Question

Find the three numbers in G.P. such that their sum is 13/3 and the sum of their squares is 91/3.

Answer 1

● LET'S THE THREE NUMBERS ARE

=> [a, ar, and ar²]
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◢NOW,

● ACCORDING TO THE QUESTION

$= > a + ar + a {r}^{2} = \frac{13}{3} \: \: \: ...Eq _{1}$

◢AND

$= > {a}^{2} + {a}^{2} {r}^{2} + {a}^{2} {r}^{4} = \frac{91}{9} \: \: \: ....Eq _{2}$
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◢NOW,

$= > a + ar + a {r}^{2} = \frac{13}{3} \\ \\ = > a(1 + r + {r}^{2} ) = \frac{13}{3} \\ \\ SQUARING \: \: BOTH \: \: SIDE \\ \\ = > {a}^{2} {(1 + r + {r}^{2}) }^{2} = \frac{169}{9} \: \: \: ....Eq _{3}$
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● WE KNOW THAT..

=> (A+B+C)²

= (A²+B²+C²+2AB+2BC+2CA)
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● USING THIS IDENTITY FOR Eq(3)

◢WE GET,

$= > {a}^{2} (1 + {r}^{2} + {r}^{4} + 2r + 2 {r}^{2} + 2 {r}^{3} ) = \frac{169}{9} \: \: \: ...Eq _{4}$
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● NOW TAKE :- [Eq(4) - Eq(2)]

$= > 2 {a}^{2} r(1 + r + {r}^{2} ) = \frac{26}{3}$
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● USING Eq(1) AGAIN

$= > 2ar \frac{13}{3} = \frac{26}{3}$

◢SO

$= > ar = 1 \\ \\ = > r = \frac{1}{a} \: \: \: \: \: Eq _{5}$
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● BY Eq(1) AND Eq(5)

$= > a + 1 + \frac{1}{a} = \frac{13}{3} \\ \\ = > \frac{ {a}^{2} + 1 }{a} = \frac{13 - 3}{3} \\ \\ = > {a}^{2} - \frac{10}{3} a + 1 = 0 \\ \\ = > {a}^{2} - 3a - \frac{1}{3} a + 1 = 0 \\ \\ = > a(a - 3) - \frac{1}{3} (a - 3) = 0 \\ \\ = > (a - \frac{1}{3} )(1 - a) = 0$
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◢THEN

$= > (a - \frac{1}{3} ) = 0 \\ \\ = > a = \frac{1}{3}$

◢OR

$= > (a - 3) = 0 \\ \\ = > a = 3$
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◢IF "a = 1/3"

◢THEN

$= > r = \frac{1}{a} \\ \\ = > r = 3$

◢and

◢IF "a = 3"

◢THEN

$= > r = \frac{1}{3}$
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◢THEN

๏ CONDITION 1ST

◢IF "a=3" and "r=1/3"

● THE NUMBER OF GP ARE

=> 3, 1, 1/3
______________[◢ANSWER]

๏ CONDITION 2ND

◢IF "a=1/3" AND "r=3"

● THE NUMBER OF GP AER

=> 1/3, 1, 3
______________[◢ANSWER]

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☆BE BRAINLY☆

Answer 2

Answer:

the answer is 3,1,1/3