Question

1. What is the sum of the geometric sequence 8, -16, 32 . if there are 15 terms? (1 point)
2. What is the sum of the geometric sequence 4, 12, 36 . if there are 9 terms? (1 point)
3. What is the sum of a 6-term geometric sequence if the first term is 11, the last term is -11,264 and the common ratio is -4? (1 point)
4. What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is
781,250?

Answer 1

For the 1st geometric sequence, the common ratio is q=-2
-16/8 = -2
32/-16=-2
The sum of 15 terms could be found using the formula:
S = 8*[(-2)^15-1]/(-2-1)

Answer 2

For the 2nd geometric series, the ratio is q = 3
s = 4*(3^9 - 1)/(3-1)
s = 2*(3^9 - 1)

Answer 3

how do you figure that out? and whats the formula?

Answer 4

Formula for calulating the sum of n terms of geometric sequence is:
Sn = b1*(q^n - 1)/(q-1), where b1 is the 1st term and q is the common ratio

Answer 5

For the 1st geometric series, the 1st term b1 is 8

Answer 6

For the 2nd geom. series, the first term is 4

Answer 7

okay, i think i understand, how would yuou solve 3 and 4 with a term?

Answer 8

Well. we notice that to apply the formula we need to know only the 1st term and the ratio, we don't need anything else.

Answer 9

S6 = 11*[(-4)^6-1]/(-4-1)
S6 = (-11/5)*(4^6 - 1)

Answer 10

For the 4th geom. series, we need to determine the common ratio.
We know the last term and the first term and the fomrula for the general term;
an = a1*q^(n-1)
781,250 = 10*q^(n-1)
q^(n-1) = 78,125