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.
The sum can be represented by :
a + (a + d) + ( 2a + d) + ( 3a + 2d) + (5a + 3d) + (8a + 5d) + (13a + 8d) + (21a + 13d) + (34a + 21d) + (55d + 34d) = 143a + 88d
And we know that the 7th term = 6, so
13a + 8d = 6 multiply through by 11
143a + 88d = 66
So....the sum is 66