need help with solution po sana​

1. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 7 \\ & a_1 = 2 \\ & a_2 = 6 \\ & a_3 = 18 \end{aligned}[/tex]Step 2: Find the common ratio.[tex]\begin{aligned} r & = \frac{a_2}{a_1} = \frac{a_3}{a_2} \\ r & = \frac{6}{2} = \frac{18}{6} \\ r & = 3 = 3 \end{aligned}[/tex]Step 3: Solve for the 7th term of a geometric sequence.[tex]\begin{aligned} a_n & = a_1r^{n - 1} \\ a_7 & = (2)(3)^{7 - 1} \\ & = 2(3)^6 \\ & = 2(729) \\ & = \boxed{1,458} \end{aligned}[/tex]Hence, the 7th term of a geometric sequence is 1,458.2. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 10 \\ & a_1 = 3 \\ & a_2 = 6 \\ & a_3 = 12 \end{aligned}[/tex]Step 2: Find the common ratio.[tex]\begin{aligned} r & = \frac{a_2}{a_1} = \frac{a_3}{a_2} \\ r & = \frac{6}{3} = \frac{12}{6} \\ r & = 2 = 2 \end{aligned}[/tex]Step 3: Solve for the 10th term of a geometric sequence.[tex]\begin{aligned} a_n & = a_1r^{n - 1} \\ a_{10} & = (3)(2)^{10 - 1} \\ & = 3(2)^9 \\ & = 3(512) \\ & = \boxed{1,536} \end{aligned}[/tex]Hence, the 10th term of a geometric sequence is 1,536.3. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 6 \\ & a_1 = 81 \\ & a_2 = 54 \\ & a_3 = 36 \end{aligned}[/tex]Step 2: Find the common ratio.[tex]\begin{aligned} r & = \frac{a_2}{a_1} = \frac{a_3}{a_2} \\ r & = \frac{54}{81} = \frac{36}{54} \\ r & = \frac{2}{3} = \frac{2}{3} \end{aligned}[/tex]Step 3: Solve for the 6th term of a geometric sequence.[tex]\begin{aligned} a_n & = a_1r^{n - 1} \\ a_6 & = (81)\left(\frac{2}{3}\right)^{6 - 1} \\ & = 81\left(\frac{2}{3}\right)^5 \\ & = 81\left(\frac{32}{243}\right) \\ & = \boxed{\frac{32}{3}} \end{aligned}[/tex]Hence, the 6th term of a geometric sequence is 32/3.4. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 8 \\ & a_1 = 64 \\ & a_2 = -32 \\ & a_3 = 16 \end{aligned}[/tex]Step 2: Find the common ratio.[tex]\begin{aligned} r & = \frac{a_2}{a_1} = \frac{a_3}{a_2} \\ r & = \frac{-32}{64} = \frac{16}{-32} \\ r & = -\frac{1}{2} = -\frac{1}{2} \end{aligned}[/tex]Step 3: Solve for the 8th term of a geometric sequence.[tex]\begin{aligned} a_n & = a_1r^{n - 1} \\ a_8 & = (64)\left(-\frac{1}{2}\right)^{8 - 1} \\ & = 64\left(-\frac{1}{2}\right)^7 \\ & = 64\left(-\frac{1}{128}\right) \\ & = \boxed{-\frac{1}{2}} \end{aligned}[/tex]Hence, the 8th term of a geometric sequence is -1/2.5. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 8 \\ & a_1 = 64 \\ & a_2 = -96 \\ & a_3 = 144 \end{aligned}[/tex]Step 2: Find the common ratio.[tex]\begin{aligned} r & = \frac{a_2}{a_1} = \frac{a_3}{a_2} \\ r & = \frac{-96}{64} = \frac{144}{-96} \\ r & = -\frac{3}{2} = -\frac{3}{2} \end{aligned}[/tex]Step 3: Solve for the 5th term of a geometric sequence.[tex]\begin{aligned} a_n & = a_1r^{n - 1} \\ a_5 & = (64)\left(-\frac{3}{2}\right)^{5 - 1} \\ & = 64\left(-\frac{3}{2}\right)^4 \\ & = 64\left(\frac{81}{16}\right) \\ & = \boxed{324} \end{aligned}[/tex]Hence, the 5th term of a geometric sequence is 324.

Answered by GreatRedSpot | 2024-09-04