1+ 3/5+9/25+37/125+........
possible answers are:
a. 4/3
b. 3/2
c. 5/3
d. 5/2
2007-06-15 08:01:07 · 5 answers · asked by Trinity 1 in Science & Mathematics ➔ Mathematics
∑1 + 3/5 + 9/25 + 37/125 + ........ =
n-1
∑ (3/5)^k = (1 + (3/5)^n)/(1 - 3/5)
k=0
lim (1 + (3/5)^n)/(1 - 3/5) = 1/(2/5) = 5/2
n→∞
2007-06-15 08:10:54 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
the formula is:
S(infinite) = (t1) / (1 - r)
t1 is the first number (1)
r is the common ratio, or the rate of decrease (3/5)
I'm assuming you mean (27/125) instead of (37/25)
S(infinite) = 1 / (1 - 3/5)
S(infinite) = 1 / (2/5)
S(infinite = 5 / 2
So the answer is [d. 5/2]
2007-06-15 15:08:23 · answer #2 · answered by eV 5 · 1⤊ 0⤋
Sâ = 1 / (1 - 3/5) = 1 / (2/5) = 5 / 2
Answer d.
2007-06-16 05:29:48 · answer #3 · answered by Como 7 · 0⤊ 0⤋
I am assuming that is supposed to be 27/125 and then the answer is b
2007-06-15 15:04:35 · answer #4 · answered by Jared D 2 · 0⤊ 2⤋
the correct option is (b)3/2
because sum=a/1-r
here a=3/5 and r=3/5.
on solving we get the answer.
2007-06-15 15:06:14 · answer #5 · answered by Chandresh 1 · 0⤊ 2⤋