In an arithmetic sequence:
Tn=t₁+(n-1)d
t₄=t₁+(4-1)d=t₁+3d
t₅=t₁+(5-1)d=t₁+4d
t₆=t₁+(6-1)d=t₁+5d
t₄+t₅+t₆=(t₁+3d) +(t₁+4d)+(t₁+5d)=3t₁+12d
Therefore:
3t₁+12d=300 (1)
t₁₅=t₁+(15-1)d=t₁+14d
t₁₆=t₁+(16-1)d=t₁+15d
t₁₇=t₁+(17-1)d=t₁+16d
t₁₅+t₁₆+t₁₇=(t₁+14d)+(t₁+15d)+(t₁+16d)=3t₁+45d
Therefore:
3t₁+45d=201 (2)
With the equations (1) and (2) we make an system of equations:
3t₁+12d=300
3t₁+45d=201
we can solve this system of equations by reduction method.
3t₁+12d=300
-(3t₁+45d=201)
-----------------------------
-33d=99 ⇒d=99/-33=-3
3t₁+12d=300
3t₁+12(-3)=300
3t₁-36=300
3t₁=300+36
3t₁=336
t₁=336/3
t₁=112
Threfore:
Tn=112+(n-1)(-3)
Tn=112-3n+3
Tn=115-3n
Now, we calculate T₁₈:
T₁₈=115-3(18)=115-54=61
Answer: T₁₈=61
