Question

A number consists of three digits which are in GP the sum of the right hand digits exceeds twice the middle digits by 1 and the sum of the left hand and middle digits is two third of the sum of the middle and right hand digits. Find the sum of digits of number?

Answer 1

Hint – Here we will proceed by using the formula for geometric progression that is $a,ar,a{r^2},a{r^3}$ and so on. Then by applying the conditions given in the question we will get our answer.

Complete Step-by-Step solution:
Let the three digits be
$a,ar,a{r^2}$
Then, the number will be
$100a + 10r + 10{r^2}$ ….. (1)
It is given that,
$a + a{r^2} = 2ar + 1$
Or $a\left( {{r^2} - 2r + 1} \right) = 1$
$a{\left( {r - 1} \right)^2} = 1$ ……(2)
Also given,
$a + ar = \dfrac{2}{3}\left( {ar + a{r^2}} \right)$ (cancel a from both the sides)
$3 + 3r = 2r + 2{r^2}$
$2{r^2} - r - 3 = 0$
$\left( {r + 1} \right)\left( {2r - 3} \right) = 0$
$r = - 1,\dfrac{3}{2}$ We have calculated the two values of $r$, now
For, $r = - 1$
$a = \dfrac{1}{{{{\left( {r - 1} \right)}^2}}} = \dfrac{1}{4}$
$\dfrac{1}{4} \notin 1$( $\dfrac{1}{4}$ is not the element of 1)
Therefore, $r$ is not equal to -1
For, $r = \dfrac{3}{2}$
$a = \dfrac{1}{{{{\left( {\dfrac{3}{2} - 1} \right)}^2}}} = 4$ …..(from 2)
As it is in the form of integer, therefore we can say that $r = \dfrac{3}{2}$
From equation (1) number is,
$400 + 10 \times 4 \times \dfrac{3}{2} + 4 \times \dfrac{9}{4}$
$ = 469$
Therefore, the sum of digits of number is 469.

Note – Whenever we come up with this type of question, one must know that G.P. that is geometric progression is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called the common ratio. Also, the nth term of a G.P. series is $Tn = a{r^{n - 1}}$, where a = first term and r = common ratio = $\dfrac{{Tn}}{{Tn - 1}}$.