If a is the first term of an AP with common difference -2, then the first several terms are
a, a - 2, a - 4, a - 6, a - 8, …
with n-th term a - 2 (n - 1).
The sum of the first n terms is equal to the sum of the first 3n terms :
[tex]\displaystyle \sum_{i=1}^n (a-2(i-1)) = \sum_{i=1}^{3n} (a-2(i-1))[/tex]
We have
[tex]\displaystyle \sum_{i=1}^N (a-2(i-1)) = (a+2)\sum_{i=1}^N - 2\sum_{i=1}^Ni = (a+2)N - 2\cdot\dfrac{N(N+1)}2 = (a+1)N-N^2[/tex]
so that in the previous equation, the sums reduce to
[tex](a+1)n-n^2 = (a+1)(3n)-(3n)^2[/tex]
Solve for a :
[tex](a+1)n-n^2 = (a+1)(3n)-(3n)^2 \\\\ (a+1)n-n^2 = (a+1)(3n)-9n^2 \\\\ 9n^2-n^2 = (a+1)(3n)-(a+1)n \\\\ 8n^2 = (a+1)(2n) \\\\ 4n = a+1 \\\\ \boxed{a=4n-1}[/tex]
Now if a = 27, we have
27 = 4n - 1
28 = 4n
n = 7
as required.
