# In a GP with alternatively positive and negative terms and any term is the AM of the next two terms. Then find the common ratio of the GP.

A. –1

B. –3

C. –2

D. $ - \dfrac{1}{2}$

Last updated date: 19th Mar 2023

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Answer

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Hint: In Geometric progression (GP), the terms are alternatively positive and negative means the common ratio will be negative. Arithmetic mean (AM) of two numbers means the average of those two numbers.

Complete step-by-step answer:

Let the first term of the GP be ‘a’ and the common ratio be ‘r’.

It is given that any term in that GP is the AM of the next two terms.

$ \Rightarrow {a_n} = \dfrac{{{a_{n + 1}} + {a_{n + 2}}}}{2}$

${n^{th}}$ term of GP is $a{r^{n - 1}}$

$ \Rightarrow a{r^{n - 1}} = \dfrac{{a{r^n} + a{r^{n + 1}}}}{2}$ [Since any term is the AM of next two terms]

On simplification of the above equation,

$ \Rightarrow 2 = r + {r^2}$

$ \Rightarrow {r^2} + r - 2 = 0$

Factorization of the above quadratic equation for finding possible ‘r’ values

$$\eqalign{

& \Rightarrow {r^2} + 2r - r - 2 = 0 \cr

& \Rightarrow r(r + 2) - 1(r + 2) = 0 \cr

& \Rightarrow (r + 2)(r - 1) = 0 \cr} $$

$ \Rightarrow r = - 2,1$

$\therefore $ The required common ratio of the given GP is $r = - 2$

So option C is correct.

Note: After solving the quadratic equation in the end, we got one positive and one negative value for ‘r’. We took the negative value of ‘r’ because it is given the GP contains the terms which are alternatively positive and negative.

Complete step-by-step answer:

Let the first term of the GP be ‘a’ and the common ratio be ‘r’.

It is given that any term in that GP is the AM of the next two terms.

$ \Rightarrow {a_n} = \dfrac{{{a_{n + 1}} + {a_{n + 2}}}}{2}$

${n^{th}}$ term of GP is $a{r^{n - 1}}$

$ \Rightarrow a{r^{n - 1}} = \dfrac{{a{r^n} + a{r^{n + 1}}}}{2}$ [Since any term is the AM of next two terms]

On simplification of the above equation,

$ \Rightarrow 2 = r + {r^2}$

$ \Rightarrow {r^2} + r - 2 = 0$

Factorization of the above quadratic equation for finding possible ‘r’ values

$$\eqalign{

& \Rightarrow {r^2} + 2r - r - 2 = 0 \cr

& \Rightarrow r(r + 2) - 1(r + 2) = 0 \cr

& \Rightarrow (r + 2)(r - 1) = 0 \cr} $$

$ \Rightarrow r = - 2,1$

$\therefore $ The required common ratio of the given GP is $r = - 2$

So option C is correct.

Note: After solving the quadratic equation in the end, we got one positive and one negative value for ‘r’. We took the negative value of ‘r’ because it is given the GP contains the terms which are alternatively positive and negative.

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