Question

3. Three numbers are in continued proportion whose
mean proportional is 12 and the sum of the remain-
ing two numbers is 26. Find the numbers.​

Answer 1

$\huge\bold\green{Solution}$

• Consider a ,b, c be the three numbers in continued proportion.

$\huge\boxed{\fcolorbox{purple}{ink}{Then:}}$

$b = 12 \: and \: a+ c = 26 \: \: (1)$

$\large \mathfrak{ \text{W}e \: \text{K}now }$

$b² = ac$

$( {12}^{2} ) = ac$

$ac= 144$

$a = \frac{144}{c} \: \: (2)$

$Put \: a = \frac{144}{c} from \: (2) \: in \\ \: a + c \: = 26 \: in \: (1)$

$\frac{144}{c} + c \: = 26$

$144 + {c}^{2} = 26c$

(by multiplying the sides by c )

$\huge\bold\green{Therefore}$

${c}^{2} - 26c + 144 = 0$

${c}^{2} - 18c - 8c + 144 = 0$

$c(c - 18) - 8(c - 8) = 0$

$(c - 8)(c - 8) = 0$

$c-18=0 \: or \: c-8 =0$

$c = 18 \: or \: c = 8$

$\huge\boxed{\fcolorbox{red}{ink}{Now:}}$

$,a+ c = 26$

$a + 18 = 26$

$a = 26 - 18$

$a = 8$

$\huge\boxed{\fcolorbox{red}{ink}{OR}}$

$a + c = 26$

$a + 8 = 26$

$a = 26 - 8$

$a = 18$

$\huge\boxed{\fcolorbox{red}{ink}{final \: SOLUTION:}}$

The required numbers are 8,12 and 18 or 18 ,12 and 8 .

$\huge\boxed{\dag\sf\red{Thanks}\dag}$