Question

Find a,b and c such that the numbers a,7,b,23 and c are in AP.

Answer 1

Hint: An AP is a series in which the difference between the successive terms of the series is equal. This difference is known as the common difference of the AP. Now, for any three numbers x,y and z to be in AP, they must have a common difference. Thus, $y-x=z-y$ which gives us the property $2y=x+z$ . now, we will take three successive numbers of the given series at a time and use this property to form different equations in a,b and c. Solving those equations, we will get our required values.

Complete step by step answer:
For the numbers to be in AP, the difference between the successive terms of the series should have the equal value.
Now, we have been given that a,7,b,23 and c are in AP.
We know that if three numbers x,y and z are in AP then they have the property:
\[2y=x+z\]
Now, from the given the series, if we take the first three numbers a,7 and b, then from the above mentioned property we will get:
$2\left( 7 \right)=a+b$
$\Rightarrow 14=a+b$ …….(i)
Now, if the take the second, third and fourth number from the given series 7,b and 23, then from the same property we will get:
$2b=7+23$
Now, we can solve this equation and get the value of ‘b’.
Solving this equation we’ll get:
 $\begin{align}
  & 2b=7+23 \\
 & \Rightarrow 2b=30 \\
 & \Rightarrow b=15 \\
\end{align}$
Thus, the value of ‘b’ is 15.
Now, we can put in the value of ‘b’ in equation (i) and we will be able to find the value of ‘a’.
Thus, putting the value of ‘b’ in equation (i) we get:
$\begin{align}
  & 14=a+b \\
 & b=15 \\
 & \Rightarrow 14=a+15 \\
 & \Rightarrow 14-15=a \\
 & \Rightarrow a=-1 \\
\end{align}$
Thus, the value of ‘a’ is -1.
Now, if take the last three numbers of the given series b,23 and c, then by the property we will get:
$2\left( 23 \right)=b+c$
$\Rightarrow 46=b+c$ ……..(ii)
Now, we know that $b=15$
Thus, putting the value of ‘b’ in equation (ii), we will get the value of ‘c’.
Thus, putting the value of ‘b’ in equation (ii) we get:
$\begin{align}
  & 46=b+c \\
 & b=15 \\
 & \Rightarrow 46=15+c \\
 & \Rightarrow 46-15=c \\
 & \Rightarrow c=31 \\
\end{align}$
Thus, the value of ‘c’ is 31.
Thus, we get:
$\begin{align}
  & a=-1 \\
 & b=15 \\
 & c=31 \\
\end{align}$

Note: An alternate method to do this question is given by the following:
We have been given the AP a,7,b,23,c.
Let the common difference of the AP be ‘d’.
Thus, the difference between any two successive terms in the AP is ‘d’.
Now, if the difference between two successive terms is ‘d’, then the difference between first and third term will be ‘2d’.
Similarly, the difference between the second and fourth term will also be ‘2d’.
Here, the second term of the AP is ‘7’ and the fourth term of the AP is ‘23’.
Thus, we get:
  $\begin{align}
  & 2d=23-7 \\
 & \Rightarrow 2d=16 \\
 & \Rightarrow d=8 \\
\end{align}$
Thus, the common difference of the AP is ‘8’.
Now, using the common difference, we will get the values of a,b and c.
Taking the first two terms we get:
$\begin{align}
  & 7-a=8 \\
 & \Rightarrow a=7-8 \\
 & \Rightarrow a=-1 \\
\end{align}$
Now, taking the second and third term we get:
$\begin{align}
  & b-7=8 \\
 & \Rightarrow b=7+8 \\
 & \Rightarrow b=15 \\
\end{align}$
Now, taking the last two terms we get:
$\begin{align}
  & c-23=8 \\
 & \Rightarrow c=23+8 \\
 & \Rightarrow c=31 \\
\end{align}$
Thus, the values of a,b and c are -1,15 and 31 respectively.