Question

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals
$(a) \sqrt{5} \\(b) \frac{1}{2} (\sqrt{5}-1)\\(c) \frac{1}{2} (1-\sqrt{5})\\(d) \frac{1}{2} \sqrt{5}.$

Answer 1

Answer:

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

option (b) is correct

Step-by-step explanation:

Formula used:

The n th term of the G.P

$a,ar,ar^2.......\:is\:t_n=ar^{n-1}$

Given:

$t_n=t_{n+1}+t_{n+2}$

$ar^{n-1}=ar^{n}+ar^{n+1}$

$ar^{n-1}=a[r^{n}+r^{n+1}]$

$r^{n}r^{-1}=r^{n}+r^n.r$

$r^{n}r^{-1}=r^n[1+r]$

$r^{-1}=1+r$

$\frac{1}{r}=1+r$

$1=r+r^2$

$r^2+r-1=0$

Then,

$r=\frac{-b+\sqrt{b^2-4ac}}{2a}\:and\:r=\frac{-b-\sqrt{b^2-4ac}}{2a}$

$r=\frac{-1+\sqrt{1^2-4(1)(-1)}}{2(1)}\:and\:r=\frac{-1-\sqrt{1^2-4(1)(-1)}}{2(1)}$

$r=\frac{-1+\sqrt{1+4}}{2}\:and\:r=\frac{-1-\sqrt{1+4}}{2}$

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

since the G.P consisting positive terms, r is positive

Therefore,

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

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

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

Answer 2

Step-by-step explanation:

r=21(5−1)

option (b) is correct

Step-by-step explanation:

Formula used:

The n th term of the G.P

a,ar,ar^2.......\:is\:t_n=ar^{n-1}a,ar,ar2.......istn=arn−1

Given:

t_n=t_{n+1}+t_{n+2}tn=tn+1+tn+2

ar^{n-1}=ar^{n}+ar^{n+1}arn−1=arn+arn+1

ar^{n-1}=a[r^{n}+r^{n+1}]arn−1=a[rn+rn+1]

r^{n}r^{-1}=r^{n}+r^n.rrnr−1=rn+rn.r

r^{n}r^{-1}=r^n[1+r]rnr−1=rn[1+r]

r^{-1}=1+rr−1=1+r

\frac{1}{r}=1+rr1=1+r

1=r+r^21=r+r2

r^2+r-1=0r2+r−1=0

Then,

r=\frac{-b+\sqrt{b^2-4ac}}{2a}\:and\:r=\frac{-b-\sqrt{b^2-4ac}}{2a}r=2a−b+b2−4acandr=2a−b−b2−4ac

r=\frac{-1+\sqrt{1^2-4(1)(-1)}}{2(1)}\:and\:r=\frac{-1-\sqrt{1^2-4(1)(-1)}}{2(1)}r=2(1)−1+12−4(1)(−1)andr=2(1)−1−12−4(1)(−1)

r=\frac{-1+\sqrt{1+4}}{2}\:and\:r=\frac{-1-\sqrt{1+4}}{2}r=2−1+1+4andr=2−1−1+4

r=\frac{-1+\sqrt{5}}{2}\:and\:r=\frac{-1-\sqrt{5}}{2}r=2−1+5andr=2−1−5

since the G.P consisting positive terms, r is positive

Therefore,

r=\frac{-1+\sqrt{5}}{2}r=2−1+5

r=\frac{\sqrt{5}-1}{2}r=25−1

r=\frac{1}{2}(\sqrt{5}-1)r=21(5