Question 662461
The first part of your work is right.
term number {{{17}}} is {{{53}}} --> {{{t[17]=53}}}
term number {{{n}}} is {{{t[n]=a+(n-1)d}}}
{{{53=a+(17-1)d}}}  --> {{{53=a+16d}}} <--> {{{a=53-16d}}}
You were on your way to finding {{{a}}} and {{{d}}}.
You just needed to use more data to get {{{a}}} and {{{d}}}.
 
Next, you misinterpreted "the 28th term is 86" (or maybe lost track of your planned steps)
You wrote {{{S[n]=(n\2)(2a+(n-1)d)}}} , the formula for sum of the first {{{n}}} terms,
followed by {{{86=(28/2)(2a+(28-1)d)}}} as if the sum of the first 28 terms was 86.
You knew that the next step would require using that formula for the sum,
but lost track, and skipped a step.
 
At that point, a bunch of people would write some of this:
term number {{{28}}} is {{{86}}} --> {{{t[28]=86}}}
term number {{{n}}} is {{{t[n]=a+(n-1)d}}}
{{{86=a+(28-1)d}}}  --> {{{86=a+27d}}}
Then, they would use {{{86=a+27d}}}, along with {{{53=a+16d}}} to find {{{a}}} and {{{d}}}.
They would write the system {{{system(86=a+27d,53=a+16d)}}}
and would solve it to get {{{highlight(a=5)}}} and {{{highlight(d=3)}}}.
 
RANT:
I may write all that to go with the flow, and please a teacher,
but I prefer to use definitions, logic and 5th grader reasoning (if possible),
trying to always understand the meaning of my calculations,
rather than using formulas.
I look at it the fifth grader way:
Going from the 17th to the 28th term, we are adding {{{d}}} from one term to the next,
and we do that {{{28-17=11}}} times,
so the difference between the 17th and the 28th terms must be {{{11}}} times {{{d}}}.
The difference is {{{86-53=33}}}, so I divide to get {{{d=33/11=highlight(3)}}}
Then, since you gave me {{{a=53-16d}}},
I use it to get {{{a=53-16*3}}} --> {{{a=53-48}}} --> {{{highlight(a=5)}}}.
 
THE SUM:
{{{S[n]=(n\2)(2a+(n-1)d)}}}
{{{S[50]=(50\2)(2*5+(50-1)*3)}}} --> {{{S[50]=(50\2)(10+49*3)}}} --> {{{S[50]=(50\2)(10+147)}}} --> {{{S[50]=(50\2)(157)}}} --> {{{highlight(S[50]=3925)}}}.