In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Find the sum of all ten terms.

\(a_{n+2} = a_{n+1}+a_n,~n\geq 3\\ s = \sum \limits_{n=1}^{10} a_n = 55 a1 + 88 a2\\ a_7 = 5a_1+8a_2 = 6\\ a_2 = \dfrac{6-5a_1}{8}\\ s = 55a_1 + 88\dfrac{6-5a_1}{8} = 55a_1 + 66-55a_1 = 66\)