Question

Find $a$, $b$ and $c$ between $2$ and $18$ such that $a+b+c=25$, and $2$, $a$, $b$ are in A.P. and $b,c,18$are in G.P.

• A. $5,8,12$
• B. $4,8,13$
• C. $3,9,13$
• D. $5,9,11$

Answer 1

The correct option is A

$5,8,12$

Step 1. Find the value of $a,b,$ and $c$:

According to question,

$a+b+c=25$ …...(1)

$2a=b+2$ …...(2) $[\because 2,a,b$are consecutive terms of A.P.$]$

and $c^2=18b$ ..…..(3) $[\because b,c,18$ are consecutive terms of G.P.$]$

Eliminating $a$ from (1) and (2),

$\Rightarrow$$b=16-\frac{2c}{3}$

Step 2. Put the value of $b$ in equation (3):

$c^2=18(16-\frac{2c}{3})$

$\Rightarrow$ $c^2+12c-18\times 16=0$

$\Rightarrow$ $(c-12)(c+24)=0$

$\Rightarrow$ $c=-24,12$

$-24$ is rejected because it does not lie between $2$ and $18$

$\therefore$$c=12$

Step 3. Put the value of $c$ in equation (3):

$\Rightarrow$${12}^2=18b$

$\Rightarrow$ $b=\frac{{12}^2}{18}$

$=8$

Step 4. Put the value of $b$ in equation (2):

$\Rightarrow$$2a=8+2$

$\Rightarrow$ $a=\frac{8+2}{2}$

$=5$
Thus, $\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}$, $\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8}$ and $\mathit{c}\mathbf{}\mathbf{=}\mathbf{}\mathbf{12}$

Hence, Option ‘A’ is Correct.