📌 Example 1: **Determining Common Ratio firstly then a Term, Given the Sum of an Infinite Geometric Series**

Find the fourth term in the geometric sequence whose first term is 6 and whose sum to infinity is 10.

✍ Solution:

(step 1) Write the formula for the sum to infinity.

(step 2) From the question it is known that the infinite sum is equal to 10 and that the first term,

*a*, is 6. Write down this information.

*a*=6;

*S*

_{∞}=10

(step 3) Substitute known values into the formula.

(step 4) Solve for

*r*.

(step 5) Write the general formula for the nth term of the geometric sequence.

*t*=

_{n}*a⋅r*

^{(n-1)}

(step 6) To find the fourth term substitute

*a*=6,

*n*=4 and

*r*=0.4 into the formula and evaluate.

*t*

_{4}=6⋅(0.4)

^{3}

*t*

_{4}=0.384

📌 Example 2: **Finding the Common Ratio**

An infinite geometric series with first term *a*_{1}=4 has a sum of 10. What is the common ratio of the series?

✍ Solution:

Write rule for sum.

Substitute for

*S*and

*a*

_{1}.

Multiply each side by 1-

*r*.

*r*)=4

Divide each side by 10.

*r*=⅖

Solve for

*r*.

*r*=⅗

The common ratio is

*r*=⅗.

📌 Ex3. Solve the following equation for *x*:

*x*+

*x*

^{2}+

*x*

^{3}+⋯

✍ Solution: It is assumed that the infinite series given in the problem is geometric since it has an indicated sum. Observe that

*a*

_{2}=

*r*=

*x*since

*r*=

*a*

_{2}/

*a*

_{1}=

*x*/1. Using the sum of an infinite geometric sequence,

📌 Ex4. In a geometric sequence of even number of terms, the sum of all terms is 5 times the sum of the odd terms. The common ratio of the geometric sequence is

(A) -⅘ (B) ⅕ (C) 4 (D) none the these

✍ Solution (C) is the correct answer.

Since common ratio of odd terms will be *r*^{2} and number of terms will be *n*), let us consider a geometric sequence *a*, *a⋅r*, *a⋅r*^{2}, … with 2*n* terms. We have

📌 Ex5. A geometric sequence consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

✍ Solution:

*S*=

_{n}*a*+

*a⋅r*+

*a⋅r*

^{2}+⋯+

*a⋅r*

^{(n-1)}

*S*

_{odd}=

*a*+

*a⋅r*

^{2}+⋯

*S*

_{even}=

*a⋅r*+

*a⋅r*

^{3}+⋯

*S*=

_{n}*S*

_{odd}+

*S*

_{even}

According to the given condition,

*S*=5⋅

_{n}*S*

_{odd}

∴

*S*

_{odd}+

*S*

_{even}=5⋅

*S*

_{odd}

Rewrite

*S*

_{even}as

*r*⋅

*S*

_{odd}.

*S*

_{odd}+

*r*⋅

*S*

_{odd}=5⋅

*S*

_{odd}

(1+

*r*)⋅

*S*

_{odd}=5⋅

*S*

_{odd}

1+

*r*=5

*r*=4

Thus, the common ratio of the geometric sequence is 4.