In an arithmetic sequence, t4 + t5 + t6 = 300, and t15 +t16+ t17 = 201. find t18.

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

T₁₈=61