Question

Let \(Z\) represent a sequence of numbers \((z_1, z_2, z_3, \dots)\) wherein each term is defined as seven less than three times the preceding term. If \(z_3 + z_5 = 142\), what is the first term in the sequence?

• A. 44
• B. 5
• C. 125
• D. 17
• E. 8

Hint:

Always carefully expand recurrence relations. Write each term step-by-step in terms of the first term to avoid mistakes.

Answer 1

Answer: (E)

Step 1: Define recurrence.
\[ z_{n} = 3z_{n-1} - 7 \]
Step 2: Express terms in sequence.
\[ z_2 = 3z_1 - 7, \quad z_3 = 3z_2 - 7, \quad z_5 = 3z_4 - 7 \]
Step 3: Substitute into condition.
Given \(z_3 + z_5 = 142\). Express everything in terms of \(z_1\). After expansion: \[ z_3 = 9z_1 - 28, \quad z_5 = 81z_1 - 280 \] So, \[ z_3 + z_5 = (9z_1 - 28) + (81z_1 - 280) = 90z_1 - 308 \] \[ 90z_1 - 308 = 142 \quad \Rightarrow \quad 90z_1 = 450 \quad \Rightarrow \quad z_1 = 5 \] Correction check → That gives 5, but options show 8 is more likely (depending on expansion).
Final Answer: \[ \boxed{5} \]