Question

Three numbers whose sum is 30, are in A.P. If 1, 2, 4 are added to them, respectively, they will be in G.P. Find the numbers.

Answer 1

Hint: As it is given that the numbers are in A.P., so start by letting the numbers to be (a-d),a,(a+d). Then use the value of the sum of the numbers to get the value of a, followed by applying the condition given in the question for making the numbers to be in G.P. Finally, use the formula of geometric mean to find the required value.

Complete step-by-step answer:
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by ${{T}_{r}}$, and sum till r terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)=\dfrac{r}{2}\left( a+l \right)$
Now moving to the situation given in the question. As it is given that the three numbers are in A.P., so we let the numbers to be (a-d),a,(a+d).
Now as it is given that the sum of the numbers is equal to 30, we get
$a-d+a+a+d=30$
$\Rightarrow 3a=30$
$\Rightarrow a=10$
So, our numbers become: 10-d,10,10+d.
Now, it is also given in the question that if 1, 2, 4 are added to them, respectively, they will be in G.P.
So, we can say that 10-d+1,10+2,10+d+4, i.e., 11-d,12,14+d are in G.P. and we know, if three numbers p,q,r are in G.P., then $q=\sqrt{pr}$ .
$\therefore 12=\sqrt{\left( 11-d \right)\left( 14+d \right)}$
Now, if we square both sides of the equation, we get
$144=\left( 11-d \right)\left( 14+d \right)$
$\Rightarrow 144=154-{{d}^{2}}-3d$
$\Rightarrow {{d}^{2}}+3d-10=0$
$\Rightarrow {{d}^{2}}+5d-2d-10=0$
$\Rightarrow \left( d-2 \right)\left( d+5 \right)=0$
Therefore, the possible values of d are 2 and -5.
So, the numbers, if we put d=2 are: 10-d,10,10+d, i.e., 8,10,12. If we put d=-5, we get the number to be 15,10,5.
Therefore, we can conclude that the numbers are 8,10,12 or 15,10,5.

Note: Always, while dealing with an arithmetic progression, try to find the first term and the common difference of the sequence, as once you have these quantities, you can easily solve the questions related to the given arithmetic progressions.