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
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 = 27 and r = –4/5?
1.) S = a/(1 - r) = 42/(1 - 6/5) = -42(5) = -210 2.) S = 1/3/(2/3) = 1/2
Find the sum of the infinite geometric series, a1=-5,r=1/6?
S = 5/(1-1/6) = 5/(5/6) = 6
What is the infinite sum of the series 1/1 + 1/3 + 1/6 + 1/10 + 1/15 +?
Let Tn represents the n-th term in this series.T1 = 1/(1)T2 =,1/(1+2)T3 = 1/(1+2+3)T4 = 1/(1+2+3+4)T5 = 1/(1+2–3+4+5) and so on-> Tn = 1/(1+2+3+……..+n)= 1/ [ n(n+1)/2 ] = 2/n(n+1)Tn = 2 / [ n(n+1) ]= 2 * [ (1/n) - 1/(n+1) ]If you add all the terms you will get >2 * [ 1/1 -1/2 + 1/2 -1/3 + 1/3 - 1/4 +1/4 - 1/5 + 1/5 - 1/6 +…….]Answer = 2
The sum of first two terms of geometric progression is 6. The sum of first 4 terms is 15/2. What is the first term?
Let a be the first term and r the ratio.a+a*r=6 or a*(1+r)=6a+a*r+a*r^2+a*r^3=15/2 or a*(1+r)+a*r^2*(1+r)=a*(1+r)*(1+r^2)=15/2Then 1+r^2=(15/2)/6=5/4 or r^2=1/4 and thus r=1/2 or -1/2a=6*2/3=4, which worksa=6*2=12, which does also work
The sum of the geometric series 42-7+7/6-...+42(-1/6)^9 is approximately? how?
collect pairs to get 35 + 35/36 + 35 / 36² + 35 / 36³ +... = 35 + 1 = 36
State if the given geometric series have sums: 6 + 42/5 + 294/25 +...?
State if the given geometric series have sums: 1) 6 + 42/5 + 294/25 +... 2) 1 - 1/5 + 1/25 - 1/125 3) -5/3 - 10/9 - 20/27 - 40/81 4) 1 + .1 + .01 + .001 + ... 5) 2 - 4 + 8 - 16 + 32 6) 6 + 2 + 2/3 + 2/9 +... 10 Points for Best answer and explaination!!!!!!!!!!!!!!!!!!!!! Thanks for helping!