Answer:
n=9 and n = 12
Explanation:
an = a1 +d(n-1) is the formula for an arithmetic series
The first term is a1 and is 2.5
The 3rd term is
2 =a1 + d(3-1)
2 = 2.5 + 2d
Solve for d
Subtract 2.5 from each side
2-2.5 = 2.5 -2.5 +2d
-.5 = 2d
Divide by 2
-.5/2 = 2d/2
-.25 =d
an = 2.5 -.25(n-1)
The sum is found by using
Sn = n/2 (a1+an)
Sn = 13.5
Substituting the formula for an)
13.5 = n/2 (2.5 + 2.5 -.25(n-1))
Multiply each side by 2 to get rid of the fraction
2*13.5 = 2* n/2 (2.5 + 2.5 -.25(n-1))
27 = n*(2.5 + 2.5 -.25(n-1))
Combine like terms
27 = n*(5- .25(n-1))
Distribute inside the parentheses
27 = n(5-.25n +.25)
Combine like terms
27 = n(5.25-.25n)
Distribute
27 = 5.25n - .25n^2
Subtract 27 on each side
27-27 = 5.25 n - .25 n^2 -27
0 = 5.25 n - .25 n^2 -27
Multiply by -4 so that the coefficient on the x^2 term is 1
0 =-21n +n^2 +108
Rewrite in standard order
0= n^2 -21n +108
Factor
0=(n-9) (n-12)
Using the zero product property
n = 9 and n=12
It Sums to 13.5 twice
When n= 9 and when n=12
Check:
a1 =2.5 a2 =2.25, a3 = 2 a4 = 1.75 a5 = 1.5 a6 = 1.25 a7 = 1 a8=.75
a9 = .5 a10 =.25 a11 =0 a12 = -.25