Find the sum of the given infinite geometric series a1=-8 r=-0.02?
sum to infinity = a/(1-r) = -8/1.02 = -7.843
What is the sum of the infinite geometric series with a1 = 42 and r = 6/5?
d. doesn't exist as r = 6/5 > 1
What is the sum of an infinite geometric series?
This is fairly straight forward: a / (1-r) = 200 a/ (1 - .15) = 200 The first term: a = 170 The second term: ar = 170*0.15 = 25.5 The third term: arr = 170 * 0.15 * 0.15 To get any term after the first we simply multiply our first term by r^(n-1) where n is the term you want.
The sume of the infinite geometric series 3/2+9/16+27/128+81/1024 + ..... is?
3/2+9/16+27/128+81/1024 + ..... Let S be the sum of the given series. Then S = (3/2) + (9/16) + (27/128) + (81/1024) + .......... inf. Here the common ratio (r) = Any term divided by the term that follows it. For example choose the term (81/1024) the following term is (27/128) hence - Common ratio (r) = (81/1024) divided by (27/128) => r = 3/8 This common ratio is less than one, hence the sum to infinite terms can be found out. ( Note that exact sum of a GP having infinite number of terms can be obtained iff, ' r ' is a fraction.) Hence the required sum : - S = [ a / ( 1 - r ) ] , where a is first term of the series, => S = [ 3/2 divided by ( 1 - 3/8 ) ] => S = ( 3/2 ) divided by ( 5/8 ) => S = 12/5 ..................................... Answer PKT
What is the sum of the infinite geometric series 5 (1/2) ^n (using calculus)?
I don’t think this can be done using calculus, but perhaps there is a clever approach.The series is geometric. If you start from n = 0, the series is 5 + 5/2 + 5/4 + … which has common ratio 1/2. The formula for the sum is 5/(1–1/2) = 10.If you start from n = 1, the sum is 5 less, i.e. 5.The derivation of the sum of a geometric series is in your text book. It doesn’t use calculus.
What is the sum of the infinite geometric series 1/3, 1/6, 1/12?
Let S = sum of infinite geometric series where a1 is the first term and r is the common ratio.If |r| < 1:S = a1/(1 - r)We know r = 0.5 since we multiply 1/3 by 0.5 to get 1/6, and we multiply 1/6 by 0.5 to get 1/12, etc.So:S = (1/3)/(1 - 0.5) = (1/3)/(1/2) = 2/3.
Help with infinite geometric series?
find the sum of the infinite geometric series 7/8 + 7/12 + 7/18 + 7/27 .......... write each decimal as a fraction in the simplest form 0.37 repeated 0.753 repeated 0.370 repeated write an infinite geometric series that converges to the given number 0.93939393939393939393... 0.358358358358358... 0.445445445445445...
Find the sum of the infinite geometric series, a1=-5,r=1/6?
The equation for the sum of an infinite geometric series that converges is: S = (a1)/(1-r) S = (-5)/(1-(1/6)) S = -6
Find the sum of the infinite geometric series: 18,12,8,...?
use the formula (for an infinite sequence) sum = a / (1 - m ) where a = first term amd m = common multiplier. In your case a = 18 and m= 12/18 or 2/3 plugging into the formula, you get 18 / (1-2/3) = 18 / (1/3) = 18 * 3 = 54