Question

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Find 3 numbers in G.P. such that their sum is 21 and sum of squares is 189.
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Answer 1

Answer:

Let three numbers in G.P be a,ar,ar

2

Since the sum of the numbers is 21

∴a+ar+ar

2=21⇒a(1+r+r

2)=21 .....(1)

Since the sum of the squares of the numbers is 189

∴a

2+a 2 r 2

+a

2

r

4

=189

⇒a

2

(1+r

2

+r

4

)=189 ......(2)

Squaring both sides of (1), we get

a

2

(1+r+r

2

)

2

=441 .......(3)

Dividing (3) by (2), we get

a

2

(1+r

2

+r

4

)

a

2

(1+r+r

2

)

2

=

189

441

⇒

(1+r

2

+r

4

)

(1+r+r

2

)

2

=

9

21

⇒

(1+2r

2

+r

4

−r

2

)

(1+r+r

2

)

2

=

3

7

⇒

(r

2

+1)

2

−r

2

(1+r+r

2

)

2

=

3

7

⇒

(r

2

+1−r)(r

2

+1+r)

(1+r+r

2

)

2

=

3

7

⇒

(r

2

+1−r)

(1+r+r

2

)

=

3

7

⇒3r

2

+3r+3=7r

2

−7r+7

⇒4r

2

−10r+4=0

⇒2r

2

−5r+2=0

⇒r=

2×2

5±

5

2

−4×2×2

=

4

5±

25−16

=

4

5±

9

=

4

5±3

=

4

8

,

4

2

=2,

2

1

When r=2 from (1)

a(1+2+2

2

)=21⇒a=3

∴ Numbers are 3,3×2,3×2

2

or 3,6,12

When r=

2

1

from (1)

a[1+

2

1

+

4

1

]=21

⇒a[

4

4+2+1

]=21

⇒a=

7

21×4

=3×4=12

∴ Numbers are 12,12×

12

1

,12×(

12

1

)

2

or 12,6,3

Thus the number is 12,6 and 3.

Hope it helpful to uh❤

Answer 2

Answer:

$\huge\boxed{\fcolorbox{black}{pink}{ANSWER:-}}$

Step-by-step explanation:

ANSWER IN THE ATTACHMENT

15❤+ FOLLOW = IBX✉