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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.

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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 $a=7.$

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  • $\begingroup$ how $1+ r+r^2 = 1 +r^2-2r$? im not getting $\endgroup$ – jasmine Jan 1 at 21:26
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    $\begingroup$ They are not equal. $28(1+r+r^2) = 196(1-r)^2$ $\endgroup$ – model_checker Jan 1 at 21:30
  • $\begingroup$ okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer $\endgroup$ – jasmine Jan 1 at 21:38
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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.

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