Let r be any real number such that |r| < 1 and let a = 6 - 6r.
Then the geometric sequence: a, ar, ar^2, ar^3, ... will converge to 6.
Since the choice of r is arbitrary within the given range, there are infinitely many possible answers.
your face thermlscghe eugbcrubah
An infinite series of geometric progressions can be summed when the common ratio ( r ) satisfies ( |r| < 1 ). In this case, the sum ( S ) of the infinite series can be calculated using the formula ( S = \frac{a}{1 - r} ), where ( a ) is the first term of the series. If ( |r| \geq 1 ), the series diverges and does not have a finite sum.
An infinite geometric series has the form ( S = \frac{a}{1 - r} ), where ( a ) is the first term and ( r ) is the common ratio. For the series to converge, the absolute value of ( r ) must be less than 1. If we consider 18 as the sum of an infinite geometric series, we can express it as ( S = \frac{18}{1 - r} ) for some ( r ) where ( |r| < 1 ). For example, if ( r = \frac{1}{2} ), the series would be ( 18 + 9 + 4.5 + 2.25 + \ldots ).
For an infinite geometric series, the sum ( S ) is given by the formula ( S = \frac{a_1}{1 - r} ), where ( a_1 ) is the first term and ( r ) is the common ratio. Given ( S = 6 ) and ( a_1 = 4 ), we can set up the equation ( 6 = \frac{4}{1 - r} ). Solving for ( r ), we get ( 6(1 - r) = 4 ), which simplifies to ( 6 - 6r = 4 ). Thus, ( 6r = 2 ) and ( r = \frac{1}{3} ).
The summation of a geometric series to infinity is equal to a/1-rwhere a is equal to the first term and r is equal to the common difference between the terms.
It depends on the series.
The geometric series is, itself, a sum of a geometric progression. The sum of an infinite geometric sequence exists if the common ratio has an absolute value which is less than 1, and not if it is 1 or greater.
your face thermlscghe eugbcrubah
An infinite series of geometric progressions can be summed when the common ratio ( r ) satisfies ( |r| < 1 ). In this case, the sum ( S ) of the infinite series can be calculated using the formula ( S = \frac{a}{1 - r} ), where ( a ) is the first term of the series. If ( |r| \geq 1 ), the series diverges and does not have a finite sum.
The sum of the series a + ar + ar2 + ... is a/(1 - r) for |r| < 1
An infinite geometric series has the form ( S = \frac{a}{1 - r} ), where ( a ) is the first term and ( r ) is the common ratio. For the series to converge, the absolute value of ( r ) must be less than 1. If we consider 18 as the sum of an infinite geometric series, we can express it as ( S = \frac{18}{1 - r} ) for some ( r ) where ( |r| < 1 ). For example, if ( r = \frac{1}{2} ), the series would be ( 18 + 9 + 4.5 + 2.25 + \ldots ).
For an infinite geometric series, the sum ( S ) is given by the formula ( S = \frac{a_1}{1 - r} ), where ( a_1 ) is the first term and ( r ) is the common ratio. Given ( S = 6 ) and ( a_1 = 4 ), we can set up the equation ( 6 = \frac{4}{1 - r} ). Solving for ( r ), we get ( 6(1 - r) = 4 ), which simplifies to ( 6 - 6r = 4 ). Thus, ( 6r = 2 ) and ( r = \frac{1}{3} ).
Eight. (8)
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The absolute value of the common ratio is less than 1.
Frederick H. Young has written: 'Summation of divergent infinite series by arithmetic, geometric, and harmonic means' -- subject(s): Infinite Series 'The nature of mathematics' -- subject(s): Mathematics
The summation of a geometric series to infinity is equal to a/1-rwhere a is equal to the first term and r is equal to the common difference between the terms.