Geometric Sequences and Sums

A comprehensive tutorial on geometric sequences, series, and summation formulas with solved examples and practice problems.

Geometric Sequences

A geometric sequence is a sequence where each term is found by multiplying the previous term by a constant called the common ratio.

Example 1: Basic Geometric Sequence

\(1, 2, 4, 8, 16, \dots\) - each term is obtained by multiplying the preceding term by \(2\).

Formulas

If \(a_1\) is the first term and \(r\) is the common ratio:

n-th term: \(a_n = a_1 \cdot r^{n-1}\)
Sum of first n terms: \(S_n = a_1 \frac{1 - r^n}{1 - r}, \quad r \neq 1\)
Infinite sum (when \(|r| < 1\)): \(S_\infty = \frac{a_1}{1 - r}\)

Examples with Detailed Solutions

Example 2: Finding a Specific Term

The 4th and 7th terms of a geometric sequence are \(\frac{1}{8}\) and \(\frac{1}{64}\) respectively. Find the 9th term.

Solution

Given: \(a_4 = a_1 r^3 = \frac{1}{8}\) and \(a_7 = a_1 r^6 = \frac{1}{64}\).

Divide the equations: \(\frac{a_7}{a_4} = r^3 = \frac{1/64}{1/8} = \frac{1}{8} \Rightarrow r = \frac{1}{2}\).

Now find \(a_9 = a_1 r^8 = a_7 \cdot r^2 = \frac{1}{64} \cdot \left(\frac{1}{2}\right)^2 = \frac{1}{256}\).

Example 3: Three Consecutive Terms

Find three consecutive numbers in a geometric sequence whose sum is 234 and product is 157,464.

Solution

Let the terms be \(x\), \(xr\), \(xr^2\).

Sum: \(x + xr + xr^2 = x(1 + r + r^2) = 234\)

Product: \(x \cdot xr \cdot xr^2 = x^3 r^3 = 157464 \Rightarrow xr = \sqrt[3]{157464} = 54\)

Substitute \(x = \frac{54}{r}\) into the sum equation: \(\frac{54}{r}(1 + r + r^2) = 234\)

Simplify: \(54(1 + r + r^2) = 234r \Rightarrow 3r^2 - 10r + 3 = 0\)

Solve: \(r = 3\) or \(r = \frac{1}{3}\)

Two solutions:

For \(r = 3\): terms are \(18, 54, 162\)
For \(r = \frac{1}{3}\): terms are \(162, 54, 18\)

Example 4: Number of Terms for a Given Sum

The first term is 2000 and the second is 1000. How many consecutive terms must be taken to obtain a sum of 3875?

Solution

Common ratio: \(r = \frac{1000}{2000} = \frac{1}{2}\)

Sum formula: \(S_n = 2000 \frac{1 - (1/2)^n}{1 - 1/2} = 3875\)

Simplify: \(1 - \left(\frac{1}{2}\right)^n = \frac{3875 \cdot 0.5}{2000} = \frac{1937.5}{2000} = \frac{31}{32}\)

Thus: \(\left(\frac{1}{2}\right)^n = \frac{1}{32} = \left(\frac{1}{2}\right)^5 \Rightarrow n = 5\)

Answer: 5 terms are needed.

Example 5: Proving Non-Geometric Sequence

Prove that \(x\), \(x^2 + 1\), and \(x^3 + x\) cannot be three consecutive terms of a geometric sequence (real numbers).

Solution

If they were geometric: \(\frac{x^2 + 1}{x} = \frac{x^3 + x}{x^2 + 1}\)

Cross-multiply: \((x^2 + 1)^2 = x(x^3 + x)\)

Expand: \(x^4 + 2x^2 + 1 = x^4 + x^2 \Rightarrow x^2 + 1 = 0\)

No real solution, hence the terms cannot form a geometric sequence.

Example 6: Finding Terms from Conditions

The first three terms of a geometric sequence are \(x\), \(x^2 + 4\), and \(16x\). Find the terms.

Solution

Using the common ratio: \(\frac{x^2 + 4}{x} = \frac{16x}{x^2 + 4}\)

Cross-multiply: \((x^2 + 4)^2 = 16x^2\)

Simplify: \(x^4 + 8x^2 + 16 = 16x^2 \Rightarrow x^4 - 8x^2 + 16 = 0\)

Factor: \((x^2 - 4)^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\)

Two solutions:

For \(x = 2\): terms are \(2, 8, 32\)
For \(x = -2\): terms are \(-2, 8, -32\)