Answer:
[tex]a_{n}[/tex] = 3 [tex](-2)^{n-1}[/tex]
Step-by-step explanation:
there is a common ratio between consecutive terms , that is
- 6 ÷ 3 = 12 ÷ - 6 = - 24 ÷ 12 = 48 ÷ - 24 = - 2
this indicates the sequence is arithmetic with explicit rule
[tex]a_{n}[/tex] = a₁ [tex](r)^{n-1}[/tex]
where a₁ is the first term and r the common ratio
here a₁ = 3 and r = - 2 , then
[tex]a_{n}[/tex] = 3 [tex](-2)^{n-1}[/tex]
Answer:
[tex]\sf \dfrac{-3}{2}(-2)^n[/tex]
Step-by-step explanation:
Explicit formulas are used to represent all the terms of the geometric sequence with a single formula.
[tex]\sf \boxed{\bf t_n = ar^{n-1}}[/tex]
a is the first term.
r is the common ratio.
r = second term ÷ first term.
3 , - 6 , 12, - 24, 48 ,........
a = 3
r = -6 ÷ 3 = -2
[tex]\sf t_n = 3*(-2)^{(n-1)}\\\\[/tex]
[tex]\sf = 3*(-2)^{n}*(-2)^{-1}\\\\ =3*(-2)^n*\dfrac{-1}{2}\\\\= \dfrac{-3}{2}(-2)^n[/tex]
Check:
[tex]\sf t_1 =\dfrac{-3}{2}*(-2)^1=\dfrac{-3}{2}*(-2) = 3\\\\\\t-2 = \dfrac{-3}{2}*(-2)^2 = \dfrac{-3}{2}*4=-6\\\\\\t_3=\dfrac{-3}{2}*(-2)^3=\dfrac{-3}{2}*(-8)=12[/tex]
