The sum of an infinite geometric series of real numbers is $14,$ and the
sum of the cubes of the terms of this series is $392$.
What is the first term of the series?
My attempt: Let the series be $\{ a ,~ ar ,~ar^2 ,\ldots \}$, then the sum is
$$ s = \frac{a}{1 – r} = 14 \tag{1}$$
When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, \ldots$
which sums to
$$ \frac{ a^3 }{ 1 – r^3 } = 392 \tag{2}$$
Now I got $\frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.
Any hints/solution is appreciated.
Best Answer
So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies $$a^2=28(1+r+r^2)=196(1+r^2-2r)$$ and so $$(2r-1)(r-2)=0\implies r=1/2$$ since $|r|<1.$ And so $a=7.$