Question

the fist term of a geometric series is 1, the nth term is 128 and the sum of the n term is 225. Find the common ratio and the number of terms?

Answer 1

If the sum of the n term = 225

your question does not make sense.

It can not be solved.


But if the sum of the n term = 255 then :


In geometric sequence :

The nth term is :

an = a1 * r ^ ( n - 1 )

Where a1 is the first term of the sequence.

r is the common ratio.

n is the number of the terms


The sum of the first n terms is given by:

S = a1 * [ ( 1 - r ^ n ) / ( 1 - r ) ]


In this case :

a1 = 1


an = a1 * r ^ ( n - 1 ) = 1 * r ^ ( n - 1 ) = r ^ ( n - 1 ) = 128

r ^ ( n - 1 ) = 128


S = a1 * [ ( 1 - r ^ n ) / ( 1 - r ) ] = 1 * [ ( 1 - r ^ n ) / ( 1 - r ) ] = ( 1 - r ^ n ) / ( 1 - r ) = 255

( 1 - r ^ n ) / ( 1 - r ) = 255


So :

r ^ ( n - 1 ) = 128 Multiplye both sides by r

r ^ ( n - 1 ) * r = 128 r

r ^ n = 128 r

Becouse r ^ ( n - 1 ) * r = r ^ n


Now :

r ^ n = 128 r

You already know :

( 1 - r ^ n ) / ( 1 - r ) = 255

( 1 - 128 r ) / ( 1 - r ) = 255 Multiply both sides by ( 1 - r )

1 - 128 r = 255 ( 1 - r )

1 - 128 r = 255 - 255 r Subtract 1 to both sides

1 - 128 r - 1 = 255 - 255 r - 1

- 128 r = 254 - 255 r Add 255 r to both sides

- 128 r + 255 r = 254

127 r = 254 Divide both sides by 127

r = 254 / 127

r = 2


Also you already know :

r ^ ( n - 1 ) = 128

In this case :

2 ^ ( n - 1 ) = 128 Take the logarithm of both sides

( n - 1 ) * log ( 2 ) = log ( 128 ) Divide both sides by log ( 2 )

n - 1 = log ( 128 ) / log ( 2 )

n - 1 = 7 Add 1 to both sides

n - 1 + 1 = 7 + 1

n = 8

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Remark:

log [ 2 ^ ( n - 1 ) ] = ( n - 1 ) * log ( 2 ) becouse

log ( a ^ x ) = x * log ( a )

In this case :

a = 2 , x = n - 1
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Solutions :


Common ratio

r = 2

Number of the terms

n = 8

Answer 2

I agree with Bosnian that the question contains a typo.
If the sum of n terms is 225 as stated we could solve for r = 2.309..
but then the solution for n is not a whole number.

Bosnian assumed correctly that sum(8) = 255