Question

How do you write the first five terms of the arithmetic sequence given `a_1=5/8, a_(k+1)=a_k-1/8` and find the common difference and write the nth term of the sequence as a function of n?

Answer 1

Common difference is `-1/8` and `n^(th)` term is `(6-n)/8`. First five terms are `{5/8,1/2,3/8,1/4,1/8}`.

Explanation:

Common difference is the difference between a term and it's immediately preceding term. In other words as first term is `a_1` and common difference is `d`, next term is `a_1+d` and then `a_1+2d`. Observe that this way `a_n=a_1+(n-1)×d`.

Let us consider a term `a_k`, ita next term would be `a_(k+1)` and common difference is

`a_(k+1)-a_k`

but as in given case `a_(k+1)=a_k-1/8`,
`a_(k+1)-a_k=-1/8`. Hence common difference is `-1/8`.

As first term is `5/8`, we have `a_2=5/8-1/8=4/8=1/2`, `a_3=4/8-1/8=3/8`, `a_4=3/8-1/8=2/8=1/4`, `a_5=2/8-1/8=1/8`. Hence first five terms are `{5/8, 1/2, 3/8, 1/4, 1/8}`.

and `n^(th)` term is given by `a_n=5/8+(n-1)×(-1/8)`

= `5/8-n/8+1/8`

= `(6-n)/8`