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?

Aug 12, 2017

Common difference is $- \frac{1}{8}$ and ${n}^{t h}$ term is $\frac{6 - n}{8}$. First five terms are $\left\{\frac{5}{8} , \frac{1}{2} , \frac{3}{8} , \frac{1}{4} , \frac{1}{8}\right\}$.

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 ${a}_{1} + 2 d$. 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 ${a}_{k + 1} = {a}_{k} - \frac{1}{8}$,
${a}_{k + 1} - {a}_{k} = - \frac{1}{8}$. Hence common difference is $- \frac{1}{8}$.

As first term is $\frac{5}{8}$, we have ${a}_{2} = \frac{5}{8} - \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$, ${a}_{3} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}$, ${a}_{4} = \frac{3}{8} - \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$, ${a}_{5} = \frac{2}{8} - \frac{1}{8} = \frac{1}{8}$. Hence first five terms are $\left\{\frac{5}{8} , \frac{1}{2} , \frac{3}{8} , \frac{1}{4} , \frac{1}{8}\right\}$.

and ${n}^{t h}$ term is given by a_n=5/8+(n-1)×(-1/8)

= $\frac{5}{8} - \frac{n}{8} + \frac{1}{8}$

= $\frac{6 - n}{8}$