Find the first six terms of the sequence. a₁ = 4, a_n = 2. a_{n - 1}

Solution:

Given, a₁ = 4

a_n = 2. a_{n - 1}

We have to find the six terms of the sequence.

The given series is a geometric sequence.

\(\\a_{2} = 2\times a_{(2 - 1)}\\a_{2} = 2\times a_{1}\\a_{2} = 2\times 4\\a_{2} = 8\)

\(\\a_{3} = 2\times a_{(3 - 1)}\\a_{3} = 2\times a_{2}\\a_{3} = 2\times 8\\a_{3} = 16\)

\(\\a_{4} = 2\times a_{(4 - 1)}\\a_{4} = 2\times a_{3}\\a_{4} = 2\times 16\\a_{4} = 32\)

\(\\a_{5} = 2\times a_{(5 - 1)}\\a_{5} = 2\times a_{4}\\a_{5} = 2\times 32\\a_{5} = 64\)

\(\\a_{6} = 2\times a_{(6 - 1)}\\a_{6} = 2\times a_{5}\\a_{6} = 2\times 64\\a_{6} = 128\)

Therefore, the six terms are 4, 8, 16, 32, 64 and 128.

Find the first six terms of the sequence. a₁ = 4, a_n = 2. a_{n - 1}

Summary:

The first six terms of the sequence. a₁ = 4, a_n = 2. a_{n - 1} are 4, 8, 16, 32, 64 and 128.