The arithmetic mean (average) of two numbers c and d is given by overline{x} = frac{c+d}{2} The value overline{x} is equidistant between c and d, so t

Question

The arithmetic mean (average) of two numbers c and d is given by

\overset{―}{x} = \frac{c + d}{2}

The value

\overset{―}{x}

is equidistant between c and d, so the sequence c,

\overset{―}{x}

, d is an arithmetic sequence. inserting k equally spaced values between c and d, yields the arithmetic sequence

c, {\overset{―}{X}}_{1}, {\overset{―}{X}}_{2}, {\overset{―}{X}}_{3}, {\overset{―}{X}}_{4}, . . ., {\overset{―}{X}}_{k}, d

. Use this information for Exercise.
Insert four arithmetic means between 19 and 64.

Answer 1

Step 1
Let the fourth arithmetic means between 19 and 64 are

{\overset{―}{X}}_{1}, {\overset{―}{X}}_{2}, {\overset{―}{X}}_{3}, {\overset{―}{X}}_{4}

therefore, 19, {\overset{―}{X}}_{1}, {\overset{―}{X}}_{2}, {\overset{―}{X}}_{3}, {\overset{―}{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 + 5 d

64 = 19 + 5 d

5 d = 64 - 19

5 d = 45

d - 9

Step 2
Now the arithmetic means can be calculated as follows

{\overset{―}{X}}_{1} = a + d = 19 + 9 = 28

{\overset{―}{X}}_{2} = a + 2 d = 19 + 2 \times 9 = 37

{\overset{―}{X}}_{3} = a + 3 d = 19 + 3 \times 5 = 46

{\overset{―}{X}}_{4} = a + 3 d = 19 + 4 \times 5 = 55

Step 3
Ans:

{\overset{―}{X}}_{1} = 28

{\overset{―}{X}}_{2} = 37

{\overset{―}{X}}_{3} = 46

{\overset{―}{X}}_{4} = 55

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