The 7th term of a g.p. is 8 times of the 4th term. find the series if its 5th term is 48.
its 5th term is 48 --- ar^4 = 48
The 7th term of a g.p. is 8 times of the 4th term
---> ar^6 = 8ar^3
ar^6 - 8ar^3 = 0
ar^3( r^3 - 8) = 0
ar^3 = 0 ---> not very likely or all terms are 0
r^3 = 8
r = 2
in ar^4 = 48
a(16) = 48
a = 3
state your conclusion
To find the series, we need to determine the common ratio of the geometric progression (g.p.). We are given that the 5th term is 48 and the 7th term is 8 times the 4th term.
Let's use the formula to find the nth term of a g.p.:
an = a1 * r^(n-1)
Let's assign variables to the given information:
a5 = 48 (5th term)
a7 = 8 * a4 (7th term is 8 times the 4th term)
We can substitute these values into the formula:
48 = a1 * r^(5-1) ...(1) (where n = 5)
8 * a4 = a1 * r^(7-1) ...(2) (where n = 7)
To solve for the common ratio (r), we can divide equation (2) by equation (1):
(8 * a4) / 48 = (a1 * r^(7-1)) / (a1 * r^(5-1))
Simplifying:
8 * a4 / 48 = r^6 / r^4
1/6 = r^2
Taking the square root of both sides:
r = (√1/6)
Note: Since we are dealing with a g.p., the common ratio (r) must be positive. Therefore, we do not need to consider the negative square root.
Now that we have found the common ratio (r), we can find the first term (a1) using the 5th term (a5):
48 = a1 * (√1/6)^(5-1)
48 = a1 * (√1/6)^4
48 = a1/36
a1 = 48 * 36
a1 = 1728
Now we have the first term (a1) and the common ratio (r), so we can write the series as follows:
Series = 1728, 1728 * (√1/6), 1728 * (√1/6)^2, 1728 * (√1/6)^3, ...
And there you have it! The series is 1728, 1728 * (√1/6), 1728 * (√1/6)^2, 1728 * (√1/6)^3, ...