what is first term of the geometric seguence if a2 is 4 and a4 is 36
Answer (1)
SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 4 \\ & m = 2 \\ & a_2 = 4 \\ & a_4 = 36 \end{aligned}[/tex]Step 2: Find the common ratio.[tex]\begin{aligned} a_n & = a_mr^{n - m} \\ a_4 & = a_2r^{4 - 2} \\ 36 & = 4r^2 \\ 4r^2 & = 36 \\ \frac{4r^2}{4} & = \frac{36}{4} \\ r^2 & = 9 \\ \sqrt{r^2} & = \sqrt{9} \\ r & = 3 \end{aligned}[/tex]Step 3: Solve for the first term.[tex]\begin{aligned} a_n & = a_1r^{n - 1} \\ a_4 & = a_1r^{4 - 1} \\ 36 & = a_1(3^3) \\ 36 & = 27a_1 \\ 27a_1 & = 36 \\ \frac{27a_1}{27} & = \frac{36}{27} \\ a_1 & = \boxed{\frac{4}{3}} \end{aligned}[/tex]Hence, the first term of the geometric sequence is 4/3.
