Question

The consecutive digits of a three digit number are in G.P. If the middle digit is increased by 2 then they form an A.P. If 792 is subtracted from this number then we get the number consisting of the same three digits but in reverse order. Find the number.

Answer 1

Hint: Firstly assume the digits to be $a,ar,a{{r}^{2}}$after that you can apply the conditions accordingly. Use the formula of A.P. given by If a,b,c are in A.P. then , $b=\dfrac{a+c}{2}$ so that you will get a simpler equation. After that use the last condition and solve it to find ‘r’ and on further solving you will get ‘a’.


Complete step-by-step answer:
To solve the given problem we have to assume the digits and as they are in G.P. therefore we have the standard consecutive numbers to be assumed which are given by,

Therefore the three consecutive digits of the number are given by,
$a,ar,a{{r}^{2}}$ …………………………………………… (1)
Where, ‘a’ is the first term and ’r’ is the common ratio of G.P.
$100a+10ar+a{{r}^{2}}$

If we take ‘a’ as common we will get,
$a\left( 100+10r+{{r}^{2}} \right)$ …………………………….. (2)

As given in the problem if the middle digit is increased by 2 then it will form an A.P.
Therefore,
$a,ar+2,a{{r}^{2}}$ form an A.P.

To proceed further we should know the formula given below,
Formula:
If a,b,c are in A.P. then , $b=\dfrac{a+c}{2}$.
By using the formula we can write,
$ar+2=\dfrac{a+a{{r}^{2}}}{2}$
$\therefore 2\left( ar+2 \right)=a+a{{r}^{2}}$
$\therefore 2ar+4=a+a{{r}^{2}}$
$\therefore 4=a{{r}^{2}}-2ar+a$
By taking ‘a’ common we will get,
$\therefore 4=a\left( {{r}^{2}}-2r+1 \right)$

To proceed further in the solution we should know the formula given below,
Formula:
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
By using the above formula we can write,
$\therefore 4=a{{\left( r-1 \right)}^{2}}$
By rearranging the above equation we will get,
$\therefore a{{\left( r-1 \right)}^{2}}=4$ …………………………………….. (3)

To proceed further we should write the number in reverse order therefore, from equation (1) we can write the number in reverse order as,
$100a{{r}^{2}}+10ar+a$
By taking ‘a’ common we will get,
$a\left( 100{{r}^{2}}+10r+1 \right)$ ………………………………. (4)

Now, by using the second condition given in the problem and using equation (2) and equation (4), we will get,
$a\left( 100+10r+{{r}^{2}} \right)-792=a\left( 100{{r}^{2}}+10r+1 \right)$

By taking 792 on the right side of the equation and reverse order number on the left hand side of the equation we will get
$a\left( 100+10r+{{r}^{2}} \right)-a\left( 100{{r}^{2}}+10r+1 \right)=792$

By taking ‘a’ common we will get,
$a\left[ \left( 100+10r+{{r}^{2}} \right)-\left( 100{{r}^{2}}+10r+1 \right) \right]=792$

By opening the parenthesis we will get,
$\therefore a\left[ 100+10r+{{r}^{2}}-100{{r}^{2}}-10r-1 \right]=792$
$\therefore a\left[ 100+{{r}^{2}}-100{{r}^{2}}-1 \right]=792$
$\therefore a\left[ 99-99{{r}^{2}} \right]=792$

By taking 99 common we will get,
$\therefore 99a\left[ 1-{{r}^{2}} \right]=792$
$\therefore a\left[ 1-{{r}^{2}} \right]=\dfrac{792}{99}$
Dividing by 9 to numerator and denominator on the right hand side of the equation we will get,
$\therefore a\left[ 1-{{r}^{2}} \right]=\dfrac{88}{11}$
$\therefore a\left[ 1-{{r}^{2}} \right]=8$
Dividing above equation by equation (3) we will get,
$\therefore \dfrac{a\left( 1-{{r}^{2}} \right)}{a{{\left( r-1 \right)}^{2}}}=\dfrac{8}{4}$
$\therefore \dfrac{\left( 1-{{r}^{2}} \right)}{{{\left( r-1 \right)}^{2}}}=2$
By taking -1 common from denominator of left hand side of the equation we will get,
$\therefore \dfrac{\left( 1-{{r}^{2}} \right)}{{{\left[ -1\left( 1-r \right) \right]}^{2}}}=2$
$\therefore \dfrac{\left( 1-{{r}^{2}} \right)}{{{\left( 1-r \right)}^{2}}}=2$

To proceed further we should know the formula given below,
Formula:
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\times \left( a-b \right)$
By using above formula we will get,
$\therefore \dfrac{\left( 1-r \right)\left( 1+r \right)}{{{\left( 1-r \right)}^{2}}}=2$
$\therefore \dfrac{\left( 1+r \right)}{\left( 1-r \right)}=2$
$\therefore \left( 1+r \right)=2\left( 1-r \right)$
$\therefore \left( 1+r \right)=2-2r$
$\therefore r+2r=2-1$
$\therefore 3r=1$
$\therefore r=\dfrac{1}{3}$ …………………………………………… (5)

By substituting the value of equation (5) in equation (3) we will get,
$\therefore a{{\left( \dfrac{1}{3}-1 \right)}^{2}}=4$
$\therefore a{{\left( \dfrac{1-3}{3} \right)}^{2}}=4$
$\therefore a{{\left( \dfrac{-2}{3} \right)}^{2}}=4$
$\therefore a\times \dfrac{4}{9}=4$
$\therefore a=4\times \dfrac{9}{4}$
$\therefore a=9$ …………………………………………. (6)

Now if we put the values of equation (5) and equation (6) in equation (2) we will get,
$\therefore $The required number $=a\left( 100+10r+{{r}^{2}} \right)$
$\therefore $The required number $=9\times \left[ 100+10\times \left( \dfrac{1}{3} \right)+{{\left( \dfrac{1}{3} \right)}^{2}} \right]$
$\therefore $The required number $=9\times 100+9\times 10\times \left( \dfrac{1}{3} \right)+9\times {{\left( \dfrac{1}{3} \right)}^{2}}$
$\therefore $The required number $=900+3\times 10\ +9\times \dfrac{1}{9}$
$\therefore $The required number $=900+30+1$
$\therefore $The required number $=931$

Therefore the number which satisfies all the conditions given in the problem is 931.

Note: Don’t assume the digits to be $\dfrac{a}{r},a,ar$, as it will become very much difficult for calculation, it will be better if we use the digits as$a,ar,a{{r}^{2}}$, as it will save your time in solving which is beneficial in competitive exams.