Question

Let $a,b,c$ be three numbers between 2 and 18 such that:
(i) their sum is 25
(ii) the numbers 2,a,b are consecutive terms of an A.P
(iii) the numbers b,c,18 are consecutive terms of a G.P.Find $a+b-c$ ?

Answer 1

Given,
$a+b+c=25\cdots \left(1\right)$
$2a=2+b\cdots \left(2\right)$
$c^2=18b\cdots \left(3\right)$
Substituting the values of $a$ and $b$ in terms of $c$ from $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$, we get
$\frac{1}{2}(2+\frac{c^2}{18})+\frac{c^2}{18}+c=25$
$\Rightarrow c^2+36+2c^2+36c=25.36$
$\Rightarrow 3c^2+36c-864=0$
$\Rightarrow c^2+12c-288=0$
$(c+24)(c-12)=0$
$\Rightarrow c=12,-24$
$\because a,b,c$ are between 2 and 18, $\therefore c\ne -24$
Hence $c=12$
from $\left(3\right)$ $b=\frac{c^2}{18}$
$=\frac{{12}^2}{18}$
$=\frac{144}{18}$
$b=8$
from $\left(2\right)$ $a=\frac{2+b}{2}=\frac{2+8}{2}=5$
Hence, $a=5,b=8,c=12$