Question 980165

Which term of the series {{{79}}},{{{77}}},{{{75}}}___ is {{{47}}}.
find out the {{{a[n]}}}

as you can see, the differences between each pair of numbers are always {{{2}}}, each next term is {{{2}}} less then previous one

    {{{a = 79}}} (the first term)
    {{{d = 2}}} (the "common difference" between terms)

And we get:

{ {{{a}}}, {{{a-d}}}, {{{a-2d}}}, {{{a-3d}}}, ... }

We can write an arithmetic sequence as a rule:

{{{a[n] = a - d(n-1)}}}

(We use "{{{n-1}}}" because {{{d }}}is not used in the 1st term).

check the formula:

{{{a[1] =79 - 2(1-1)=79}}}
{{{a[2] =79 - 2(2-1)=79 - 2(1)=79-2=77}}}
{{{a[3] =79 - 2(3-1)=79 - 2(2)=79-4=75}}}

calculate which term is equal to {{{47}}}:

{{{47= 79 - 2(n-1)}}}
{{{47= 79 - 2n+2}}}
{{{47= 81 - 2n}}}
{{{2n= 81 - 47}}}
{{{2n= 34}}}
{{{n= 34/2}}}
{{{n= 17}}}

so, {{{47}}} is {{{17th}}} term

check:

{{{79}}},{{{77}}},{{{75}}},{{{73}}},{{{71}}},{{{69}}},{{{67}}},{{{65}}},{{{63}}},{{{61}}},{{{59}}},{{{57}}},{{{55}}},{{{53}}},{{{51}}},{{{49}}},{{{highlight(47)}}}