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}$
Answer
Verified
504.9k+ views
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.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success
Master Class 10 Computer Science: Engaging Questions & Answers for Success
Master Class 10 Science: Engaging Questions & Answers for Success
Master Class 10 Social Science: Engaging Questions & Answers for Success
Master Class 10 Maths: Engaging Questions & Answers for Success
Master Class 10 English: Engaging Questions & Answers for Success
Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
The capital of British India was transferred from Calcutta class 10 social science CBSE
Explain the Treaty of Vienna of 1815 class 10 social science CBSE