Question

If the sum of infinite gp is 9/4 and the second term is 1/2 find the series

Answer 1

Step-by-step explanation:

hope u will understand

Answer 2

Answer:

The series is either  $\frac{3}{2}$ , $\frac{1}{2}$  , $\frac{1}{6}$ , . . . . or  $\frac{3}{4}$ , $\frac{1}{2}$  , $\frac{1}{3}$ , . . . .

Step-by-step explanation:

Given that the sum of infinite GP = $\frac{9}{4}$

and the second term = $\frac{1}{2}$

We know that sum of an infinite GP = $\frac{a}{1-r}$

where a = First term

           r = Common ratio , r < 1

On substituting the values, we get,

$\frac{9}{4}$ =  $\frac{a}{1-r}$

=> 9(1 - r) = 4a

=> 9 - 9r = 4a

=> a =  $\frac{9-9r}{4}$

We know that the nth term of a GP = a rⁿ⁻¹

where a = First term

           r = Common ratio

           n = Number of terms

On substituting the values, we get,

$\frac{1}{2}$ = a * r²⁻¹

=> $\frac{1}{2}$ = a * r

=> 1 = 2ar

Substituting the value of a above,

=> 1 = 2 $\frac{9-9r}{4}$ *r

=> 2 = r(9-9r)

=> 2 = 9r - 9r²

=> 9r² - 9r + 2 = 0

=> 9r² - 6r - 3r + 2 = 0

=> 3r(3r - 2) -1(3r - 2) = 0

=> (3r - 1) (3r - 2) = 0

either 3r - 1 = 0  or 3r - 2 = 0

=> r = 1/3 or r = 2/3

Case 1 : if r = 1/3,

a =  $\frac{9-9r}{4}$ =  $\frac{9-9(\frac{1}{3} )}{4}$ =  $\frac{9-3}{4}$ =  $\frac{6}{4}$ =  $\frac{3}{2}$

Then the GP =  a, ar , ar² , . . . .

                     =  $\frac{3}{2}$ , $\frac{3}{2}$ * $\frac{1}{3}$ , $\frac{3}{2}$ * ( $\frac{1}{3}$ )²

                     = $\frac{3}{2}$ , $\frac{1}{2}$  , $\frac{1}{6}$ , . . . .

Case 2 : if r = 2/3,

a =  $\frac{9-9r}{4}$ =  $\frac{9-9(\frac{2}{3} )}{4}$ =  $\frac{9-6}{4}$ =  $\frac{3}{4}$

Then the GP =  a, ar , ar² , . . . .

                     =  $\frac{3}{4}$ , $\frac{3}{4}$ * $\frac{2}{3}$ , $\frac{3}{4}$* ( $\frac{2}{3}$ )²

                     = $\frac{3}{4}$ , $\frac{1}{2}$  , $\frac{1}{3}$ , . . . .

Therefore, the series is either  $\frac{3}{2}$ , $\frac{1}{2}$  , $\frac{1}{6}$ , . . . . or  $\frac{3}{4}$ , $\frac{1}{2}$  , $\frac{1}{3}$ , . . . .