The first three terms of a geometric sequence are given: \( \quad 81 ; m ; \frac{m}{3} ; \ldots \ldots \) 3.1.1 \( \quad \) Determine the value of \( m \). 3.1.2 \( \quad \) Calculate: \( \sum_{t=1}^{9} 81\left(\frac{1}{3}\right)^{t-1} \) Given the arithmetic sequence: \( \frac{12}{5} ; 3 ; \frac{18}{5} ; \ldots . . . . . . . . . . . . . . . . ; \frac{333}{5} \) 3.2.1 \( \quad \) Calculate the number of terms in this sequence. 3.2.2 How many terms of this sequence are integers?

We are given a geometric sequence with the first three terms:   81 ; m ; m/3 ; … Step 1: Find m (3.1.1) In any geometric sequence, the ratio between consecutive terms is constant. Let r be the common ratio. Then we have:   1st term: a₁ = 81   2nd term: a₂ = 81 · r = m   3rd term: a₃ = 81 · r² = m/3 Using the fact that m = 81r, substitute into the expression for a₃:   81 · r² = (81r)/3 Divide both sides by 81 (since 81 ≠ 0):   r² = r/3 If r ≠ 0, divide both sides by r:   r = 1/3 Then m = 81 · (1/3) = 27. So, the value of m is 27. Step 2: Calculate the sum (3.1.2) We need to calculate:   S = Σₜ₌₁⁹ 81 · (1/3)^(t–1) This is a geometric series with first term a = 81 and common ratio r = 1/3. The sum of the first n terms of a geometric series is given by:   Sₙ = a · (1 – rⁿ) / (1 – r) For n = 9:   S₉ = 81 · (1 – (1/3)⁹) / (1 – 1/3) Since 1 – 1/3 = 2/3, we have:   S₉ = 81 · (1 – (1/3)⁹) / (2/3) = 81 · (3/2) · (1 – (1/3)⁹) Thus, the sum is:   S₉ = (243/2) · [1 – (1/3)⁹] You can leave the answer in this form or, if desired, express (1/3)⁹ = 1/19683. Step 3: Work with the arithmetic sequence We are given the arithmetic sequence:   12/5 ; 3 ; 18/5 ; … ; 333/5 Part 3.2.1: Find the number of terms. Write the first term in fifths:   a₁ = 12/5   Second term, 3 = 15/5 Thus, the common difference d = 15/5 – 12/5 = 3/5. Let n be the number of terms. The nth term of an arithmetic sequence is given by:   aₙ = a₁ + (n – 1)d The last term is given as 333/5, so:   12/5 + (n – 1)·(3/5) = 333/5 Multiply through by 5 to eliminate fractions:   12 + 3(n – 1) = 333 Solve for n:   3(n – 1) = 333 – 12 = 321   n – 1 = 321/3 = 107   n = 107 + 1 = 108 So, there are 108 terms in the arithmetic sequence. Part 3.2.2: Find how many terms are integers. A general term of the sequence is:   aₙ = 12/5 + (n – 1)·(3/5) = [12 + 3(n – 1)] / 5 Simplify the numerator:   12 + 3(n – 1) = 3n + 9     (since 3(n – 1) = 3n – 3, and 12 – 3 = 9) Thus:   aₙ = (3n + 9) / 5 = 3(n + 3)/5 For aₙ to be an integer, the fraction must simplify to an integer. Since 3 and 5 are relatively prime, 5 must divide (n + 3). That is:   n + 3 ≡ 0 (mod 5)   n ≡ 2 (mod 5) Now we need to count the number of integers n in the range 1 ≤ n ≤ 108 such that n ≡ 2 (mod 5). Write n in the form:   n = 5k + 2, where k is an integer. Find the smallest and largest k:   For the smallest term, when k = 0, n = 2 (which is ≥ 1)   For the largest term, n = 5k + 2 ≤ 108 ⇒ 5k ≤ 106 ⇒ k ≤ 106/5 = 21.2, so the largest integer k is 21. The values of k range from 0 to 21 inclusive, which gives:   Number of terms = 21 – 0 + 1 = 22 Thus, 22 terms of the arithmetic sequence are integers. Final Answers: • 3.1.1: m = 27. • 3.1.2: The sum is (243/2) · [1 – (1/3)⁹] or equivalently (243/2)[1 – 1/19683]. • 3.2.1: The number of terms is 108. • 3.2.2: The number of integer terms is 22.