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 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.
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
The sum of the infinite geometric series?
S=t1/(1-r) r must be 3/8 because 3/8 is multiplied by t each time S=(3/2)/(1-(3/8)) S=2.4 I hope that this helps.
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...
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.
Consider the infinite geometric series?
a. a(n) = -4(⅓) ^ (n - 1) a(1) = -4 a(2) = -4 / 3 a(3) = -4 / 9 a(4) = -4 / 27 a(5) = -4 / 81 b. The sum converges because -1 < r < 1. c. Use the sum to infinity formula to find the sum: S = a / (1 - r) S = -4 / (1 - ⅓) S = -4 / ⅔ S = -4 ˣ 3 / 2 S = -6
Find the sum of the infinite geometric series 80+40+20+..?
160
Please help with geometric & infinite series?
1) These terms do not constitute a geometric sequence: 4x / 1 = 4x 16x^2 / 4x = 4x 48 x^2 / 16x^2 = 3 ... not 4x The 4th term probably should be 64x^3 to make the common ratio 4x Then for it to be convergent, 4x < 1 or x < 1/4 2) 26 + .70 * 26 + .70^2 * 26 + .70^3 * 26 + .... = 26 * (1 + .70 + .70^2 + .70^3 + ...) The infinite sum of the series = 1 /(1 - .7) = 1 / .3 = 3.33333 3.333 * 26 = 86.66666 meters distance in infinite time. The depth of 86 is just short of that, so how much time is required to do that last 2/3 m ? From here it's possible to determine where an infinite sum that covers the last 2/3 m starts, and then figure out which term of the original sequence that is, but it was easier to just throw it into a spreadsheet to get these results: time .... distance ... total distance at end of time period - - - -.... - - - - - - -. . - - - - - - - - - - 1 ... 26.0000 ... 26.0000 2 ... 18.2000 ... 44.2000 3 ... 12.7400 ... 56.9400 4 ... 8.9180 ... 65.8580 5 ... 6.2426 ... 72.1006 6 ... 4.3698 ... 76.4704 7 ... 3.0589 ... 79.5293 8 ... 2.1412 ... 81.6705 9 ... 1.4988 ... 83.1694 10 ... 1.0492 ... 84.2185 11 ... 0.7344 ... 84.9530 12 ... 0.5141 ... 85.4671 13 ... 0.3599 ... 85.8270 14 ... 0.2519 ... 86.0789 Answer: 14 minutes 3) If the common ratio is 1, then all the terms are the same: a, a, a, a, a, a, .... and their sum just keeps growing: a, 2a, 3a, 4a, .... Only way that is finite is if a = 0 and then the sums are 0, 0+0, 0+0+0, ... all of them, including the infinite sum, equal to 0. 4) Day 1: 1/4 eaten, 3/4 remain Day 2: 3/16 eaten, 9/16 remain Day 3: 9/64 eaten, 27/64 remain ← Answer: 9/64 Day 4: 27/256 eaten, 81/256 remain.