Question

Please help me in solving this 30 question.
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Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Let us consider positive terms of GP series as

$\rm :\longmapsto\:a_1,a_2,a_3,a_4, - - - -$

Let further assume that first term of GP series is a and Common ratio of GP series is r and ( r > 0 )

Further it is given that each term is equal to sum of next two terms.

So, it implies its an decreasing GP series with common ratio, r < 1

Now,

According to statement, each term is equal to sum of next two terms.

$\rm \implies\:a_1 = a_2 + a_3$

$\rm \implies\:a = ar + {ar}^{2}$

$\rm \implies\:1 = r + {r}^{2}$

$\rm \implies\:{r}^{2} + r - 1 = 0$

So, its a quadratic equation in 'r', So by using quadratic formula, we have

$\rm :\longmapsto\:\boxed{ \tt{ \: r = \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}$

So, here

$\red{\rm :\longmapsto\:a = 1}$

$\red{\rm :\longmapsto\:b = 1}$

$\red{\rm :\longmapsto\:c = \: - \: 1}$

So, on substituting the values in above formula, we get

$\rm :\longmapsto\:r \: = \: \dfrac{ - 1 \: \: \pm \: \sqrt{ {1}^{2} - 4(1)( - 1)} }{2(1)}$

$\rm :\longmapsto\:r \: = \: \dfrac{ - 1 \: \: \pm \: \sqrt{ 1 + 4} }{2}$

$\rm :\longmapsto\:r \: = \: \dfrac{ - 1 \: \: \pm \: \sqrt{5} }{2}$

$\rm :\longmapsto\:r \: = \: \dfrac{ - 1 \: + \:\sqrt{5} }{2} \: \: \: or \: \: \: \dfrac{ - 1 \:- \:\sqrt{5} }{2} \: \{rejected \}$

$\rm \implies\:\boxed{ \tt{ \: \: r \: \: = \: \: \frac{ \sqrt{5} - 1 }{2} \: \: }}$

• Hence, Option (b) is correct

Basic Concept Used :-

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an geometric sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\: {r}^{n \: - \: 1} }}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

r is the common ratio.