Question

If in a G.P. of positive terms every term is sum of two preceding terms, then its common ratio is
A. $ \dfrac{{1 - \sqrt 5 }}{2} $
B. $ \dfrac{{\sqrt 5 + 1}}{2} $
C. $ \dfrac{{\sqrt 5 - 1}}{2} $
D.1

Answer 1

Hint: In the question, we are given that every positive term of a GP is the sum of two preceding. For that firstly, consider any GP series. Then use the value for the nth term of a GP and then put the value into the formed equation. After solving we will get the common ratio.

Complete step by step solution:
In the question, we are given the G.P. series whose full form is geometric progression which means if in a sequence of terms, each succeeding term is generated by multiplying each preceding term with the constant value and that constant value is known as the common ratio.It can be written in the form of $ a,ar,a{r^2},a{r^3} \ldots \ldots $ where r is the common ratio.
According to the question, “every positive term is the sum of two preceding terms”.
Then firstly, considering the G.P. terms be
 $ {a_n},{a_{n + 1}},{a_{n + 2}} $
Now, writing the terms in the form of G.P.,
 $ {a_n} = a{r^{n - 1}} $
Similarly, the other two $ {a_{n + 1}} = a{r^n} $ and $ {a_{n + 2}} = a{r^{n + 1}} $
Using the given statement, writing
 $ {a_n} = {a_{n + 1}} + {a_{n + 2}} $
Putting the values
 $ a{r^{n - 1}} = a{r^n} + a{r^{n + 1}} $
On solving,
 $ 1 = r + {r^2} $ or can be written as
$ {r^2} + r - 1 = 0 $
Solving the quadratic equation,
 $
  r = \dfrac{{ - 1 \pm \sqrt {1 - 4( - 1)} }}{2} \\
   = \dfrac{{ - 1 \pm \sqrt 5 }}{2} \;
  $
But $ r > 0 $
Hence, the value $ r = \dfrac{{ - 1 - \sqrt 5 }}{2} $ is rejected.
The common ratio for the above question is $ r = \dfrac{{ - 1 + \sqrt 5 }}{2} $
So, the correct answer is “Option C”.

Note: Read the question carefully as to which series (AP OR GP) we have to use. Mostly students got confused while selecting the series for the question. If the series is not given then make sure to consider the nth term for the series. Consecutive terms are taken as $ \left( {n + 1} \right),\left( {n + 2} \right), \ldots \ldots $