Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Question :-

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then, the common ratio of this progression is equal to ______ .

Solution :-

General series of a Geometric Progression :-

a, ar, ar², ar³, . . . . . . ., arⁿ⁻¹

where

a = First term of the Geometric Progression

r = Common ratio of the Geometric Progression

n = Order of term

It is given that,

Each term equals the sum of the next two terms.

So, according to statement,

First term = Second term + Third term

a = ar + ar²

Taking 'a' common,

a(1) = a(r + r²)

1 = r + r²

Rearranging terms,

r² + r - 1 = 0

Using Quadratic formula,

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

$\sf{r=\dfrac{-1\pm\sqrt{1+4}}{2}}$

$\sf{r=\dfrac{-1\pm\sqrt{5}}{2}}$

Since, geometric progression consists of positive terms, we will reject negative value of 'r'.

Thus,

$\sf{r=\dfrac{\sqrt{5}-1}{2}}$

Answer :-

The common ratio of this geometric progression is equal to (√5 - 1)/2.

Hence, option D is correct option.

Answer 2

Step-by-step explanation:

Let the G.P. be a,ar,ar²,ar³, . . . . . . .

acc to ques ,,

→ a = ar+ ar²

→ a = ar(1+r)

→ 1 = r(1+r)

→ r(r+1) = 1

→ r²+r-1 = 0

$= > r \: = \frac{ - 1± \sqrt{ {1}^{2}- 4 \times 1 \times ( - 1) } }{2 \times 1}$

$= > r = \frac{ - 1± \sqrt{1 + 4} }{2}$

$= > r = \frac{ - 1± \sqrt{5} }{2}$

Since, The G.P. consist of positive terms.

so, we will take positive value of 'r'.

$= > r = \frac{ - 1 + \sqrt{5} }{2}$

hence,

$r = \frac{ \sqrt{5} - 1}{2}$