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Problem:

The second, third and sixth terms of an arithmetic progression are consecutive terms of a geometric progression. Find the common ratio of the geometric progression.

My attempt:

I thought of rewriting the terms as $a+d,a+2d,a+5d$ where $a$ is the first term and $d$ is the common difference. Now, since these are consecutive terms of a GP, $$\Longrightarrow \dfrac{a+2d}{a+d}=\dfrac{a+5d}{a+2d}=r$$ where $r$ is the common ratio. However, when I cross-multiplied the terms$\dfrac{a+2d}{a+d}=\dfrac{a+5d}{a+2d},$ I got $$a^2+4ad+4d^2=a^2+6ad+5d^2$$ I'm unable to proceed any further. Any help would be really appreciated. Many thanks!

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  • $\begingroup$ Drop the first term and you have 'the first, second and fifth term'. That doesn't essentially change the problem, but simplifies calculations a little bit. $\endgroup$
    – CiaPan
    Jul 5 '15 at 18:25
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$$a^2+4ad+4d^2=a^2+6ad+5d^2$$

$$2ad+d^2=0$$

$$d(2a+d)=0$$

$$d=0 \vee 2a+d=0$$

If $d$ is zero, then the ratio is clearly 1.

If $2a+d=0$ is zero, you have $d=-2a$. Substitute this in the first ratio we get $r=\frac{-3a}{-a}=3$.

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  • $\begingroup$ You are welcome. $\endgroup$
    – wythagoras
    Jul 5 '15 at 17:48
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$$a^2+4ad+4d^2=a^2+6ad+5d^2$$

This gives you $$d(2a+d)=0$$ so that either $d=0$ or $d= -2a$.

Your common ratio is given by $$\frac{a+2d}{a+d}$$ Substituting $d=0$ into that gives you the common ratio as $1$. This would make both the geometric and the arithmetic series constant, so we discard that solution.

Substitute $d=-2a$ into the common ratio equation to get $$r = 3.$$

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Note that $$a^2+4ad+4d^2=a^2+6ad+5d^2$$ $$\iff 2ad+d^2=0\iff d(2a+d)=0\iff d=0\ \ \text{or}\ \ 2a+d=0$$

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More generally, if $i < j < k$, suppose the $i$-th, $j$-th, and $k$-th terms in an arithmetic progression are in a geometric progression with ratio $r$.

I will show that (1) if $ik \ne j^2$ then $r=\frac{k-j}{j-i}$; (2) if $ik = j^2$ then $r=\frac{j}{i}$.

We have $r =\frac{a+jd}{a+id} =\frac{a+kd}{a+jd} $. Cross-multiplying, $(a+id)(a+kd) =(a+jd)^2 $ so $a^2+(i+k)ad+ikd^2 =a^2+2ajd+j^2d^2 $ or $d(d(j^2-ik)+a(2j-(i+k))) = 0 $.

Therefore either $d = 0$ or $d(j^2-ik)+a(2j-(i+k))=0$.

As a check, with $i=1, j=2, k=5$ in the OP, this is $-d-2a=0$, which matches.

So we must have $d(j^2-ik)+a(2j-(i+k))=0$.

There are two cases, depending if $ik \ne j^2$ or $ik = j^2$.

If $ik \ne j^2$, $d =\frac{a(2j-(i+k))}{ik-j^2} $. Then,

$\begin{array}\\ r &=\dfrac{a+jd}{a+id}\\ &=\dfrac{a+j\frac{a(2j-(i+k))}{ik-j^2}}{a+i\frac{a(2j-(i+k))}{ik-j^2}}\\ &=\dfrac{a(ik-j^2)+j(a(2j-(i+k)))}{a(ik-j^2)+i(a(2j-(i+k)))}\\ &=\dfrac{ik-j^2+2j^2-j(i+k)}{ik-j^2+2ij-i(i+k)}\\ &=\dfrac{j^2-ij-jk+ik}{-j^2+2ij-i^2}\\ &=\dfrac{j(j-i)-k(j-i)}{-(i-j)^2}\\ &=\dfrac{(k-j)(j-i)}{(j-i)^2}\\ &=\dfrac{k-j}{j-i}\\ \end{array} $

In the OP's case, we get $\frac{5-2}{2-1} =\frac{3}{1} = 3 $, which, again, matches.

If $ik = j^2$, then $a(2j-(i+k))=0 $.

If $a=0$, then $r = \frac{j}{i} = \frac{k}{j} $ and any $d$ works.

If $a \ne 0$, then $2j = i+k$. Then $(i+k)^2 = 4j^2 =4ik $ so $i^2+2ik+k^2 =4ik$ or $(i-k)^2 = 0$. This contradicts our original assumption of $i < j < k$, so this can not happen.

Therefore we can always get the ratio, without knowing the initial or increment values.

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let 2nd, 3rd and 6th term of AP are a+d, a+2d and a+5d. a+d = A, a+2d = Ar and a+5d =ar^2 a+2d = Ar or (a+d) +d = Ar A + d = Ar A=d/(r-1)-----------1) a+5d =Ar^2 a+2d + 3d = Ar^2 Ar + 3d = A*r^2 A = 3d/r(r-1) ----------2) by eq 1) and 2) d/(r-1) = 3d/r(r-1) 1 = 3/r r =3

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