evaluate each infinite geometric series
3+2+(4/3)+(8/9)

Question

evaluate each infinite geometric series
3+2+(4/3)+(8/9)

apoorvk · Answer

you're basically multiplying each term by 2/3 - so the common ratio 'r' is 2/3

now for a GP, where the common ration is between 0 and 1, the sum upto infinite term is a finite no.

S = a/(1-r)

S = sum
a=first term
r = common ratio.

Evaluate. can you?

anonymous · Answer

2/3
(4/3)/2=2/3
(8/9)/(4/3) = 2/3
therefore, it is a G.S. with common ratio = 2/3
Sum of G.S.= a(1-r^n) / (1-r)
=(3)(1-(2/3)^4) /(1-2/3)
=9(1-(2/3)^4) =.....

anonymous · Answer

that will give me the answer?

anonymous · Answer

65/9

apoorvk · Answer

How we arrive at "S = a(1-r)" is this way

Sum of any GP upto 'n' terms is:

$$S = \frac {a(1-r^n)}{(1-r)}$$

When you put in n=infinity in "r^n", and r is a positive fraction less than 1, r multiplied by itself infinite times will render it so small, that r^n can almost be neglected.  So, you are left with:

S = a/(1-r)