Question

In a GP of alternately positive and negative terms, any term is the AM of the next two terms. Then the common ratio is, choose the correct option.
A. -1
B. -3
C. -2
D. \[-\dfrac{1}{2}\]

Answer 1

Hint: Find the nth term, (n+1)th term and (n+2)th term of the GP and then find the AM (Arithmetic Mean) of the nth term with other two terms. The nth term of the GP is given by \[{{T}_{n}}={{(-1)}^{n-1}}a{{r}^{n-1}}\]. Here, r is the modulus of common ratio and a is the first term of the GP (Geometric Progression).

Complete step-by-step answer:
In the problem, we have to find the common ratio if in the GP of alternately positive and negative terms, any term is the AM of the next two terms.
Now for the alternatively positive and negative term of the GP, we have the first term positive and then negative in the alternate manner. So if the common ratio of the GP is r and the first term is a, then the second term is (-ar). Similarly, the third term is \[a{{r}^{2}}\], fourth term is \[-a{{r}^{3}}\]. So the nth terms will be given by the formula \[{{T}_{n}}={{(-1)}^{n-1}}a{{r}^{n-1}}\].
Next, it is given that any term is the AM Arithmetic Mean) of the next two terms. So we will find the (n+1)th term and (n+2)th term of the GP (Geometric Progression). Which is given as \[{{T}_{n+1}}={{(-1)}^{n}}a{{r}^{n}}\] and \[{{T}_{n+2}}={{(-1)}^{n+1}}a{{r}^{n+1}}\]respectively.
Now, the Arithmetic Mean (AM) of the (n+1)th term and (n+2)th term is the average of these two terms and that will be equal to the nth term, as shown below;
\[\begin{align}
  & \Rightarrow {{T}_{n}}=\dfrac{{{T}_{n+1}}+{{T}_{n+2}}}{2} \\
 & \Rightarrow {{(-1)}^{n-1}}a{{r}^{n-1}}=\dfrac{{{(-1)}^{n}}a{{r}^{n}}+{{(-1)}^{n+1}}a{{r}^{n+1}}}{2} \\
 & \Rightarrow \dfrac{{{(-1)}^{n-1}}}{r}\left( a{{r}^{n}} \right)=\left( a{{r}^{n}} \right)\dfrac{\left[ {{(-1)}^{n}}+{{(-1)}^{n+1}}r \right]}{2} \\
\end{align}\]
\[\begin{align}
  & \text{Cancelling}\,\,\text{the}\,\,\text{common}\,\,\text{term}\,\left( a{{r}^{n}} \right)\text{, we get} \\
 & \Rightarrow \dfrac{{{(-1)}^{n}}}{(-1)r}={{(-1)}^{n}}\dfrac{\left[ 1+(-1)r \right]}{2} \\
 & \Rightarrow \dfrac{1}{(-1)r}=\dfrac{\left[ 1+(-1)r \right]}{2} \\
 & \text{Cross}\,\,\text{multiply, to get:} \\
 & \Rightarrow 2=-r(1-r) \\
 & \Rightarrow {{r}^{2}}-r-2=0 \\
\end{align}\]
Now, we will solve this quadratic equation using the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a},where\,\,a{{x}^{2}}+bx+c=0\] is the quadratic equation. So comparing the above equation with the general form, we get, \[a=1,b=-1,c=-2\]. So, the value of r (common ratio) will be:
\[\begin{align}
  & \Rightarrow r=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\cdot \;1\left( -2 \right)}}{2\cdot \;1} \\
 & \Rightarrow r=\dfrac{1\pm \sqrt{9}}{2} \\
 & \Rightarrow r=2,r=-1 \\
\end{align}\]
Since r is modulus so it’s value cannot be negative.
So out of the given options, we have option C. -2
Hence, option C is the correct answer.

Note: Make sure that we find the AM of (n+1)th term and (n+2)th term of the GP and this is equal to the Arithmetic mean of the nth term. Not the Arithmetic mean of nth term and the (n+2)th term. Also, here n is not known so here we can’t say that \[{{(-1)}^{n}}\] is negative or positive.