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An arithmetic progression is given with a common difference $\ne0.$ The $2nd$, the $1st$ and the $3rd$ term of the ap form a geometric progression. Find the common ratio.

So we have the ap: $a_1,a_1+d,a_1+2d$ and the gp: $a_1+d,a_1,a_1+2d.$ How can we find $q?$ Thank you in advance!

I have tried to solve the problem using a system, but I am not sure which equations we can use. For example, $a_1^2=(a_1+d)(a_1+2d)$ and $\dfrac{a_1+d}{2}=a_1+(a_1+2d)$ hold, but I don't see how to use them.

Why is it important for the common difference of the AP to be $\ne0?$ Then we will have a constant sequence in which all terms are equal, right?

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5 Answers 5

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For the geometric progression, we have $$a_1^2=(a_1+d)(a_1+2d)=a_1^2+3a_1d+2d^2.$$ Cancelling $a_1^2$ from both sides, we can see that $$d(3a_1+2d)=0.$$ If $d=0$, then $q=1$.

If $d=-\dfrac{3a_1}{2}$, then $a_1+d=-\dfrac{a_1}{2}$, $a_1+2d=-2a_1$. The common ratio $q=-2$.

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  • $\begingroup$ Thank you for the response! Why $d=-\dfrac{3a_1}{2}$ in the second case? $\endgroup$ Commented Dec 28, 2020 at 17:57
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    $\begingroup$ Either $d=0$ or $3a_1+2d=0$ in the second formula. The equation $3a_1+2d=0$ gives $d=-3a_1/2$. $\endgroup$ Commented Dec 28, 2020 at 17:57
  • $\begingroup$ Thank you! We have from the text of the problem that $d\ne0,$ but if it is allowed for it to be $0,$ how did you get that $q=1?$ $\endgroup$ Commented Dec 28, 2020 at 18:00
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    $\begingroup$ If $d=0$, then $a_1=a_1+d=a_1+2d$. It is a constant sequence, hence $q=1$. $\endgroup$ Commented Dec 28, 2020 at 18:01
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    $\begingroup$ Yes, that is the case! $\endgroup$ Commented Dec 28, 2020 at 18:04
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This implies that $$a_1+d=\frac1qa_1$$$$a_1+2d=qa_1$$So, we have $$d=\left(q-\frac1q\right)a_1$$$$a_1\cdot\left(1+q-\frac1q\right)=\frac1qa_1$$Assuming $a_1\neq0$, which is pretty obvious, we have $$1+q=\frac2q$$You can solve from here.

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  • $\begingroup$ Thank you for the response! I don't see why $d=(q-\dfrac{1}{q})a_1.$ Isn't it $d=(\dfrac{1}{q}-1)a_1?$ $\endgroup$ Commented Dec 28, 2020 at 17:51
  • $\begingroup$ I really don't understand how you got the second group of equalities. I see why $a_1+d=\dfrac{1}{q}a_1$ and $a_1+2d=qa_1,$ but I don't see what happens after. $\endgroup$ Commented Dec 28, 2020 at 17:56
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    $\begingroup$ I subtracted eq 2 - eq 1 to get the next equation, and then substituted $d$ into equation $1$ to get the 4th equation. $\endgroup$ Commented Dec 28, 2020 at 19:03
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You see, the basic idea for an AP is the common difference . Similarly, for the GP(geometric progression) the basic idea is the common ratio .

Now, If you assume an AP (a, a+d, a+2d) then you get the GP as- (a+d, a, a+2d) . Here the common ratio will always be same.... So, Your first approach may be to equate the terms, find a relation and get the values and this completely works!

Now coming to the method you used............ (a+d)²=a(a+2d) is the correct and the same equation which you get if you apply the method I suggested above. This is because of the fact that the equation you used is the formula for the geometric mean which is derived from the way I suggested...... But you see, The second equation you wrote is not correct. The equation should have been this- 2(a+d)=a+(a+2d) and let me clear you that this formula won't help you do this question as it is always true for any a and d .

Now if you ask how to solve this question, then you should understand that we have one equation and two variables, so there are infinite solutions of this equation(infinite values of a and d). So you just need to find a relation between a and d(which you will get from the equation) and then form the GP which involves only one variable (either a or d)....... Then find the common ratio of that GP...... As easy as that........ :)

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  • $\begingroup$ Hi! Also If d=0, then there would be infinite number of GP's among which is 0,0,0.....but here the problem is that you cannot determine the common ratio...... $\endgroup$ Commented Dec 28, 2020 at 18:17
  • $\begingroup$ So I guess that's why d≠0......if I am not correct then please feel free to correct me :) $\endgroup$ Commented Dec 28, 2020 at 18:18
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The three terms of a geometric series $a_0, a_1= a_0 + d, a_2 = a_0+2d$ or if it is easier $a_0= a_1-d, a_1, a_2= a_1 + d$.

The three terms of a geometric series is $b_0, b_1=b_0\cdot r, b_2=b_0\cdot r^2$ or if it is easier $b_0=\frac {b_1}r, b_1, b_2= b_1\cdot r$.

We are told that $a_1, a_1 -d, a_1 +d$ form a geometric series. so

$\frac {b_1}r = a_1$. $b_1 = a_1-d$, $b_1r = a_1 + d$.

So Solve for $r$.

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substitute $b_1$ with $a_1 -d$ and

$\frac {a_1 -d}r = a_1$ and $(a_1 -d)r = a_1 + d$.

So $r = \frac {a_1 -d}{a_1} = 1 - \frac d{a_1}$ and $r = \frac {a_1 + d}{a_1-d}=1 + \frac {2d}{a_1-d}$

So $-\frac d{a_1} = \frac {2d}{a_1-d}$ so $-\frac 1{a_1} = \frac 2{a_1-d}$ so

$-(a_1 -d) = 2a_1$ so $d = 3a_1$.

Substituting $3a_1$ for $d$ we get

$r = \frac {a_1 -d}{a_1} = \frac {-2a_1}{a_1} = -2$ and $r = \frac {a_1 + d}{a_1-d}= \frac {4a_1}{-2a_1} = -2$.

So $r = -2$.

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Furthermore the three terms as an arithmetic series $a_0 = a_1-d =-2a_1; a_1 = a_1; a_2= a_1 + d = 4a_1$.

And as a geometric series they are $b_0 = a_1; b_1 =-2a_1; b_2 = 4a_1$.

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$a_1 = q(a_1 + d)$ and $(a_1 + 2d) = qa_1 = q^2 (a_1 + d)$

(So $q = \pm \sqrt{\frac{a_1 + d}{a_1 + 2d}}$.)

We can remark that $a_1 (1-q) = qd$ and $a_1 (1-q) = -2d$ so $-2d = qd$, so $q=-2$ if $d\neq 0$.

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