# math

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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 -

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.

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 -

In first question:

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

• math -

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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