The sum of the last three terms of a G.P having n terms is 1024 times the sum of the first 3 terms of G.P if the third term id 5 find the last term
Let the common ratio of the G.P be denoted by r.
Given that the third term is 5, we can write the terms of the G.P as follows:
First term = a
Second term = ar
Third term = ar^2
It is also given that the sum of the last three terms is 1024 times the sum of the first three terms.
So,
a + ar + ar^2 = 1024(a + ar + ar^2)
Expanding both sides and simplifying, we get:
a(1+r+r^2) = 1024a(1+r+r^2)
1 + r + r^2 = 1024(1 + r + r^2)
1023 = 1023(1 + r + r^2)
1 = r(1 + r + r^2)
Since the third term is ar^2 = 5, we have:
r^2 = 5/a
1 = r(1 + r + 5/a)
1 + r + 5/a = r + r^2 + 5r/a
Substitute r^2 = 5/a, we get:
r + 5/a + 5/a = r + 5(5/a)/a
r + 10/a = r + 25/a^2
10/a = 25/a^2
a = 2.5
Now we can find the common ratio:
r^2 = 5/2.5
r^2 = 2
r = √2
The last term of the G.P can be found as:
Last term = ar^(n-1) = 2.5 * √2^(n-1)