Question

Find three numbers in G.P. such that their
sum is 21 and sum of their squares is 189.​

Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Questions}}}}$

Assume,

Common ratio of GP be t

Common ratio between these terms be (t)^a

(t)^a = b

Assume,

${\boxed{\sf\:{Required\;Terms\;are=\dfrac{a}{b},a\;and\;ab}}}$

k = Common ratio

Given,

Sum of terms = 21

Sum of squares = 189

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{a}{b}+a+ab=21.....(1)$

Also,

$\rm \: \: \: \: \: \: \: \: \:=(\dfrac{a}{b})^2+a^2+(ab)^2=189.....(2)$

Squaring both sides of (1)

$\rm \: \: \: \: \: \: \: \: \:=(\dfrac{a}{b}+a+ab)^2=(21)^2$

$\rm \: \: \: \: \: \: \: \: \:=(\dfrac{a}{b})^2+a^2+(ab)^2+2[(\dfrac{a}{b}\times a)+(a\times ab)+(ab\times\dfrac{a}{b})]=441$

$\rm \: \: \: \: \: \: \: \: \:=189+2(\dfrac{a^2}{b}+a^2 b+a^2=441$

$\rm \: \: \: \: \: \: \: \: \:=2(\dfrac{a^2}{b}+a^2 b+a^2=441-189$

$\rm \: \: \: \: \: \: \: \: \:=a^2(\dfrac{1}{b}+b+1=\dfrac{252}{2}$

$\rm \: \: \: \: \: \: \: \: \:=a^2(\dfrac{1}{b}+b+1=126$

$\rm \: \: \: \: \: \: \: \: \:=a(\dfrac{a}{b}+ab+a=126$

From eq (1) we have

a(21) = 126

$\rm \: \: \: \: \: \: \: \: \:a=\dfrac{126}{21}$

a = 6

Putting the value of a in (1) we get :-

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{6}{b}+6+6b=21$

$\rm \: \: \: \: \: \: \: \: \:=6(\dfrac{1}{b}+1+b)=21$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{1}{b}+b+1=\dfrac{21}{6}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{1+b^2}{b+1}=\dfrac{7}{2}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{1+b^2}{b}=\dfrac{5}{2}$

2b² - 5b + 2 = 0

2b² - (4 + 1)b + 2 = 0

2b² - 4b - b + 2 = 0

2b(b - 2) - (b - 2) = 0

(b - 2)(2b - 1) = 0

$\rm \: \: \: \: \: \: \: \: \:b=2\;or\;\dfrac{1}{2}$

When, b = 2

$\textbf{\underline{Tems\;are:-}}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{a}{b}=\dfrac{6}{2}=3$

a = 6

ab = 6 × 2 = 12

$\textbf{\underline{Numbers\;are\;3\;and\;6\;and\;12}}$

$\rm \: \: \: \: \: \: \: \: \:When,\;b=\dfrac{1}{2}$

$\textbf{\underline{Tems\;are:-}}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{a}{b}=\dfrac{6}{1/2}$

= 6 × 2

= 12

a = 6

$\rm \: \: \: \: \: \: \: \: \:ab=6\times\dfrac{1}{2}$

= 3

$\textbf{\underline{Numbers\;are\;12\;and\;6\;and\;3}}$

Therefore,

$\huge{\boxed{\bigstar{{Required\;numbers=3,6\;and\;12}}}}$

Answer 2

Answer:

HI MATE.

Assume,

Common ratio of GP be t

Common ratio between these terms be (t)^a

(t)^a = b

Assume,

{\boxed{\sf\:{Required\;Terms\;are=\dfrac{a}{b},a\;and\;ab}}}

RequiredTermsare=

b

a

,aandab

k = Common ratio

Given,

Sum of terms = 21

Sum of squares = 189

\rm \: \: \: \: \: \: \: \: \:=\dfrac{a}{b}+a+ab=21.....(1)=

b

a

+a+ab=21.....(1)

Also,

\rm \: \: \: \: \: \: \: \: \:=(\dfrac{a}{b})^2+a^2+(ab)^2=189.....(2)=(

b

a

)

2

+a

2

+(ab)

2

=189.....(2)

Squaring both sides of (1)

\rm \: \: \: \: \: \: \: \: \:=(\dfrac{a}{b}+a+ab)^2=(21)^2=(

b

a

+a+ab)

2

=(21)

2

\rm \: \: \: \: \: \: \: \: \:=(\dfrac{a}{b})^2+a^2+(ab)^2+2[(\dfrac{a}{b}\times a)+(a\times ab)+(ab\times\dfrac{a}{b})]=441=(

b

a

)

2

+a

2

+(ab)

2

+2[(

b

a

×a)+(a×ab)+(ab×

b

a

)]=441

\rm \: \: \: \: \: \: \: \: \:=189+2(\dfrac{a^2}{b}+a^2 b+a^2=441=189+2(

b

a

2

+a

2

b+a

2

=441

\rm \: \: \: \: \: \: \: \: \:=2(\dfrac{a^2}{b}+a^2 b+a^2=441-189=2(

b

a

2

+a

2

b+a

2

=441−189

\rm \: \: \: \: \: \: \: \: \:=a^2(\dfrac{1}{b}+b+1=\dfrac{252}{2}=a

2

(

b

1

+b+1=

2

252

\rm \: \: \: \: \: \: \: \: \:=a^2(\dfrac{1}{b}+b+1=126=a

2

(

b

1

+b+1=126

\rm \: \: \: \: \: \: \: \: \:=a(\dfrac{a}{b}+ab+a=126=a(

b

a

+ab+a=126

From eq (1) we have

a(21) = 126

\rm \: \: \: \: \: \: \: \: \:a=\dfrac{126}{21}a=

21

126

a = 6

Putting the value of a in (1) we get :-

\rm \: \: \: \: \: \: \: \: \:=\dfrac{6}{b}+6+6b=21=

b

6

+6+6b=21

\rm \: \: \: \: \: \: \: \: \:=6(\dfrac{1}{b}+1+b)=21=6(

b

1

+1+b)=21

\rm \: \: \: \: \: \: \: \: \:=\dfrac{1}{b}+b+1=\dfrac{21}{6}=

b

1

+b+1=

6

21

\rm \: \: \: \: \: \: \: \: \:=\dfrac{1+b^2}{b+1}=\dfrac{7}{2}=

b+1

1+b

2

=

2

7

\rm \: \: \: \: \: \: \: \: \:=\dfrac{1+b^2}{b}=\dfrac{5}{2}=

b

1+b

2

=

2

5

2b² - 5b + 2 = 0

2b² - (4 + 1)b + 2 = 0

2b² - 4b - b + 2 = 0

2b(b - 2) - (b - 2) = 0

(b - 2)(2b - 1) = 0

\rm \: \: \: \: \: \: \: \: \:b=2\;or\;\dfrac{1}{2}b=2or

2

1

When, b = 2

\textbf{\underline{Tems\;are:-}}

Temsare:-

\rm \: \: \: \: \: \: \: \: \:=\dfrac{a}{b}=\dfrac{6}{2}=3=

b

a

=

2

6

=3

a = 6

ab = 6 × 2 = 12

\textbf{\underline{Numbers\;are\;3\;and\;6\;and\;12}}

Numbersare3and6and12

\rm \: \: \: \: \: \: \: \: \:When,\;b=\dfrac{1}{2}When,b=

2

1

\textbf{\underline{Tems\;are:-}}

Temsare:-

\rm \: \: \: \: \: \: \: \: \:=\dfrac{a}{b}=\dfrac{6}{1/2}=

b

a

=

1/2

6

= 6 × 2

= 12

a = 6

\rm \: \: \: \: \: \: \: \: \:ab=6\times\dfrac{1}{2}ab=6×

2

1

= 3

\textbf{\underline{Numbers\;are\;12\;and\;6\;and\;3}}

Numbersare12and6and3

Therefore,

\huge{\boxed{\bigstar{{Required\;numbers=3,6\;and\;12}}}}

★Requirednumbers=3,6and12

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