Question

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Break the numbers 155 into three parts so that
the obtained numbers form a GP, the first term
being less than the third one by 120​

Answer 1

Answer:

The required three parts are 5, 25 , 125

Step-by-step explanation:

Let the three parts in G.P be $\frac{a}{r},a,ar$

As per given data,

$\frac{a}{r}+a+ar=155$

$a(\frac{1+r+r^2}{r})=155$.......(1)

Also,

$ar-\frac{a}{r}=120$

$a(\frac{r^2-1}{r})=120$............(2)

Divide (1) by (2)

$\frac{1+r+r^2}{r^2-1}=\frac{155}{120}$

$\frac{1+r+r^2}{r^2-1}=\frac{31}{24}$

$24+24r+24r^2=31r^2-31$

$7r^2-24r-55=0$

$7r^2-35r+11r-55=0$

$7r(r-5)+11(r-5)=0$

$(7r+11)(r-5)=0$

$r=5,\frac{-11}{7}$

But r cannot be negative

$\implies\:r=5$

put r=5 in (2) we get

$a(\frac{25-1}{5})=120$

$a(\frac{24}{5})=120$

$a(\frac{1}{5})=5$

$a=25$

The required 3 parts are

$\frac{25}{5},25,25(5)$

$5,25,125$