Chapter 12.2, Problem 68E

a.

To determine

Find two arithmetic means between $10and18$ .

Answer to Problem 68E

The two arithmetic mean are $\frac{38}{3}and\frac{46}{3}$ .

Explanation of Solution

Given:

It is given in the question that the numbers are $10and18$ .

Concept Used:

In this, use the concept of nth term of sequence $a_n=a+(n-1)d$ and the formula for arithmetic mean is $m=\frac{a+b}{2}$ .

Calculation: Let a , b and c be two arithmetic mean between the numbers.

So, the numbers be $10,a,b,18$

Then , common difference be like that $a-10=18-b$

So, nth term of sequence

  $\begin{array}{l}{a}_n=a+(n-1)d\\ 18=10+(4-1)d\\ 18=3d\\ d=\frac{8}{3}\end{array}$

Now, $a=10+\frac{8}{3}=\frac{38}{3},b=10+\frac{16}{3}=\frac{46}{3}$

Conclusion:

The two arithmetic mean are $\frac{38}{3}and\frac{46}{3}$ .

b.

To determine

Find three arithmetic means between $10and18$ .

Answer to Problem 68E

The three arithmetic mean are $12,14and16$ .

Explanation of Solution

Given:

It is given in the question that two numbers are $10and18$ .

Concept Used:

In this, use the concept of nth term of sequence $a_n=a+(n-1)d$ and the formula for arithmetic mean is $m=\frac{a+b}{2}$ .

Calculation: Let a , b and c be two arithmetic mean between the numbers.

So, the numbers be $10,a,b,c,18$ .

Then , common difference be like that $a-10=c-b=18-c$ .

So, nth term of sequence

  $\begin{array}{l}{a}_n=a+(n-1)d\\ 18=10+(5-1)d\\ 8=4d\\ d=2\end{array}$

Hence, $a=10+2=12,b=12+2=14,c=14+2=16$ .

Conclusion:

The three arithmetic mean are $12,14and16$ .