# What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250?

What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is 781,250?

Name:

Recall that the sum of the first n terms of a geometric sequence (which I will denote as S(n)) with a first term of a₁ and a common ratio of r is:
S(n) = a₁(1 - r^n)/(1 - r).
Now, notice that we have a₁ (the first term) and n (the number of terms), but we do not have r; we can find r by using the fact that the last term is 781250 and the formula:
a(n) = a₁*r^(n - 1) (a(n) is the n-th term of the sequence).
Since the last term corresponds to n = 8 (as there are 8 terms), we have:
a₁ = 10, n = 8, and a₈ = 781250.
Plugging these values into the above formula gives:
a₈ = a₁*r^(8 - 1) ==> 781250 = 10r^7
==> r = (781250/10)^(1/7) = 5.
----
Now that we have r, we can now apply the summation formula to yield:
S₈ = 10(1 - 5^8)/(1 - 5) = 976560,
to be the required sum.
I hope this helps!
Let the first term be a.
Let r be the common ratio.
We have:
a = 10
ar^7 = 781250
Thus, r^7 = 781250 / 10 = 78125
r = 5
Hence, we have:
Summation of 10(5)^n from n = 0 to n = 7
= 10 [Summation of 5^n from n = 0 to n = 7]
= 10 ((5^8 - 1) / (5 - 1))
= 976560
Hope that helped! :D