What is a30 for the arithmetic sequence presented in the table below?
n 6 11
an 50 35
Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference.
a30 = −42
a30 = −38
a30 = −28
a30 = −22
+129845 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
+26387 What is \(a_{30}\)
\(\small{ \boxed{~ \text{arithmetric sequence: }\quad a_k = a_i \frac{j-k}{j-i} + a_j\frac{k-i}{j-i} ~} }\\ \small{ \boxed{~ \text{geometric sequence: } \quad a_k = a_i^{ \frac{j-k}{j-i} }\cdot a_j^{ \frac{k-i}{j-i} } ~} }\)
\(a_6 = a_i = 50 \qquad i = 6\\ a_{11}= a_j =35 \qquad j = 11\\ a_{30}= a_k = \ ? \qquad k = 30\)
\(a_{30} = 50 \cdot ( \frac{11-30}{11-6} )+35\cdot ( \frac{30-6}{11-6} ) \\ a_{30}=50 \cdot ( \frac{-19}{5} ) +35\cdot ( \frac{24}{5} ) \\ a_{30}=-19\cdot 10 + 7 \cdot 24\\ a_{30}=-190+168\\ a_{30}=-22\)
