# Find the value of S8, the 8th partial sum, as a mixed number?

32,16,8,4,...

### 3 Answers

- PhilipLv 63 weeks ago
Series is a GP of form a,r,ar^2,...,ar^(n-1),... where a = t(1), t(n) = ar^(n-1), n = 1,2,3,... and r = common between consecutive terms {= t(n+1)/t(n), n = 1,2,3,...};

Sum, S(n), of first n terms = a[(1-r^n)/(1-r)] = [a/(1-r)][(1-r^n)].;

Here, a = 32, r = (1/2) & r^n = (1/2)^n.;

For n = 8, S(8) =[32/{1-(1/2)}][1-(1/2)^8] =64[1 -1/(256)] =64 -64/256 = 64 -1/4;

= 63 3/4.

- llafferLv 73 weeks ago
If you mean to find the sum of the first 8 terms of this geometric sequence, I'll start with this generic form for the n'th term of a geometric sequence:

a(n) = ar^(n - 1)

Where a is the first term (32) and r is the common ratio (1/2)

The sum of the first "n" terms of a geometric sequence can be found with:

S(n) = a(1 - r^n) / (1 - r)

We have values for a, r, and n (8) so we can substitute and simplify:

S(n) = 32(1 - (1/2)^n) / (1 - 1/2)

S(8) = 32(1 - (1/2)^8) / (1/2)

S(8) = 32(1 - 1/256) / (1/2)

S(8) = 32(255/256) / (1/2)

S(8) = (8160/256) / (1/2)

S(8) = (255/8) / (1/2)

Division is the same as the multiplication of the reciprocal:

S(8) = (255/8) * 2

S(8) = 255/4

As a mixed number:

63 3/4