Find a, b such that 7.2,a, b,3 are in A. P.

Step-by-step explanation:

As

7.2, a, b, 3 are in A.P

As we know common difference of the arithmetic sequence is the difference between the successive term and its preceding term, which is always constant

so

a -7.2 = b - a

a + a = b + 7.2

2a = b + 7.2

b = 2a - 7.2

also

a, b, 3 are in A.P

b - a = 3 - b

b + b = 3 + a

2b = 3 + a

Putting b = 2a - 7.2 in the equation 2b = 3 + a to find the value of 'a'.

2b = 3 + a

2 (2a - 7.2) = 3 + a

4a - 14.4 = 3 + a

3a = 3 + 14.4

3a = 17.4

Dividing the equation by 3

3a ÷ 3 = 17.4 ÷ 3

a = 5.8

Putting a = 5.8 in the equation b = 2a - 7.2

b = 2a - 7.2

  = 2(5.8) - 7.2

  = 11.6 - 7.2

b = 4.4

Therefore,

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )

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