Question 1110298
In an A.P. the {{{n^th}}} term is {{{a[n]=a+(n-1)d}}} , where
{{{a}}}= the first term of the A.P. and
{{{d}}}= the common difference of the A.P.
 
So, the second, third, fourth, fifth and eight terms are respectively
{{{a[2]=a+d}}} ,
{{{a[3]=a+2d}}} ,
{{{a[4]=a+3d}}} ,
{{{a[5]=a+4d}}} and
{{{a[8]=a+7d}}} .
 
The sum of the third and fifth terms is
{{{a[3]+a[5]=a+2d+a+4d}}} , which simplifies to
{{{a[3]+a[5]=2a+6d}}} .
For this problem, that sum is {{{20}}} , so
{{{2a+6d=20}}} <--> {{{highlight(a+3d=10)}}} is one of our equations.
 
As the ratio of consecutive terms in a G.P. is always the same number,
if the  second, fourth and eight terms of an A.P.
form the first three consecutive terms of a G.P. ,
then {{{a[4]/a[2]}}}{{{"="}}}{{{a[8]/a[4]}}}=the common ratio of the G.P.
So, {{{(a+3d)/(a+d)}}}{{{"="}}}{{{(a+7d)/(a+3d)}}} ,
{{{(a+3d)^2}}}{{{"="}}}{{{(a+d)(a+7d)}}} ,
{{{a^2+6ad+9d^2}}}{{{"="}}}{{{a^2+8ad+7d^2}}} ,
{{{6ad+9d^2}}}{{{"="}}}{{{8ad+7d^2}}} ,
{{{9d^2-7d^2=8ad-6ad}}} ,
{{{2d^2=2ad}}} ,
{{{highlight(d=a)}}} .
 
Now, we have
{{{system(a+3d=10,a=d)}}} --> {{{system(d+3d=10,a=d)}}} --> {{{system(4d=10,a=d)}}} --> {{{system(d=2.5,a=d)}}} --> {{{highlight(system(d=2.5,a=2.5))}}} .
 
(a) Knowing the first term, {{{a}}} , and the common difference, {{{d}}} ,
we can write any terms of the A.P.
The first 4 terms are
{{{highlight("2.5 , 5 , 7.5 , 10")}}} .
 
(b) You may have been taught a "formula" for the sum the first {{{n}}} terms of an A.P.
that could be {{{S[n]=(2a[1]+(n-1)d)n/2}}} or something like that.
If a teacher likes to see formulas,
it is a good idea to write the formula the teacher prefers.
Using that formula, the sum of the first 10 terms calculates as
{{{S[10]}}}{{{"="}}}{{{(2*2.5+(10-1)2.5)10/2}}}{{{"="}}}{{{(5+9*2.5)10/2}}}{{{"="}}}{{{(5+22.5)10/2}}}{{{"="}}}{{{27.5*5}}}{{{"="}}}{{{highlight(137.5)}}} .
 
I do not memorize formulas, but I remember that
the average of a number of consecutive terms of an A.P. is the average of the first and last of those terms.
I can easily list the second , fourth, sixth, eighth, and ten terms as
5, 10, 15, 20, 25,
so I can calculate the average of the first 10 terms as 
{{{(2.5+25)/2}}} ,
and the sum of those {{{10}}} terms as
{{{(2.5+25)*10/2}}}{{{"="}}}{{{27.5*10/2}}}{{{"="}}}{{{275/2}}}{{{"="}}}{{{137.5}}} .