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In Geometric Progression, first member is equal to 7 and the common ratio is equal to 4. How many members should be taken in to the sequence that it would amount to 7340025?

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7*(4^(10)-1)=7340025 and the sum of the n first terms of the sequence is 7*(4^n-1)/3 hence no partial sum equals 7340025, the closest one being the sum of the 10 first terms 7*(4^(10)-1)/3=2446675 and the sum of the 11 first terms 7*(4^(11)-1)/4=9786707. (No member of the sequence is 7340025 either, the closest one being the eleventh term 7*4^(10)=7340032.) –  Did Aug 25 '11 at 10:31

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up vote 3 down vote accepted

If the answer is given, I bet it is $10$. But in fact there is no answer, one number in the problem is wrong.

The sum of the first $n$ terms of the generic geometric progression is given by $$a+ar+ar^2+\cdots+ar^{n-1}=\frac{a(r^n-1)}{r-1}.$$

Put $a=7$, and $r=4$. We want to find $n$ such that $$\frac{7(4^n-1)}{3}=7340025.$$

Easily we find that $4^n=(3/7)(7340025)+1$.

Let's do the arithmetic. We get a number which is not an integer power of $4$. Since $n$ must be an integer, the problem cannot be solved, there is no geometric progression with the desired property. (This was pointed out in the comment by Didier Piau.)

But now in the calculation, let's do the division by $7$, but forget to multiply by $3$. Equivalently, let's use the wrong formula for the sum of a geometric progression, and forget about the $r-1$ in the denominator. Then we get $4^n=1048576$. Note that $1024^2=2^{20}=4^{10}=1048576$. So the conclusion is that $n=10$.

Or else everything was done right, except that $r=2$ was used in the calculation.

Whoever made up the problem naturally started from the desired answer of $n=10$. Then the person worked backwards using the formula. So first (s)he subtracted $1$, then multiplied by $7$, then forgot to divide by $3$, or worked from the wrong formula. Or else the backwards calculation started from $2^{20}$, which is $4^{10}$, and the person didn't realize that summing consecutive powers of $4$ is not the same as summing consecutive powers of $2$.

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I can't believe this has happened to me twice (Error/Mistake in the problem itself). Thank you for answering Andre. –  BeatShot Aug 25 '11 at 11:32
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@BeatShot: it is not so rare-mistakes creep in. If you think you can solve the problem, see if you can check it. In this case, a quick spreadsheet would show the behavior André Nicolas describes and would give confidence that you are right and the book is wrong. –  Ross Millikan Aug 25 '11 at 15:56

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