Question 1087344


 
recall:
If the common difference is {{{d}}}, then to get the next term, 
we just add d each time:

{{{3/5}}}, {{{3/5+d}}}, {{{3/5+2d}}}, {{{5/3}}}

use given terms and general formula to find {{{d}}}:

{{{a[n]=a[1]+(n-1)d}}}........since first term is {{{3/5}}} and fourth term is {{{5/3}}}

{{{a[4]=a[1]+(4-1)d}}}

{{{5/3=3/5+3d}}}

{{{3d=5/3-3/5}}}

{{{3d=25/15-9/15}}}

{{{3d=16/15}}}

{{{d=16/45}}}


{{{3/5}}},{{{ 3/5+16/45}}},{{{3/5+2(16/45)}}}, {{{5/3}}}

{{{3/5}}}, {{{43/45}}},{{{59/45}}},{{{ 5/3}}}