sum of a geometric sequence whose first term is 150, last (n^th) term is 4,common ratio is 1/2?

First, we need to figure out how many terms there are.

The n-th term of a sequence (a_n) given the first term (a₁) and the common ratio (r) is:

a_n = a₁*r^(n - 1).

With a_n = 4, a₁ = 150, and r = 1/2, we have:

4 = 150(1/2)^(n - 1)

==> (1/2)^(n - 1) = 2/75.

This doesn't have an integer solution as:

(1/2)^(6 - 1) < 2/75 < (1/2)^(7 - 1).

So that means that 4 is not a term of the sequence. Are you sure you have this down right?

There's something wrong with this problem statement. If the common ratio is 1/2 and the first term is 150, then 4 is not a term in the series. The terms are 150, 75, 37.5, 18.75, 9.375, 4.688, 2.344, ...

The sum of a geometric series is a * (1 - r^(n+1))/(1 - r) where r = common ratio, a = first term, n = number of terms.

You know that a=150, r=1/2. If the nth term is 4, then ar^(n-1) = 4. Use that to find n. Then use that n in the partial sum formula for geometric sequences, which should be in your book.

the following we've, a = fifty 4, r = a million/3 Tn = fifty 4(a million/3)^(n -a million) = 2 OR. . (a million/3)^(n -a million) = 2/fifty 4 = a million/27 (a million/3)^(n -a million) = (a million/3)^3 as a outcome n -a million = 3 n = 4 Sn = fifty 4(a million -(a million/3)^4) / (a million -a million/3) Sn = fifty 4(80/80 one) / 2/3 Sn = 80 A n s w e r