Posted by **la bellgoss** on Saturday, March 19, 2011 at 2:18pm.

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

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

- math -
**Anonymous**, Saturday, March 19, 2011 at 4:26pm
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

- math -
**Anonymous**, Saturday, March 19, 2011 at 4:51pm
In first question:

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

- math -
**bobby**, Monday, February 16, 2015 at 10:50am
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

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