# find the first term of the series?

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.

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 $$a=7.$$
• how $1+ r+r^2 = 1 +r^2-2r$? im not getting – jasmine Jan 1 at 21:26
• They are not equal. $28(1+r+r^2) = 196(1-r)^2$ – model_checker Jan 1 at 21:30
You started fine. This leads you to the system$$\left\{\begin{array}{l}\dfrac a{1-r}=14\\\dfrac{a^3}{1-r^3}=392.\end{array}\right.$$This system has two solutions: $$(a,r)=(-14,2)$$ and $$(a,r)=\left(7,\frac12\right)$$. But only the second one leads to convergent series.