There are two answers
The first answer is 8
The second answer is 15
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Work Shown:
t_1 = first term, t_2 = second term, etc etc
s_n = sum of the first n terms
t_1 = 11/3 ... is given
t_2 = t_1 + d ... add 'd' to the first term to get the second
t_3 = t_2 + d ... repeat last step to second term
t_3 = (t_1+d) + d ... substitution
t_3 = t_1+2d
t_3 = (11/3)+2d ... plug in t_1 = 11/3
t_3 = 3 ... this is given
The third term t_3 is both (11/3)+2d and 3. Equate the two expressions and solve for d.
(11/3)+2d = 3
2d = 3 - (11/3)
2d = (9/3) - (11/3)
2d = (9-11)/3
2d = -2/3
d = (-2/3)*(1/2)
d = -1/3
Turn to the nth term series formula for arithmetic sequences. Plug in the given and found values and isolate n
s_n = (n/2)*(t_1+t_n)
s_n = (n/2)*(t_1+t_1+d(n-1))
s_n = (n/2)*(2*t_1+d(n-1))
20 = (n/2)*(2*(11/3)+(-1/3)*(n-1))
20 = (n/2)*( (22/3)+(-1/3)(n-1))
2*20 = n*( (22/3)+(-1/3)(n-1) )
40 = n*( (22/3)+(-1/3)n + (-1/3)(-1) )
40 = n*( (22/3)+(-1/3)n + (1/3) )
40 = n*( (23/3)+(-1/3)n )
40 = (23/3)n+(-1/3)n^2
3*40 = 3*(23/3)n+3*(-1/3)n^2
120 = 23n - n^2
0 = 23n - n^2 - 120
-n^2 + 23n - 120 = 0
n^2 - 23n + 120 = 0
(n - 8)(n - 15) = 0
n-8 = 0 or n-15 = 0
n = 8 or n = 15
If you were to add up the first n = 8 terms, then s_n = 20
If you were to add up the first n = 15 terms, then s_n = 20