Question

Three numbers, the third of which is 4 form a decreasing G.P. If the last term is replaced by 3, the three numbers form an A.P., then the first number of the G.P. is

Answer 1

Hint: At first we will assume the first and the second number of series be x and y respectively, then we will form an equation using the statement that they are in G.P., now it is given that replacing the third number i.e. 4 by 3 the three numbers will form an A.P. from which we will get another equation. Now using the substitution method we will find the value of the first two numbers by solving these two equations.

Complete step by step answer:

Given data: Three numbers form a G.P., the third of the three numbers\[ = \]4
If the last term is replaced by 3, the three numbers form an A.P.
Let the first and second number x and y respectively
It is given that x, y and, 4 are in GP, therefore the common ratio will be constant
i.e. $\dfrac{y}{x} = \dfrac{4}{y}$
On cross multiplication we get,
$ \Rightarrow {y^2} = 4x........(i)$
Now if 4 is replaced by 3 then they will form an A.P., i.e. the common difference will be constant
i.e. $y - x = 3 - y$
On rearranging we get,
$ \Rightarrow 2y - 3 = x$
On Substituting the value of x in equation (i), we get,
$ \Rightarrow {y^2} = 4(2y - 3)$
On simplifying the brackets we get,
$ \Rightarrow {y^2} - 8y + 12 = 0$
Now splitting the coefficient of y using the factors of the product of the other two coefficients
$ \Rightarrow {y^2} - 6y - 2y + 12 = 0$
Taking common from first two and last two terms
$ \Rightarrow y\left( {y - 6} \right) - 2\left( {y - 6} \right) = 0$
Taking (y-6) common we get,
$ \Rightarrow \left( {y - 6} \right)\left( {y - 2} \right) = 0$
I.e. $y - 6 = 0$ or$y - 2 = 0$
$\therefore y = 6$ or $y = 2$
Since it is given that x, y and 4 form a decreasing G.P. i.e. $x > y > 4$
Therefore, $y = 6$
Substituting $y = 6$ in equation(i)
$ \Rightarrow {6^2} = 4x$
Dividing both sides by 4
$\therefore x = 9$
Therefore the first number 9.

Note: On solving the quadratic equation in y we get the two values of y as 2 and 6, some of the students take both the cases that if $y = 6$or$y = 2$have two values of A.P. like
 If $y = 6$
Substituting $y = 6$ in equation(i)
$ \Rightarrow {6^2} = 4x$
Dividing both sides by 4
$\therefore x = 9$
Therefore the first number 9
And if $y = 2$
Substituting $y = 2$ in equation(i)
$ \Rightarrow {2^2} = 4x$
Dividing both sides by 4
$\therefore x = 1$
Therefore the first number 1
And include this answer also which is not correct as 1, 2, 4 is not a decreasing G.P. as mentioned so avoid doing this, learn to apply all the data given to us.