Find the first six terms of the sequence. a₁ = -6, a_n = 4.a_{n - 1}

Solution:

Given, the series is in geometric progression

First term, a₁ = -6

a_n = 4.a_{n - 1}

We have to find the six terms of a finite sequence.

Geometric progression can be represented by the formula,

\(f(n) = a_{1}(r)^{n-1}\)

\(a_{2}=4\times a_{(2-1)}\\a_{2}=4\times a_{1}\\a_{2}=4\times (-6)\\a_{2}=-24\)

\(a_{3}=4\times a_{(3-1)}\\a_{3}=4\times a_{2}\\a_{3}=4\times (-24)\\a_{3}=-96\)

\(a_{4}=4\times a_{(4-1)}\\a_{4}=4\times a_{3}\\a_{4}=4\times (-96)\\a_{4}=-384\)

\(a_{5}=4\times a_{(5-1)}\\a_{5}=4\times a_{4}\\a_{5}=4\times (-384)\\a_{5}=-1536\)

\(a_{6}=4\times a_{(6-1)}\\a_{6}=4\times a_{5}\\a_{6}=4\times (-1536)\\a_{6}=-6144\)

Therefore, the six terms are -6, -24, -96, -384, -1536 and -6144.

Find the first six terms of the sequence. a₁ = -6, a_n = 4.a_{n - 1}

Summary:

The first six terms of the sequence. a₁ = -6, a_n = 4.a_{n - 1} are -6, -24, -96, -384, -1536 and -6144.