helpppppp

a(30) = 65-3 (30-1), solve for a(30)

a(30) = -22

50 = a₁ +  d(5)       →  50  = a₁ +    5d    (1)

35 = a₁ + d(10)      →  35  = a₁  + 10d     (2)

Subtract (2)  from (1)

15  =  -5d      divide both sides by -5

-3 = d

And....solving for a₁, we have

50  = a₁ - 3(5)

50 = a₁ - 15

65  = a₁

So

a₃₀  = 65  -3(29)  =   -22

an = a1 + d(n − 1)

50 =a1 + -3(6 - 1) solve for a1

a1=65

a30=65 + -3(30 - 1)

a30=-22

What is $a_{30}$

$\small{ \boxed{~ \text{arithmetric sequence: }\quad a_k = a_i \frac{j-k}{j-i} + a_j\frac{k-i}{j-i} ~} }\\ \small{ \boxed{~ \text{geometric sequence: } \quad a_k = a_i^{ \frac{j-k}{j-i} }\cdot a_j^{ \frac{k-i}{j-i} } ~} }$

$a_6 = a_i = 50 \qquad i = 6\\ a_{11}= a_j =35 \qquad j = 11\\ a_{30}= a_k = \ ? \qquad k = 30$

$a_{30} = 50 \cdot ( \frac{11-30}{11-6} )+35\cdot ( \frac{30-6}{11-6} ) \\ a_{30}=50 \cdot ( \frac{-19}{5} ) +35\cdot ( \frac{24}{5} ) \\ a_{30}=-19\cdot 10 + 7 \cdot 24\\ a_{30}=-190+168\\ a_{30}=-22$