Question

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Topic - Geometric Progression

$\large \color{red} \pmb{ \mathbb{QUESTION}} \: \: \: \: \longmapsto$
If the sum and product of three numbers in GP are 31 and 125 respectively.

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Answer 1

Answer:

Let the three terms in GP a/r , a , ar .

Product of three terms = 125

a/r×a×ar =125

a cube = 125

a = cube root of 125

a= 5

• sum of three terms is 31
• a/r+a+ar=31
• 5/r +5+5r =31, 5+5r^2/r =26
• 5+5r ^2=26r
• 5r^2- 26r+5=0

By factorising we get r= 1/5 or r=5

By using the answer we get the three terms are 25,5,1or 1,5,25.

Hope it helps u mate

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Answer 2

Answer:

$let \: the \: 3 \: terms \: in \: g.p \: be \frac{a}{r} = 125$

Product of 3 terms =125

$\frac{a}{r} \times a \times ar = 125$

${a}^{3} = 125$

$a = \sqrt[3]{125} = 5$

Sum of 3 terms =31

$\frac{a}{r} + a + ar = 31$

$\frac{5}{r} + 5 + 5r = 31 \: , \frac{5 + 5r {}^{2} }{r} = 26$

$5 + 5 {r}^{2} + 26r$

${5r}^{2} - 26r + 5 = 0$

$(r - 5)(5r - 1) = 0$

$5r - 1 = 0 \: \: \: or \: \: \: r - 5 = 0$

$r \frac{1}{5} \: \: \: or \: \: \: r = 5$

$if \: a = 5 \: and \: r = \frac{1}{5},the \: terms \: are \: \frac{a}{r} = \frac{5}{1} = 5 \times 5 = 25$

$or = 5 \times \frac{1}{5} = 1$

$5 \times \frac{1}{5} = 25, \: a = 5 \: ar \: 5 \times \frac{1}{5} = 1$

$if \: a = 5 = 25, \: r= 25, \: 5 \times \frac{1}{5} = 1$

$if \: a = 5 \: and \: r = 5 \: or \: \: \frac{a}{r} = \frac{5}{5} = 1, \: a = 5$

Ar=5×5=25

The three terms are 25, 5, 1 or 1, 5, 25