Question

the mean and median of the data x+a,x+b,x+c are equal and a<b<c then find the value of a and b​

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• We are given three values, x+a, x+b, x+c where a < b < c.

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• The value of a and b.

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

Arrange the terms in ascending or descending order, but for us, it is already arranged.

In the given data x+a, x+b, x+c, we can find mean:

$\boxed{\sf{\green{\overline{x}=\dfrac{Sum \ of \ terms}{Number \ of \ terms} }}}$

According to the above data, mean:

$\sf \to \overline{x}=\dfrac{x+a+x+b+x+c}{3}$

$\sf \to \overline{x}=\dfrac{3x+a+b+c}{3}$

Now median is the "middle most term" of the given sequence. The middle most term in x+a, x+b, x+c is x+b. Therefore, median is x+b.

It is given that:

$\red\bullet \sf Median=Mode$

Therefore,

$\sf \to x+b=\dfrac{3x+a+b+c}{3}$

$\sf \to 3x+3b=3x+a+b+c$

$\sf \to 2b=a+c$

$\leadsto \sf{\red{b=\dfrac{a+c}{2} }}$

Now, finding a is simple. Use the same equation:

$\sf \to x+b=\dfrac{3x+a+b+c}{3}$

$\sf \to 3x+3b=3x+a+b+c$

$\sf \to 3b=a+b+c$

Here rearrange the terms.

$\sf \to -a=-3b+b+c$

$\sf \to a=-(-3b+b+c)$

$\to \sf a=3b-b-c$

$\leadsto \sf{\red{a=2b-c}}$

Therefore,

★ a = 2b - c.

★ b = (a+c)/2