Question

the first three terms of a G.p are x+10,x-2,x-0 respectively claculate the vale of x and find the sum of infinity of this series​

Answer 1

Answer:

Hii

Step-by-step explanation:

Let the G.P be

r

a

,a,ar.

So,

r

a

+a+ar=13

And

r

a

×a×ar=27

a

3

=27

a=3

Therefore,

r

3

+3+3r=13

3+3r+3r

2

=13r

3r

2

−10r+3=0

3r

2

−9r−r+3=0

3r(r−3)−1(r−3)=0

(3r−1)(r−3)=0

r=

3

1

,3

So, r=3

So, the G.P is 1,3,9.

Now, the sum of first five terms

=

3−1

3(3

5

−1)

=

2

3(243−1)

=

2

3(242)

=3(121)=363

Hence, this is the answer.

Answer 2

Let the G.P be

r

a

,a,ar.

So,

r

a

+a+ar=13

And

r

a

×a×ar=27

a

3

=27

a=3

Therefore,

r

3

+3+3r=13

3+3r+3r

2

=13r

3r

2

−10r+3=0

3r

2

−9r−r+3=0

3r(r−3)−1(r−3)=0

(3r−1)(r−3)=0

r=

3

1

,3

So, r=3

So, the G.P is 1,3,9.

Now, the sum of first five terms

=

3−1

3(3

5

−1)

=

2

3(243−1)

=

2

3(242)

=3(121)=363

I HOPE ITS HELPFUY