Question

An infinite geometric series with second term 9/4 has a sum of 81/8. Determine the possible values for the first term.

I understand the formula for an infinite geometric series is \[S=\frac{a}{1-r}\]
Given that, I need to find a and r. I thought I could find r by using \[t_n=ar^{n-1}\] but I still do not know r or a. I would know how to do this if I was given two terms, but not when I am given only one.

Answer 1

Well. s = 81/8
=> 8a = 81 - 81r --1
And, ar = 9/4

Answer 2

You have two equations. Two unkowns. What is the problem?

Answer 3

i am also confused, and i think the problem is ill posed. it could be
\[\sum_{k=0}^{\infty}ar^k\] or
\[\sum_{k=1}^{\infty}ar^k\]

Answer 4

@satellite73 - I am not sure. It does not give an equation in sigma notation. I have basically typed out the question as it is.

@siddhantsharan - What are you doing on the second line? => 8a = 81 - 81r --1

Answer 5

Why?
From the sum, thing, we have the first equations. 8a = 81 - 81r
Now the second term is First term * Common Ratio. = ar = 9/4

Answer 6

The sum of the infinite series is,
S = a/ (1-R)
I just plugged in s.

Answer 7

Oh, I guess the --1 part was a mistake then.

Answer 8

Oh yeahh. I was just writing ------equation(1)
And ar = 9/4 as equation 2.

Answer 9

27/8 and 27/4 are the possible values for the first term?

Answer 10

Wouldn't 8a = 81-81r simplify to 8a = 81(1-r), then 8a/81=(1-r), then (1-r)/a?

Answer 11

It would.
My original equation was :
\[81/8 = a/(1-r)\]

Answer 12

Wait a sec. Can you list both the equations you used again? I thought the second one was 8a = 81 - 81r and first ar=9/4.

Answer 13

I plugged in ar=9/4, a=27/8 to find r = 2/3. Putting that in S=a/(1-r) gave me -27/10.

Answer 14

\[Equation 1 : 81/8 = a/1-r\]
\[Equation 2 = ar = 9 /4\]

Answer 15

If. ar = 9/4, a = 27/8, r = 2/3

S = 27/8 / 1/3 = 81/8

Answer 16

Which clearly satisfies the condition!

Answer 17

Woops, plugged in the wrong number. Still don't understand how you found a though. Did you just solve
\[\frac{a}{1-r}-\frac{81}{8}=ar-\frac{9}{4}\]

Answer 18

No.
From ar = 9/4 .....
=> r = 9/4a
Substituted r in the first equation.
Got a quadratic in a.
Solved for 2 values.

Answer 19

\[Quadratic was : 32a^2 - 324a + 729 = 0\]

Answer 20

Look.
\[r = 9/4a\]
\[8a = 81 - 81(9/4a)\]

Multiply the equation by a. Take the LCM.
You would get a quadratic.

Answer 21

I see. Thank you very much!