The first term of an infinite GP is 1 and each term is twice the sum of the succeeding terms, then the sum of the series is

Question

The first term of an infinite GP is 1 and each term is twice the sum of the succeeding terms, then the sum of the series is (A)  2  (B)  (5/2)  (C)  (7/2)  (D)  (3/2)

Tardigrade Admin · Accepted Answer

According to question 1=2(r+r2+r3+....) (1/2)=(r/1-r) ⇒ r=(1/3) ∴ The series is 1,(1/3),(1/9),(1/27),...... Required sum =1+(1/3)+(1/9)+..... =(1/1-(1)3)=(3/2)

Q. The first term of an infinite $GP$ is $1$ and each term is twice the sum of the succeeding terms, then the sum of the series is

$2$

$\frac{5}{2}$

$\frac{7}{2}$

$\frac{3}{2}$

$\frac{9}{2}$

According to question $1 = 2 (r + r^{2} + r^{3} + ....)$ $\frac{1}{2} = \frac{r}{1 - r}$
$\Rightarrow$ $r = \frac{1}{3}$
$∴$ The series is $1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, ...... < b r >$ Required sum
$= 1 + \frac{1}{3} + \frac{1}{9} + .....$
$= \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}$

Q. The first term of an infinite GP is 1 and each term is twice the sum of the succeeding terms, then the sum of the series is

2

25​

27​

23​

29​

According to question 1=2(r+r2+r3+....) 21​=1−rr​ ⇒ r=31​ ∴ The series is 1,31​,91​,271​,......<br> Required sum =1+31​+91​+..... =1−31​1​=23​