sjeikdom0

2020-10-31

The arithmetic mean (average) of two numbers c and d is given by $\stackrel{―}{x}=\frac{c+d}{2}$
The value $\stackrel{―}{x}$ is equidistant between c and d, so the sequence c,$\stackrel{―}{x}$, d is an arithmetic sequence. inserting k equally spaced values between c and d, yields the arithmetic sequence $c,{\stackrel{―}{X}}_{1},{\stackrel{―}{X}}_{2},{\stackrel{―}{X}}_{3},{\stackrel{―}{X}}_{4},...,{\stackrel{―}{X}}_{k},d$. Use this information for Exercise.
Insert four arithmetic means between 19 and 64.

Latisha Oneil

Step 1
Let the fourth arithmetic means between 19 and 64 are ${\stackrel{―}{X}}_{1},{\stackrel{―}{X}}_{2},{\stackrel{―}{X}}_{3},{\stackrel{―}{X}}_{4}$
$\text{therefore},19,{\stackrel{―}{X}}_{1},{\stackrel{―}{X}}_{2},{\stackrel{―}{X}}_{3},{\stackrel{―}{X}}_{4},64$ are in A.P with first term 19 and sixth term 64
$a=19,{a}_{6}=64$
We know that first term of A.P is ${a}_{n}=a+\left(n-1\right)d$
${a}_{6}=a+\left(6-1\right)d$
${a}_{6}=a+5d$
$64=19+5d$
$5d=64-19$
$5d=45$
$d-9$
Step 2
Now the arithmetic means can be calculated as follows
${\stackrel{―}{X}}_{1}=a+d=19+9=28$
${\stackrel{―}{X}}_{2}=a+2d=19+2×9=37$
${\stackrel{―}{X}}_{3}=a+3d=19+3×5=46$
${\stackrel{―}{X}}_{4}=a+3d=19+4×5=55$
Step 3
Ans: ${\stackrel{―}{X}}_{1}=28$
${\stackrel{―}{X}}_{2}=37$
${\stackrel{―}{X}}_{3}=46$
${\stackrel{―}{X}}_{4}=55$

Do you have a similar question?