+1124 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!
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
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)
+1124 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?
