Question 551085
My best guess is that your textbook or class notes tell you that the nth term of an arithmetic sequence, {{{a[n]}}}, can be calculated based on the first term, {{{a[1]}}}, and the common difference, {{{d}}}, as
{{{a[n]=a[1]+(n-1)*d}}}.
Then they must have told you that the sum of the first n terms can be calculated as
{{{S[n]=n*(a[n]+a[1])/2}}} or
{{{S[n]=n*a[1]+n(n-1)*d/2}}}
SOLUTIONS FOR THE FIRST TWO PROBLEMS
45. 3+8+13+18+23+...
{{{a[1]=3}}} and the common difference is
{{{d=8-3=5}}}
a. n=20 Since you are not given {{{a[20]}}}, use the second formula
{{{S[20]=20*3+20(20-1)*5/2=60+20*19*5/2=60+950=1010}}}
b. Sn=366 I don't think I ever did this kind of problem. 
I suppose they could expect you to use a guess and check strategy, and you could try n=10, and figure out that the answer is between 10 and 20.
Otherwise, it gets complicated, but here it goes:
With {{{S[n]=n*a[1]+n(n-1)*d/2}}} plus {{{S[n]=366}}} for this sequence, it means
{{{366=n*3+n(n-1)*5/2}}}, and multiplying both sides times 2
{{{732=6n+n(n-1)*5}}} --> {{{732=6n+(n^2-n)*5}}} --> {{{732=6n+5n^2-5n}}} --> {{{5n^2+n-732=0}}}
Applying the quadratic formula:
 {{{n = (-1 +- sqrt( 1^2-4*5*(-732) ))/(2*5)=(-1 +- sqrt( 1+14640))/10=(-1 +- sqrt(14641))/10=(-1 +- 121)/10}}}
That gives us {{{n=12}}}, our answer.
We disregard the other solution because it's negative, and we were looking for a positive integer for {{{n}}}.
46. 50+42+34+26+18+...
has {{{a[1]=50}}} and {{{d=42-50=-8}}}.
That means {{{a[7]=a[1]+(7-1)*(-8)=50-48=2}}}, and {{{a[8]=a[7]-8=-6}}}, with all the terms after that being negative.
What's more the sum of the positive terms (the first 7 terms) is
{{{S[7]=7*(a[7]+a[1])/2=7*(2+50)/2=7*52/2=182}}}, which answers part b by lucky guess. 
a. n=40
{{{S[40]=40*50+40(40-1)*(-8)/2=40*50+40*39*(-8)/2=2000-6240=-4240}}}
b. Sn=182 We answered it by guess above, but if we need to solve quadratic equations again:
{{{S[n]=n*a[1]+n(n-1)*d/2}}}, so {{{182=n*50+n(n-1)*(-8)/2}}}
So, {{{182=50n-4n^2+4n}}} --> {{{182=50n-4n^2+4n}}} --> {{{182=54n-4n^2}}} --> {{{4n^2-54n+182=0}}} --> {{{2n^2-27n+91=0}}}
If you are good at factoring you see that  {{{2n^2-27n+91=(2n-13)(n-7)}}} to find the solutions to the equation. Otherwise, you apply the quadratic formula. Either way, you find the answer to part b {{{n=7}}},
and another solution to the equation, {{{13/2}}}, that we discard because it's not a positive integer.
FOR THE REST
I hope you know how to do it now. I also hope that guess and check is an accepted and viable alternative to find the answer to part b.
Here are the answers I calculated for problems 47-50
47. -10+(-5)+0+5+10+...
a. n=19 {{{S[19]=665}}}
b. Sn=375  {{{n=15}}} {{{S[15]=375}}}
48. 34+31+28+25+22+...
a. n=32 {{{S[32]=-400}}}
b. Sn=-12 {{{n=24}}} {{{S[24]=-12}}}
49. 2+9+16+23+30+...
a. n=68 {{{S[68]=16082}}}
b. Sn=1661 {{{n=22}}} {{{S[22]=1661}}}
50. 2+16+30+44+58+...
a. n=24 {{{S[32]=3912}}}
b. Sn=2178 {{{n=18}}} {{{S[18]=2178}}}