Math, asked by dasaatreyee01, 4 months ago

show that all the values of i^i are real and in GP

Answers

Answered by Anonymous
0

Step-by-step explanation:

Answer

Let the three terms in GP be x,xr, xr

2

x+xr+xr

2

=aS

x

2

+x

2

r

2

+x

2

r

4

=S

2

Therefore,

1+r

2

+r

4

(1+r+r

2

)

2

=a

2

(1+r+r

2

)×(1−r+r

2

)

(1+r+r

2

)

2

=a

2

(1−r+r

2

)

(1+r+r

2

)

=a

2

(1−a

2

)r

2

+(1+a

2

)r+(1−a

2

)=0

Since r is real, (1+a

2

)

2

−4(1−a

2

)

2

≥0

Let a

2

=t

(1+t)

2

−4(1−t)

2

≥0

(3t−1)(t−3)≤0

From the equation, we get

3

1

≥a

2

≤3

But for a

2

=1, r=0 and GP will not be defined.

Hence,

3

1

≥a

2

<1 and 1>a

2

≤3

Answered by Legend42
17

Answer:

Let the three terms in GP be x,xr, xr

2

x+xr+xr

2

=aS

x

2

+x

2

r

2

+x

2

r

4

=S

2

Therefore,

1+r

2

+r

4

(1+r+r

2

)

2

=a

2

(1+r+r

2

)×(1−r+r

2

)

(1+r+r

2

)

2

=a

2

(1−r+r

2

)

(1+r+r

2

)

=a

2

(1−a

2

)r

2

+(1+a

2

)r+(1−a

2

)=0

Since r is real, (1+a

2

)

2

−4(1−a

2

)

2

≥0

Let a

2

=t

(1+t)

2

−4(1−t)

2

≥0

(3t−1)(t−3)≤0

From the equation, we get

3

1

≥a

2

≤3

But for a

2

=1, r=0 and GP will not be defined.

Hence,

3

1

≥a

2

<1 and 1>a

2

≤3

Similar questions