An explicit rule for the n - th term of a geometric sequence where the second and fifth terms are -6 and 162, respectively is b. Tn = 2 • (-3)ⁿ⁻¹
Further explanation
Firstly , let us learn about types of sequence in mathematics.
Arithmetic Progression is a sequence of numbers in which each of adjacent numbers have a constant difference.
\(\boxed {T_n = a + (n-1)d}\)
\(\boxed {S_n = \frac{1}{2}n ( 2a + (n-1)d )}\)
Tn = n-th term of the sequence
Sn = sum of the first n numbers of the sequence
a = the initial term of the sequence
d = common difference between adjacent numbers
Geometric Progression is a sequence of numbers in which each of adjacent numbers have a constant ratio.
\(\boxed {T_n = a ~ r^{n-1}}\)
\(\boxed {S_n = \frac{a( 1 - r^n ) }{1 - r}}\)
Tn = n-th term of the sequence
Sn = sum of the first n numbers of the sequence
a = the initial term of the sequence
r = common ratio between adjacent numbers
Let us now tackle the problem!
Given:
T₂ = -6
T₅ = 162
Unknown:
Tn = ?
Solution:
\(T_2 = a ~ r^{2-1}\)
\(-6 = a ~ r\)
\(a = \frac{-6}{r}\) → Equation ( 1 )
\(T_5 = a ~ r^{5-1}\)
\(162 = a ~ r^4\)
\(162 = \frac{-6}{r} \times r^4\) ← Equation ( 1 )
\(162 = \frac{-6}{r} \times r^4\)
\(\frac{162}{-6} = r^3\)
\(r^3 = -27\)
\(r = \sqrt [3]{-27}\)
\(r = -3\)
\(a = \frac{-6}{r}\) ← Equation ( 1 )
\(a = \frac{-6}{-3}\)
\(a = 2\)
Finally , the explicit rule for the n - th term is :
\(T_n = a ~ r^{n-1}\)
\(\large { \boxed {T_n = 2 \cdot (-3)^{n-1} } }\)
Learn more
- Geometric Series : https://brainly.com/question/4520950
- Arithmetic Progression : https://brainly.com/question/2966265
- Geometric Sequence : https://brainly.com/question/2166405
Answer details
Grade: High School
Subject: Mathematics
Chapter: Arithmetic and Geometric Series
Keywords: Arithmetic , Geometric , Series , Sequence , Difference , Term
