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2020-10-31

The arithmetic mean (average) of two numbers c and d is given by $\stackrel{\u2015}{x}=\frac{c+d}{2}$

The value$\stackrel{\u2015}{x}$ is equidistant between c and d, so the sequence c,$\stackrel{\u2015}{x}$ , d is an arithmetic sequence. inserting k equally spaced values between c and d, yields the arithmetic sequence $c,{\stackrel{\u2015}{X}}_{1},{\stackrel{\u2015}{X}}_{2},{\stackrel{\u2015}{X}}_{3},{\stackrel{\u2015}{X}}_{4},...,{\stackrel{\u2015}{X}}_{k},d$ . Use this information for Exercise.

Insert four arithmetic means between 19 and 64.

The value

Insert four arithmetic means between 19 and 64.

Latisha Oneil

Skilled2020-11-01Added 100 answers

Step 1

Let the fourth arithmetic means between 19 and 64 are${\stackrel{\u2015}{X}}_{1},{\stackrel{\u2015}{X}}_{2},{\stackrel{\u2015}{X}}_{3},{\stackrel{\u2015}{X}}_{4}$

$\text{therefore},19,{\stackrel{\u2015}{X}}_{1},{\stackrel{\u2015}{X}}_{2},{\stackrel{\u2015}{X}}_{3},{\stackrel{\u2015}{X}}_{4},64$ 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${a}_{n}=a+(n-1)d$

${a}_{6}=a+(6-1)d$

${a}_{6}=a+5d$

$64=19+5d$

$5d=64-19$

$5d=45$

$d-9$

Step 2

Now the arithmetic means can be calculated as follows

${\stackrel{\u2015}{X}}_{1}=a+d=19+9=28$

${\stackrel{\u2015}{X}}_{2}=a+2d=19+2\times 9=37$

${\stackrel{\u2015}{X}}_{3}=a+3d=19+3\times 5=46$

${\stackrel{\u2015}{X}}_{4}=a+3d=19+4\times 5=55$

Step 3

Ans:${\stackrel{\u2015}{X}}_{1}=28$

${\stackrel{\u2015}{X}}_{2}=37$

${\stackrel{\u2015}{X}}_{3}=46$

${\stackrel{\u2015}{X}}_{4}=55$

Let the fourth arithmetic means between 19 and 64 are

We know that first term of A.P is

Step 2

Now the arithmetic means can be calculated as follows

Step 3

Ans:

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