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

Answer 1

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)