Question 1102476
WITHOUT MEMORIZED FORMULAS:
{{{a[n]=a[n-1]-3}}} or {{{highlight(a[n+1]=a[n]-3)}}}
would be the recursive rule spelled out in
"each term is 3 less than the previous term."
Then {{{n-5}}} terms after the 5th term, we would have
subtracted from 25 {{{n-5}}} times 3, so
{{{a[n]=25-3(n-5)}}} , which simplifies to
{{{a[n]=25-3n-5}}} and to the simplest explicit rule
{{{a[n]=20-3n}}} .
 
Your teacher may have in mind a lot of formula memorizing and writing, like
{{{d=-3}}}= common difference.
{{{a[n+1]=a[n]+d}}} is the memorizable recursive rule formula for any arithmetic sequence,
so in this case {{{a[n+1]=a[n]+(-3)}}} ,
or {{{highlight(a[n+1]=a[n]-3)}}} .
For all arithmetic sequences,
{{{a[n]=a[1]+(n-1)d}}} is the memorizable explicit rule for any arithmetic sequence.
For {{{n=5}}} , with {{{a[5]=25}}} and {{{d=-3}}}, we can find {{{a[1]}}} :
{{{a[5]=a[1]+(5-1)(-3)}}}
{{{25=a[1]+4(-3)}}}
{{{25=a[1]-12}}}
{{{25+12=a[1]}}}
{{{a[1]=37}}} .
Plugging {{{a[1]=37 and {{{d=-3}}}
into the memorized {{{a[n]=a[1]+(n-1)d}}} , we get explicit rule
{{{highlight(a[n]=37-3(n-1))}}} ,
not that simple, but maybe the expected answer.