2 = 12 + 2 = 14$ $a_5=a_4+2=12+2=14$

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

As $a_{k + 1} = a_{k} + 2$ $a_(k+1)=a_k+2$ , each term is $2$ $2$ more than previous term

it is an arithmetic sequence with first term as $a_{1}$ $a_1$ and common difference $d$ $d$ and hence $n^{t h}$ $n^(th)$ term is

$a_{n} = a_{1} + (n - 1) d$ $a_n=a_1+(n-1)d$ and hence $n^{t h}$ $n^(th)$ term of the sequence is

$a_{n} = 6 + (n - 1) \times 2 = 6 + 2 n - 2 = 2 n + 4$ $a_n=6+(n-1)xx2=6+2n-2=2n+4$

How do you write the first five terms of the sequence defined recursively a_1=6, a_(k+1)=a_k+2a1=6,ak+1=ak+2a_1=6, a_(k+1)=a_k+2, then how do you write the nth term of the sequence as a function of n?