2. In an exponential sequence, the sixth term is eight times the third term and the sum of the seventh and eighth term is 192 . Find a. the common ratio. b. the first term. c. the sum of the fifth term to the eleventh term, inclusive.

Let the exponential (geometric) sequence have the first term a and common ratio r. Then, the nth term is represented by:   Term n = a · r^(n–1) Step 1. Use the condition that the 6th term is eight times the 3rd term.   6th term = a · r^(6–1) = a · r⁵   3rd term = a · r^(3–1) = a · r² The condition gives:   a · r⁵ = 8 · (a · r²) Assuming a ≠ 0 and r ≠ 0, cancel a and simplify:   r⁵ = 8r²   r^(5–2) = 8  ⟹  r³ = 8   r = ∛8 = 2 So, the common ratio is 2. Step 2. Use the condition that the sum of the 7th and 8th terms is 192.   7th term = a · r^(7–1) = a · r⁶ = a · 2⁶ = 64a   8th term = a · r^(8–1) = a · r⁷ = a · 2⁷ = 128a Their sum is:   64a + 128a = 192a Setting the sum equal to 192:   192a = 192  ⟹  a = 1 So, the first term is 1. Step 3. Compute the sum of the terms from the 5th to the 11th term, inclusive. The 5th term is:   5th term = a · r^(5–1) = 1 · 2⁴ = 16 The 11th term is:   11th term = a · r^(11–1) = 1 · 2¹⁰ = 1024 There are 7 terms in total (from 5th through 11th). The sum of a geometric series is given by:   S = first term · (r^n – 1) / (r – 1) Here, first term = 16, r = 2, and n = 7. Calculate:   S = 16 · (2⁷ – 1) / (2 – 1) = 16 · (128 – 1) / 1 = 16 · 127 = 2032 Thus, the sum from the 5th to the 11th term is 2032. Summary: a. The common ratio is 2. b. The first term is 1. c. The sum of the 5th to 11th terms is 2032.