The $2$nd, the $1$st and the $3$rd term of an arithmetic progression form a geometric progression

Question

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?

Answer 1

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 $q=-2$.

Answer 2

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.

Answer 3

$a_1 = q(a_1 + d)$ 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 $d\neq 0$.

Answer 4

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........ :)

Answer 5

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 $r = -2$.

......

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 $b_0 = a_1; b_1 =-2a_1; b_2 = 4a_1$.