Question

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

Answer 1

In this question, we have to find the common ratio of the given term in GP. We will use the general definition of GP to find the solution. We proceed by considering the first term be $a$ and the common ratio be $r$. We then use the given fact that every term is the sum of two previous terms to find a relation between common ratios and then solve to get the required result.
We will also use the formula for finding any terms of GP. i.e. ${T_n} = a{r^{n - 1}}$, where ${n^{th}}$ term in the G.P.

Complete step by step answer:
Consider the given question,
Let us consider the first terms of the given G.P be $a$ and the common ratio be $r$.
We are given that the terms of GP are positive.
Therefore, $a > 0$ and $r > 0$.
Now we are given that every term is the sum of two previous terms.
i.e. ${T_n} = {T_{n - 1}} + {T_{n - 2}}$
We also know that ${T_n} = a{r^{n - 1}}$, where ${n^{th}}$ term in the G.P.
Therefore, ${T_{n - 1}} = a{r^{n - 2}}$ and ${T_{n - 2}} = a{r^{n - 3}}$
Putting the value we have
$\Rightarrow {T_n} = {T_{n - 1}} + {T_{n - 2}}$
$\Rightarrow a{r^{n - 1}} = a{r^{n - 2}} + a{r^{n - 3}}$
On simplifying, we have
$\Rightarrow \dfrac{{a{r^n}}}{r} = \dfrac{{a{r^n}}}{{{r^2}}} + \dfrac{{a{r^n}}}{{{r^3}}}$
Taking $a{r^n}$common from both side, we have
$\Rightarrow a{r^n}\left( {\dfrac{1}{r}} \right) = a{r^n}\left( {\dfrac{1}{{{r^2}}} + \dfrac{1}{{{r^3}}}} \right)$
Cancelling $a{r^n}$ both side, we get
$\Rightarrow \left( {\dfrac{1}{r}} \right) = \left( {\dfrac{1}{{{r^2}}} + \dfrac{1}{{{r^3}}}} \right)$
Taking LCM on RHS we get,
$\Rightarrow \left( {\dfrac{1}{r}} \right) = \left( {\dfrac{{r + 1}}{{{r^3}}}} \right)$
Cancelling $r$both side and the cross multiplying we have
$\Rightarrow {r^2} = r + 1$
Transferring all terms to LHS with change in sign, we have
$\Rightarrow {r^2} - r - 1 = 0$
This is a quadratic equation of the form $a{x^2} + bx + c = 0$, where $a = 1$, $b = - 1$ and $c = - 1$.
Its solution is given by $r = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
Substituting the values we have,
$r = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4(1)( - 1)} }}{{2(1)}} = \dfrac{{1 \pm \sqrt {1 + 4} }}{2}$
On solving, we get
$r = \dfrac{{1 \pm \sqrt 5 }}{2}$
Since we are given that every terms of GP is positive (i.e. $a > 0$ and $r > 0$)
Therefore, neglecting $r = \dfrac{{1 - \sqrt 5 }}{2}$ as it gives a negative value.
Hence, $r = \dfrac{{1 + \sqrt 5 }}{2}$
Therefore, the common ratio of the series is $r = \dfrac{{1 + \sqrt 5 }}{2}$. So, option (D) is correct.

Note:
i. The standard quadratic equation is of the form $a{x^2} + bx + c = 0$, where $a$, $b$and $c$are real numbers.
The solution of the quadratic equation is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
ii. We use the formula, ${T_n} = a{r^{n - 1}}$ to find the ${n^{th}}$ term of the G.P.
For example, the ${5^{th}}$ term of the GP $2,{\text{ }}4,8 \ldots \ldots$ is given by
Here the first term $a = 2$ and common ratio $r = \dfrac{4}{2} = 2$
Therefore ${5^{th}}$ term is given by ${T_5} = a{r^{5 - 1}}$.
Substituting the values we have,
$\Rightarrow {T_5} = 2 \times {2^{5 - 1}} = 2 \times {2^4} = 2 \times 16 = 32$
Hence ${5^{th}}$ term is $32$