Question 983172

First find a formula for the general term.


{{{1}}},{{{4/5}}},{{{2/3}}},{{{4/7}}},... {{{n=12}}}

you can write first term {{{1}}} as {{{4/4}}} and third term {{{2/3}}} as {{{2/6}}}, so your sequence will be:

{{{4/4}}},{{{4/5}}},{{{4/6}}},{{{4/7}}}



in your case, numerators are same and equal to {{{4}}}
as denominators, you have {{{4}}, {{{5}}}, {{{6}}}, {{{7}}},...

to get first term

{{{4/(n+d)=1}}} where {{{n=1}}} you have {{{4/(1+d)=1}}}=> {{{d}}} must be equal to {{{3}}}
{{{4/(1+3)=1}}}=>{{{4/4=1}}}

so, general formula will be:

{{{a[n]= 4/(n+3)}}} (for all terms given) where {{{n}}}={{{1}}},{{{2}}},{{{3}}},.....{{{n}}}

{{{a[n ]= 4/(n+3)}}} ...........if {{{n=12}}} we have

{{{a[12 ]= 4/(12+3)}}}

{{{a[12 ]= 4/15}}}