1)Find the sum of the first eight terms of the Geometric progression 256,128,64,32

Question

2)How many terms should be taken from the Geometric progression 4,12,36 for the sum to be 2188

Answer 1

1)
The sum of n numbers in Geometric progression is:

Sn=a1*[(1-q^n)/(1-q)]

Where:
a1 is first number in progresion
q is the common ratio.

In your case:

a1=32
q=2

Sn=S8=32*[(1-2^8)/81-2]

S8=32*[(1-256)/(1-2)]

S8=32*( -255)/( -1)

S8=32*255

S8=8160

2)
I am not shure that this question have solution.

Geometric progression in this case:

Six terms:
4,12,36,108,324,972

4+12+36+108+324+972=1456

Seven terms:
4,12,36,108,324,972,2916

4+12+36+108+324+972+2916=4372

Answer 2

In first question:

Sn=S8=32*[(1-2^8)/(1-2)]

Answer 3

In a geometric progression,the product of the 2nd and 4th terms is double the 5th terms and the sum of the first four terms is 80.find the gp