Given that the first and forth terms of G.P are 6 and 162 respectively, find the sum of the first three terms of the progression
To find the sum of the first three terms of a geometric progression (G.P.) where the first term \( a \) and the fourth term \( ar^3 \) are given, we set up the following equations:
Given:
- \( a = 6 \)
- \( ar^3 = 162 \)
Find the common ratio \( r \)
From the equation for the fourth term:
\(ar^3\) = 162
Substituting \( a \):
\(6r^3\) = 162
Dividing both sides by 6:
\(r^3 = \frac{162}{6}\) = 27
Taking the cube root:
r = \(\sqrt[3]{27}\) = 3
Find the first three terms
The first three terms of the G.P. are:
- First term: \( a = 6 \)
- Second term: \( ar = 6 \cdot 3 = 18 \)
- Third term: \( ar^2 = 6 \cdot 3^2 = 6 \cdot 9 = 54 \)
Calculate the sum of the first three terms
The sum \( S \) of the first three terms is:
S = a + ar + ar\(^2\) = 6 + 18 + 54
Calculating this:
S = 6 + 18 + 54 = 78
The sum of the first three terms of the progression is \( \boxed{78} \).
There is an explanation video available below.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}