Question

The sum of three numbers in GP is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in AP. Find the largest of those numbers - smallest of those numbers?

Answer 1

Hint: In this question, we are given a statement about three numbers and we need to find the difference between the largest and the smallest of those numbers. For this, we will first suppose their numbers in the form of GP. Then we will form an equation using the given statement. Solving these equations will help us in finding our numbers. Comparing them will give us our largest number and smallest number.

Complete step by step answer:
Here we are given that the three numbers are in geometric progression. Let us suppose that one of the numbers is a and the common ratio is r. So our rest two numbers will be ar and $ a{{r}^{2}} $ .
(Because a, ar, $ a{{r}^{2}} $ are in GP with common ratio r).
So our three numbers become equal to a, ar and $ a{{r}^{2}} $ . Now we are given that their sum is 14. So we can write it in mathematical form as, $ a+ar+a{{r}^{2}}=14 $ .
Taking a common we get, \[a\left( 1+r+{{r}^{2}} \right)=14\cdots \cdots \cdots \left( 1 \right)\].
We are given that if the first two terms are increased by 1 and the third term is decreased by 1, then they form an arithmetic progression.
Our three numbers become,
First number: a+1
Second number: ar+1
Third number: $ a{{r}^{2}}-1 $ .
Hence we can say that a, ar+1, $ a{{r}^{2}}-1 $ are in arithmetic progression.
Now let us use the arithmetic mean property on these numbers according to which the sum of the first term and the third term is equal to twice of the second term. So we can say that, $ a+1+a{{r}^{2}}-1=2\left( ar+1 \right) $ .
Simplifying we get, $ a+a{{r}^{2}}=2ar+2\Rightarrow a+a{{r}^{2}}-2ar=2 $ .
Taking a common from the left side we get, \[a\left( 1+{{r}^{2}}-2r \right)=2\cdots \cdots \cdots \left( 2 \right)\].
We have got two equations (1) and (2). Let us solve them to find the value of a and r.
Dividing (1) by (2) we get,
\[\dfrac{a\left( {{r}^{2}}+r+1 \right)}{a\left( {{r}^{2}}-2r+1 \right)}=\dfrac{14}{2}\]
Cancelling a on the left side we get, \[\dfrac{\left( {{r}^{2}}+r+1 \right)}{\left( {{r}^{2}}-2r+1 \right)}=7\].
Cross multiplying we get,
\[\begin{align}
  & {{r}^{2}}+r+1=7\left( {{r}^{2}}-2r+1 \right) \\
 & \Rightarrow {{r}^{2}}+r+1=7{{r}^{2}}-14r+7 \\
 & \Rightarrow {{r}^{2}}+r+1-7{{r}^{2}}+14r-7=0 \\
\end{align}\]
Simplifying we get, $ -6{{r}^{2}}+15r-6=0 $ .
Dividing by -3 on both sides we get, $ 2{{r}^{2}}-5r+2=0 $ .
Using quadratic formula for $ a{{x}^{2}}+bx+c $ we know that $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
Comparing with $ 2{{r}^{2}}-5r+2=0 $ we get,
\[\begin{align}
  & x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 2 \right)\left( 2 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25-16}}{4} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{9}}{4} \\
\end{align}\]
We know that, $ {{3}^{2}}=9 $ so we get $ \sqrt{9}=3 $ .
\[\begin{align}
  & \Rightarrow x=\dfrac{5\pm 3}{4} \\
 & \Rightarrow x=\dfrac{5+3}{4}\text{ and }x=\dfrac{5-3}{4} \\
 & \Rightarrow x=\dfrac{8}{4}\text{ and }x=\dfrac{2}{4} \\
 & \Rightarrow x=2\text{ and }x=\dfrac{1}{2} \\
\end{align}\]
Therefore, the value of r is \[\dfrac{1}{2}\text{ and }2\].
Putting the values in equation (1) we get,
Putting $ r=\dfrac{1}{2} $ we get,
\[\begin{align}
  & a\left( 1+\left( \dfrac{1}{2} \right)+{{\left( \dfrac{1}{2} \right)}^{2}} \right)=14 \\
 & \Rightarrow a\left( 1+\dfrac{1}{2}+\dfrac{1}{4} \right)=14 \\
\end{align}\]
Taking LCM on 4 on the left side we get,
\[\begin{align}
  & a\left( \dfrac{4+1+2}{4} \right)=14 \\
 & \Rightarrow a\left( \dfrac{7}{4} \right)=14 \\
\end{align}\]
Taking $ \dfrac{7}{4} $ to the other side we get,
\[\begin{align}
  & a=\dfrac{14\times 4}{7} \\
 & \Rightarrow a=2\times 4 \\
 & \Rightarrow a=8 \\
\end{align}\]
Hence the value of a is 8.
Putting r = 2, we get,
\[\begin{align}
  & a\left( 1+2+{{\left( 2 \right)}^{2}} \right)=14 \\
 & \Rightarrow a\left( 1+2+4 \right)=14 \\
 & \Rightarrow a\left( 7 \right)=14 \\
\end{align}\]
Dividing both sides by 7, we get,
\[\begin{align}
  & a=\dfrac{14}{7} \\
 & \Rightarrow a=2 \\
\end{align}\]
By putting a = 8 and $ r=\dfrac{1}{2} $ we have three numbers as a, ar, $ a{{r}^{2}} $ .
\[\begin{align}
  & \Rightarrow \left( 8 \right),\left( 8 \right)\left( \dfrac{1}{2} \right),8{{\left( \dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow 8,4,\dfrac{8}{4} \\
 & \Rightarrow 8,4,2 \\
\end{align}\]
Hence the numbers are 8, 4, 2.
By putting a = 2 and r = 2 we have,
\[\begin{align}
  & a,ar,a{{r}^{2}} \\
 & \Rightarrow 2,2\left( 2 \right),2{{\left( 2 \right)}^{2}} \\
 & \Rightarrow 2,4,2\times 4 \\
 & \Rightarrow 2,4,8 \\
\end{align}\]
Hence the three numbers are 2, 4, 8.
By both values of a and r, the three numbers are 2, 4, 8.
Largest number = 8.
Smallest number = 2.
The difference between the largest and the smallest number becomes equal to 8-2 = 6.
Hence the largest number - smallest number = 6.

Note:
Students should take care of the signs while solving this sum. Make sure to put both values of a and r one by one. Students should know the properties of geometry mean and arithmetic mean to solve the problem.