Question

the sum of first three terms in a GP is 38 their product is 1728 find the values of the terms step by step solutions​

Answer 1

PG: $(a_1, a_2, a_3, a_4, ..., a_n)$

common ratio:  $q= \displaystyle(\frac{a_2}{a_1}=\frac{a_3}{a_2} =\frac{a_4}{a_3} , ...)$

$a_1 = a\\a_2 = aq\\a_3 = aq^2$

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$a_1 * a_2 * a_3 = 1728$

$a * aq * aq^2 = 1728$

$a^3q^3 = 2^6*3^3=12^3$

$aq=12$

$q= \displaystyle\frac{12}{a}$

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$a_1 + a_2 + a_3 = 38$

$a + aq + aq^2 = 38$

$a + a \displaystyle\frac{12}{a} + a( \displaystyle\frac{12}{a})^2 = 38$

$a + 12 + \displaystyle\frac{12^2}{a} = 38$

$a + \displaystyle\frac{12^2}{a} = 26$

$a^2 + 12^2 = 26a$

$a^2 -26a =-12^2$

$(a-13)^2-13^2=-144$

$(a-13)^2=-144+169$

$(a-13)^2=25$

$a-13=\pm5$

$a=\pm5+13$

$a'=+5+13=18$

$a"=-5+13=8$

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$q= \displaystyle\frac{12}{a}$

$q'= \displaystyle\frac{12}{a'}=\frac{12}{18}=\frac{2}{3}$

$q"= \displaystyle\frac{12}{a"}=\frac{12}{8}=\frac{3}{2}$

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Then,

$a_1 = a\\a_2 = aq\\a_3 = aq^2$

Answer 1:

$a'_1 = a'=18\\\\a'_2 = a'q'=18 (\displaystyle\frac{2}{3})=12\\\\a'_3 =a'(q')^2= 18(\frac{2}{3})^2=8$

Answer 2:

$a_1 =a"=8\\\\a_2 = a"q"=8 (\displaystyle\frac{3}{2})=12\\\\a_3 =a"(q")^2= 8(\frac{3}{2})^2=18$

See you later ^^)

Study hard!