What is the 7th term of the geometric sequence where a₁ = 1,024 and a₄ = -16?

Solution:

A geometric sequence or progression abbreviated as G.P. is of the form \(a_{1}, a_{2}, a_{3}, .....a_{n}, .....\)

\(a_{1}\) = the first term;\(\frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}}\)= common ratio.

A sequence \(a_{1}, a_{2}, a_{3}, .....a_{n}, .....\) is called Geometric Progression if each term is non-zero and

\(\frac{a_{k+1}}{a_{k}}\) = r(constant) for k ≥ 1

In the problem statement above

a₁ = 1024 (The first term)

The first term is 1024.

The fourth term is given by:

a₃ = -16

⇒ 1024r³ = -16

⇒ r³ = -16/1024

⇒ r³ = -1/64

Common Ratio = r = \(\sqrt[3]{\frac{-1}{64}}\) = -1/4

Therefore the seventh term is given as :

a₇ = ar⁶

= 1024 × (-1/4)⁶

= 1024 × (1/(16 × 16 × 16))

= 1/4

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What is the 7th term of the geometric sequence where a₁ = 1,024 and a₄ = -16?

Summary:

Given a₁ = 1,024 and a₄ = -16 the value of the 7th term of the geometric series is 1/4.