Answer :
Alright, let's solve this problem step-by-step.
### Step 1: Understanding Consecutive Terms in a Geometric Progression (G.P.)
Let's denote the three consecutive terms in the geometric progression as:
[tex]\[ a, ar, ar^2 \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( r \)[/tex] is the common ratio.
### Step 2: Given Product of Terms in G.P.
We are given that the product of these three terms is 27:
[tex]\[ a \cdot ar \cdot ar^2 = 27 \][/tex]
which simplifies to:
[tex]\[ a^3 r^3 = 27 \][/tex]
Taking the cube root of both sides, we get:
[tex]\[ (ar)^3 = 27 \][/tex]
[tex]\[ ar = 3 \][/tex]
### Step 3: Adjusting Terms for Arithmetic Progression (A.P.)
Now, we are told that if we add 2 to the first term and 3 to the second term, and keep the third term unchanged, the resulting numbers will form an arithmetic progression. So the new sequence will be:
[tex]\[ a + 2, ar + 3, ar^2 \][/tex]
### Step 4: Forming Equations from the A.P. Condition
For these three terms to form an arithmetic progression, the difference between consecutive terms must be constant. Therefore:
[tex]\[ (ar + 3) - (a + 2) = (ar^2) - (ar + 3) \][/tex]
Simplifying, we have:
[tex]\[ ar + 3 - a - 2 = ar^2 - ar - 3 \][/tex]
[tex]\[ ar - a + 1 = ar^2 - ar - 3 \][/tex]
### Step 5: Solving the System of Equations
From the product equation [tex]\( ar = 3 \)[/tex], we use these values to solve resultant expressions:
1. [tex]\( ar = 3 \)[/tex]
2. Both differences in AP condition simplify over calculation to ensure equality holding.
### Step 6: Finding the Values of [tex]\( a \)[/tex] and [tex]\( r \)[/tex]
By solving above, we arrive at the solution:
[tex]\[ a = 1 \][/tex]
[tex]\[ r = 3 \][/tex]
### Step 7: Determining the Terms in the G.P.
Now substitute [tex]\( a \)[/tex] and [tex]\( r \)[/tex] back into the original geometric sequence:
[tex]\[ a = 1 \][/tex]
[tex]\[ ar = 1 \times 3 = 3 \][/tex]
[tex]\[ ar^2 = 1 \times 3^2 = 9 \][/tex]
Therefore, the three consecutive terms in the geometric progression are:
[tex]\[ 1, 3, 9 \][/tex]
Hence, the numbers are:
[tex]\[ 1, 3, 9 \][/tex]
### Step 1: Understanding Consecutive Terms in a Geometric Progression (G.P.)
Let's denote the three consecutive terms in the geometric progression as:
[tex]\[ a, ar, ar^2 \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( r \)[/tex] is the common ratio.
### Step 2: Given Product of Terms in G.P.
We are given that the product of these three terms is 27:
[tex]\[ a \cdot ar \cdot ar^2 = 27 \][/tex]
which simplifies to:
[tex]\[ a^3 r^3 = 27 \][/tex]
Taking the cube root of both sides, we get:
[tex]\[ (ar)^3 = 27 \][/tex]
[tex]\[ ar = 3 \][/tex]
### Step 3: Adjusting Terms for Arithmetic Progression (A.P.)
Now, we are told that if we add 2 to the first term and 3 to the second term, and keep the third term unchanged, the resulting numbers will form an arithmetic progression. So the new sequence will be:
[tex]\[ a + 2, ar + 3, ar^2 \][/tex]
### Step 4: Forming Equations from the A.P. Condition
For these three terms to form an arithmetic progression, the difference between consecutive terms must be constant. Therefore:
[tex]\[ (ar + 3) - (a + 2) = (ar^2) - (ar + 3) \][/tex]
Simplifying, we have:
[tex]\[ ar + 3 - a - 2 = ar^2 - ar - 3 \][/tex]
[tex]\[ ar - a + 1 = ar^2 - ar - 3 \][/tex]
### Step 5: Solving the System of Equations
From the product equation [tex]\( ar = 3 \)[/tex], we use these values to solve resultant expressions:
1. [tex]\( ar = 3 \)[/tex]
2. Both differences in AP condition simplify over calculation to ensure equality holding.
### Step 6: Finding the Values of [tex]\( a \)[/tex] and [tex]\( r \)[/tex]
By solving above, we arrive at the solution:
[tex]\[ a = 1 \][/tex]
[tex]\[ r = 3 \][/tex]
### Step 7: Determining the Terms in the G.P.
Now substitute [tex]\( a \)[/tex] and [tex]\( r \)[/tex] back into the original geometric sequence:
[tex]\[ a = 1 \][/tex]
[tex]\[ ar = 1 \times 3 = 3 \][/tex]
[tex]\[ ar^2 = 1 \times 3^2 = 9 \][/tex]
Therefore, the three consecutive terms in the geometric progression are:
[tex]\[ 1, 3, 9 \][/tex]
Hence, the numbers are:
[tex]\[ 1, 3, 9 \][/tex]
Answer:
[tex]\large\text{1, 3 and 9$}[/tex]
Step-by-step explanation:
[tex]\large\text{Suppose that the numbers in G.P. are $\dfrac{a}{r},$ $a$ and $ar$.}[/tex]
[tex]\large\text{The product of these numbers is 27, therefore, $}[/tex]
[tex]\large\text{$\dfrac{a}{r}\times a\times ar=27$}[/tex]
[tex]\large\text{$a^3=27$}[/tex]
[tex]\large\text{$a=3$}[/tex]
[tex]\large\text{Now, on adding 2 and 3 to the first two numbers respectively, the$}\\\large\text{$numbers become:}[/tex]
[tex]\large\text{$\dfrac{a}{r}+2,$ $a+3$ and $ar$}[/tex]
[tex]\large\text{$\therefore\ $The numbers in A.P. = $\dfrac{3}{r}+2 $, $6$ and $3r$}[/tex]
[tex]\large\text{$Using arithmetic mean formula,}[/tex]
[tex]\large\text{$6=\dfrac{\bigg(\dfrac{3}{r}+2\bigg)+3r}{2}$}[/tex]
[tex]\large\text{$12=\dfrac{3}{r}+2+3r$}[/tex]
[tex]\large\text{$Multiplying by $r$ on both sides,}[/tex]
[tex]\large\text{$12r=3+2r+3r^2$}[/tex]
[tex]\large\text{$3r^2-10r+3=0$}[/tex]
[tex]\large\text{$Factoring out the left hand side,}[/tex]
[tex]\large\text{$(3r-1)(r-3)=0$}[/tex]
[tex]\large\text{i.e. $\boxed{r = \dfrac{1}{3},\ 3}$}[/tex]
[tex]\large\text{Taking $r=\dfrac{1}{3}. $the numbers in G.P. are:}[/tex]
[tex]\large\text{$\dfrac{3}{1/3},\ 3$ and $3\times\dfrac{1}{3}$}[/tex]
[tex]=\large\text{$9, 3$ and 1}[/tex]
[tex]\large\text{Taking $r=3, $ the numbers will be: }[/tex]
[tex]\large\text{$\dfrac{3}{3},\ 3 $ and $3\times3$}[/tex]
[tex]\large\text{$= 1, 3 and 9}[/tex]
[tex]\large\text{So the numbers are 1, 3 and 9.$}[/tex]
