infinite geometric series: sum is 10, sum of squares is 25

[leilsilver]

The sum of an infinite geometric series is 10 and the sum of the squares of the terms of this series is 25. What is the first term of the series?

Can anyone help me with this?

I know that the first one is S=a/(1-b). So it's 10=a/(1-b)

But the second one, is it 25= a^2/(1-b^2)?

 

[tkhunny]

You have it. What are you waiting for? Solve for 'a' and 'b'.

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\frac{a}{{1 - r}} = 10\quad \& \quad \frac{{a^2 }}{{1 - r^2 }} = 25 \\
a = 10(1 - r)\quad \Rightarrow \quad \frac{{\left( {10(1 - r)} \right)^2 }}{{1 - r^2 }} = 25 \\
\end{array}\)

Now you solve for ratio, r; then a.

 

[leilsilver]

So if I did all the work right, would the answer be a = 2.229?

25 = [(10 - 10b)(10 - 10b)] / (1 - b^2)

25 = (100 - 20b + 100b^2) / (1 - b^2)

25 - 25b^2 = 100 - 20b + 100b^2

0 = 95b^2 + 20b + 75

b = -20 ± sqrt 400 - 28500 / 190

b = -20 ± sqrt -28100 / 190

b = -20 + sqrt 28100 / 190

b = 0.777003

10 - 10b = a

10 - 10(0.777003) = a

2.229 = a

How is that?

 

[stapel]

You can check the answer to any "solving" problem by plugging it back into the original exercise. What did you get when you did this?

Meanwhile, why make the computations so hard? Why not take what you've been given, simplify a bit first, and then solve for "r" and then "a"? You were given the following:

. . . . .\(\displaystyle \L \frac{\left(10(1\,-\,r)\right)^2}{1\,-\,r^2}\, =\, 25\)

Since 1 - r² = (1 - r)(1 + r), then:

. . . . .\(\displaystyle \L \frac{100(1\,-\,r)(1\,-\,r)}{(1\,-\,r)(1\,+\,r)}\, =\, 25\)

Cancel the common factor, cross-multiply, and solve the resulting linear equation for the exact value of r. Then plug into the "a = 10(1 - r)" equation you were given, and solve for the exact value of "a".

Eliz.