Lesson Example Discussion Quiz: Class Homework |
Step-4 |
Title: Geometric progression |
Grade: 9-a Lesson: S4-L7 |
Explanation: Hello Students, time to practice and review the steps for the problem. |
Lesson Steps
| Step | Type | Explanation | Answer |
|---|---|---|---|
1 |
Problem |
Find the sum of the geometric series \$2,4/3,8/9,16/27,32/81,64/243\$ |
|
2 |
Step |
The given geometric series is \$2,4/3,8/9,16/27,32/81,64/243\$ |
|
3 |
Step |
We know that geometric progression is a series of numbers in which each term is obtained by multiplying the previous term by a fixed number, known as the common ratio. |
|
4 |
Formula: |
common ratio of the given geometric series is given by |
\$ r = (r_2)/(r_1) = (4/3)/2 = 2/3\$ |
5 |
Step |
From the given data |
\$a = 2 ,n =6\$ |
6 |
Formula: |
Sum of the first n terms \$S_n = (a(1-r^n))/(1-r)\$ |
|
7 |
Step |
Substitute the values in the formula |
\$S_6 = (2(1-(2/3)^6))/(1-(2/3))\$ |
8 |
Step |
Simplification |
\$S_6 = (2(1-(64/729)))/(1/3)\$ |
9 |
Step |
Simplification |
\$S_6 = (1330/729)/(1/3) \$ |
10 |
Step |
Simplification |
\$S_6 = (1330 × 3 )/729 = 3990/729\$ |
11 |
Step |
After simplification |
\$S_6 = 1330/243\$ |
12 |
Step |
Sum of the 6 terms of geometric series is \$1330/243\$ |
|
13 |
Answer |
A |
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