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# Infinity Sigma

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Hi good friends,

I am quite stuck with this one, I do not really know HOW to even start off with this. Please advise?..

a Convergent Geometric series consisting of only positive terms has first term a, constant ratio r and nth term Tn, such that   $$\sum_{n=3}^{\infty} Tn={1 \over 4}$$

1). If T1 + T2 =2, write down an expression for a in terms of r

2). Calculate the values of a and r....

I am not sure but this is what I think: I know

$$(2+\sum_{n=3}^{\infty} Tn={1 \over 4})=(\sum_{n=1}^{\infty} Tn={1 \over 4})$$

so $$(\sum_{n=3}^{\infty} Tn={1 \over 4})=(\sum_{n=1}^{\infty} Tn={1 \over 4}-2)$$

$$\sum_{n=1}^{\infty} Tn=-{7 \over 4}$$

I believe this will be the nth term?..however this is not a positve term?

am I even on the right track here??? ..please help...Thank you all very much, it is always appreciated!

Apr 2, 2023

#1
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a + a*r ==2,

1/4 ==a*r^2 / [1 - r], solve for a, r

a ==3/2 ==1.5

r ==1/3

So, your GS looks like this:

3 / 2 , 1 / 2 , 1 / 6 , 1 / 18 , 1 / 54 , 1 / 162 , 1 / 486 , 1 / 1458 , 1 / 4374 , 1 / 13122 , 1 / 39366 , 1 / 118098 , 1 / 354294 , 1 / 1062882 , 1 / 3188646 , 1 / 9565938 , 1 / 28697814 , 1 / 86093442 , 1 / 258280326 , 1 / 774840978 , >>Number of terms = 20>>Total Sum ==2.249999999==~2.25

Apr 2, 2023
#2
+1

Guest,

please, how do you get a=3/2?

juriemagic  Apr 3, 2023
#3
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a + ar==2................................(1)
1/4 ==a*r^2 / [1 - r]...............(2), solve for a, r
r ==2/a - 1 [From (1) above. Sub r into (2) above]
a *(2/a - 1)^2 /[1 - (2/a - 1)]==1/4............(3)
Cross multiply (3) above.
4a(2/a - 1)^2==(1 - (2/a  -  1))
4a^2(- 1  +  2/a)^2 ==2(a  -  1)
4a^2 - 16a + 16==2a - 2
4a^2  - 18a  +  18 ==0
2(a - 3)(2a - 3) ==0
(a - 3)(2a  - 3) ==0
a ==3/2  and  a ==3 [discard this one because it gives negative terms)

Guest Apr 3, 2023
#4
+2

Hello again guest,

I sent response yesterday, but must have "forgotten" to actually publish my response.....

I was saying the following:

You indicated that $$r={2 \over{a-1}}$$

I do not get that, because this is how I see it...please advise:

$$a+ar=2\Rightarrow ar=2-a\Rightarrow r={{2-a} \over a}$$

How is my calculation incorrect?

juriemagic  Apr 3, 2023
#5
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r ==(2 - a) /a ==2/a - 1

They are exactly the same. 2/a - 1 is easier for calculating things because you only have one "a".

Your (2 - a) / a ==2/a - a/a ==2/a  -  1 which is what I did. OK?

Guest Apr 4, 2023
#6
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Ok, got it, thank you guest,...stay well

juriemagic  Apr 4, 2023