Divide 155 into three parts such that the thr | vedclass

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(C) Let the three terms in $GP$ be $\frac{a}{r}, a, ar$. Given that their sum is $155$: $\frac{a}{r} + a + ar = 155$  $(1)$ Also,the first term is $12

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$5, 25, 125$

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(C) Let the three terms in $GP$ be $\frac{a}{r}, a, ar$. Given that their sum is $155$: $\frac{a}{r} + a + ar = 155$  $(1)$ Also,the first term is $120$ less than the third term: $ar - \frac{a}{r} = 120$  $(2)$ From $(2)$,$a(r - \frac{1}{r}) = 120 \implies a(\frac{r^2 - 1}{r}) = 120$. From $(1)$,$a(\frac{1 + r + r^2}{r}) = 155$. Dividing the two equations: $\frac{r^2 - 1}{r^2 + r + 1} = \frac{120}{155} = \frac{24}{31}$. $31r^2 - 31 = 24r^2 + 24r + 24$. $7r^2 - 24r - 55 = 0$. Solving the quadratic equation: $7r^2 - 35r + 11r - 55 = 0 \implies 7r(r - 5) + 11(r - 5) = 0$. So,$r = 5$ or $r = -\frac{11}{7}$. If $r = 5$,then $a(\frac{1}{5} + 1 + 5) = 155 \implies a(\frac{31}{5}) = 155 \implies a = 25$. The terms are $\frac{25}{5}, 25, 25 	imes 5$,which are $5, 25, 125$.

Divide $155$ into three parts such that the three numbers are in a Geometric Progression $(GP)$ and the first term is $120$ less than the third term.

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