In the sequence s of numbers, each term after the first two terms is the sum of the two immediately preceding terms. what is the 5th term of s ? the 6th term of s minus the 4th term equals 5. the 6th term of s plus the 7th term equals 21.

Sn=n/2 (2a1 +(n-1)d)

[s5= 5/2 (2a1 +(5-1)d) = 5a1+10d]

s6=3(2a1+5d) = 6a1+15d
s4=2(2a1+3d) = 4a1+6d

6a1+15d -(4a1+6d) = 5 (1)

s7=7/2(2a1+6d) =7a1+21d

6a1+15d + 7a1+21d = 21 (2)

(1) 2a1+9d=5 (3)
(2) 13a1+36d=21 (4)

(3)*4 ; 8a1+36d=20 (5)
(4)-(5) 5a1 = 1
so a1 = 1/5

(3) 2a1+9d=5
2(1/5)+9d = 5
9d = 5-2/5 =23/5
d = 23/45

so [s5= 5/2 (2a1 +(5-1)d) = 5a1+10d = 5(1/5) + 10(23/45) = 1 + 5.11 = 6.11 ]

i'm not sure about the answer but the process is correct