The second term of a geometric progression is 12 more than the first term given that the common the ratio is half of the first term. Find the third term of the Geometric progression
Solution.Tn=ar'n-1 where n=3
ar = 12 + a (i)
ar^2 = x (ii)
subtract equ (i) from equ (ii)
a = 6
substituting a = 6 into equ (i)
ar = 12 + a
6r = 12 + 6=18
r = 3.
ar'2 = 6 * 3'2
ar'2 = 54
65
ar'2=54
To find the third term of a geometric progression, we can start by identifying the first term, the common ratio, and the relationship between the terms.
Let's denote:
- The first term as 'a'.
- The common ratio as 'r'.
From the problem statement, we have two pieces of information:
1. The second term is 12 more than the first term, which can be expressed as:
a + 12 = (a) * r
2. The common ratio is half of the first term, which can be expressed as:
r = a/2
By substituting the second equation into the first equation, we can solve for 'a':
a + 12 = (a) * (a/2)
2a + 24 = a^2
Rearranging the equation:
a^2 - 2a - 24 = 0
Now we can factorize or use the quadratic formula to solve for 'a'. Factoring the equation, we get:
(a - 6)(a + 4) = 0
So, a can be equal to either 6 or -4. Since the first term of a geometric progression cannot be negative, we take a = 6.
Now we can find the common ratio 'r' using the second equation:
r = a/2 = 6/2 = 3
Finally, we can find the third term of the geometric progression by multiplying the second term by the common ratio:
Third term = (First term) * (Common ratio)^2 = a * r^2 = 6 * 3^2 = 6 * 9 = 54
Therefore, the third term of the geometric progression is 54.