Question

find three number in g.p such that their sum is 21 and sum of their squares is 189

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Answer 1

Answer:

Required numbers are 3 , 6 and 12.

Step-by-step explanation:

Let,

the common ratio of this GP be r.

the common ratio between these terms be r^a.

r^a = k .

Let the required terms are a / k , a and ak , where a represents the first terms and k represents the common ratio between these terms.

Given,

Sum of these terms is 21.

Sum of their squares is 189.

Case 1 : Sum of these terms is 21 :

= > a / k + a + ak = 21 ... ( 1 )

Case 2 : Sum of their squares is 189 :

= > ( a / k )^2 + a^2 + ( ak )^2 = 189 ... ( 2 )

Square on both sides of ( 1 ) :

= > ( a / k + a + ak )^2 = 21^2

= > ( a / k )^2 + a^2 + ( ak )^2 + 2[ ( a / k x a ) + ( a x ak ) + ( ak x a / k ) ] = 441 { using ( a + b + c )^2 = a^2 + b^2 + c^2 + 2( ab + bc + ca )

= > 189 + 2( a^2 / k + a^2 k + a^2 ) = 441

= > 2( a^2 / k + a^2 k + a^2 ) = 441 - 189

= > a^2( 1 / k + k + 1 ) = 252 / 2

= > a^2( 1 / k + k + 1 ) = 126

= > a( a / k + ak + a ) = 126

= > a( 21 ) = 126 { from ( 1 ) }

= > a = 126 / 21

= > a = 6

Substituting the value of a in ( 1 ) :

= > 6 / k + 6 + 6k = 21

= > 6( 1 / k + 1 + k ) = 21

= > 1 / k + k + 1 = 21 / 6

= > ( 1 + k^2 ) / k + 1 = 7 / 2

= > ( 1 + k^2 ) / k = 5 / 2

= > 2k^2 - 5k + 2 = 0

= > 2k^2 - ( 4 + 1 )k + 2 = 0

= > 2k^2 - 4k - k + 2 = 0

= > 2k( k - 2 ) - ( k - 2 ) = 0

= > ( k - 2 )( 2k - 1 ) = 0

= > k = 2 or 1 / 2

Case 3 : If k = 2

Terms are :

• a / k = 6 / 2 = 3
• a = 6
• ak = 6 x 2 = 12

Result : Numbers are 3 , 6 and 12.

Case 4 : If k = 1 / 2

Terms are

• a / k = 6 / ( 1 / 2 ) = 6 x 2 = 12
• a = 6
• ak = 6 x 1 / 2 = 3

Result : Numbers are 12 , 6 and 3.

Hence the required numbers are 3 , 6 and 12.