Express 203 as a sum of four terms which are in a Geometric Progression. Then what is the common ratio ?

Hence, there by arrive at a common formula for deriving four numbers which are in a G.P.

Hint #1:

How does factoring of 203 helps here ? Also, the four numbers are integers...but not necessarily the case with the common ratio !!

Hint #2:

203 = 7 * 29 = (5 + 2) * (25 + 4) that is (5 + 2) ( (5^2) + (2^2) )...there by leading to a G.P. with 4 terms ... which are ... and the common ratio is ____. So the generic formula to arrive at such four terms is (a+b) (a^2 + >! b^2)... which gives four terms which are in a G.P. with a common ratio of a/b >! (assuming a > b )

  • $\begingroup$ If you don’t want to accept 13.533, 27.066, 54.133, 108.266 as a solution, you must specify that the four “terms” are integers as part of the puzzle.  Tacking on constraints like that in a hint is unacceptable. $\endgroup$ – Peregrine Rook Jul 19 '17 at 7:06
  • $\begingroup$ Well, @PeregrineRook, is it possible to express your values as fractions (not decimals) ? I am interested to see which four fractions (which are in GP of course) when added result in 203. Just a matter of curiosity ... do not mistake me. $\endgroup$ – Mea Culpa Nay Jul 19 '17 at 8:15
  • $\begingroup$ That’s easy: they are $\frac{203}{15}$,  $\frac{406}{15}$,  $\frac{812}{15}$,  and $\frac{1624}{15}$.   As proper fractions: 13$\frac{8}{15}$,  27$\frac{1}{15}$,  54$\frac{2}{15}$,  and 108$\frac{4}{15}$. $\endgroup$ – Peregrine Rook Jul 20 '17 at 16:14
  • $\begingroup$ And, in case you don’t get it, there is an infinite number of such solutions. Pick any non-zero ratio $r$, and then the first term is $$203 \over (1+r+r^2+r^3)$$You can use $r=1$ and get the trivial sequence 50.75,  50.75,  50.75,  50.75. Another example is −40.6,  81.2,  −162.4,  324.8. $\endgroup$ – Peregrine Rook Jul 24 '17 at 1:48
  • $\begingroup$ Well, thanks @PeregrineRook, got it...as it a class 9th/10th mathematics problem. However, placing a condition such as all terms should be integers brings in some puzzling elelment in it, I guessed and hence this puzzle. $\endgroup$ – Mea Culpa Nay Jul 24 '17 at 8:08

Let the terms be $a$, $ak$, $ak^2$, $ak^3$. Their sum is:


This should be equal to $203=7*29$. None of the non-trivial divisors of $203$ is one more than a square, so $k$ cannot be an integer. Suppose then that $k=m/n$. This means that $a$ is divisible by $n^3$, say $a=bn^3$. We then have

$b(m+n)(m^2+n^2) = 7*29$

At least one of the factors must be $1$, so clearly $b=1$. That leaves $m+n=7$ and $m^2+n^2=29$. Therefore $\{m,n\}=\{2,5\}$. This gives the solution $8, 20, 50, 125$.

P.S. My first attempt was wrong. This solution has now been corrected.

  • $\begingroup$ It seems a vital point is being missed out....@Jaap Scherphuis. Perhaps I can provide a hint. $\endgroup$ – Mea Culpa Nay Jul 18 '17 at 15:00
  • $\begingroup$ Yes, @Jaap S...s, you got the first part correct! $\endgroup$ – Mea Culpa Nay Jul 19 '17 at 5:26

There are infinitely many real numbers satisfying that. Just choose any k greater than 1, and choose m as 203(k-1)/(k^5-1).

  • $\begingroup$ Good catch @Fhccgn, so let us take integers as of now. $\endgroup$ – Mea Culpa Nay Jul 18 '17 at 14:24
  • 1
    $\begingroup$ Please don't be rude. $\endgroup$ – Rand al'Thor Jul 18 '17 at 14:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.