Question

The sum of the series `(1)/(3) (1)/(9) (1)/(27)...oo` is
a (2)/(3)
b (1)/(3)
c (1)/(2)
d (1)/(6)​

Answer 1

Answer:

1/2

Step-by-step explanation:

The given series is a GP,  in this GP:

     a = 1/3     ;     r = (1/9) / (1/3) = 1/3

Using,

      Sum of ∞ terms(GP) = a/(1 - r)

⇒ S = (1/3) / (1 - 1/3)

       = (1/3) / (2/3)

       = 1/2

Correct option is (c)

Answer 2

Given :-

1/3, 1/9, 1/27

To Find :-

Sum of series

Solution :-

We know that

r = a₂/a₁ = a₃/a₂

r = (1/9)/(1/3) = (1/27)/(1/9)

r = 1/9 × 3/1 = 1/27 × 9/1

r = 3/9 = 9/27

r = 1/3 = 1/3

Now

Sum of series = a/(1 - r)

Where

a = first term of the GP = 1/3

r = common ratio of the GP = 1/3

$\sf \implies Sum= \dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}$

$\sf \implies Sum= \dfrac{\dfrac{1}{3}}{\dfrac{3-1}{3}}$

$\sf \implies Sum= \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}$

$\sf \implies Sum= \dfrac{1}{3}\times\dfrac{3}{2}$

$\sf \implies Sum= \dfrac{1}{2}$