If a, b, 72, (a+b) are consecutive terms of an A. P., find the values of a and b.

a = 36, b = 54

Step-by-step explanation:

Since the terms are in arithmetic progression then there is a common difference d between consecutive terms , that is

b - a = 72 - b ( add b to both sides )

2b - a = 72 → (1)

and

a + b - 72 = b - a ( subtract b - a from both sides )

2a - 72 = 0 ( add 72 to both sides )

2a = 72 ( divide both sides by 2 )

a = 36

Substitute a = 36 into (1) and evaluate for b

2b - 36 = 72 ( add 36 to both sides )

2b = 108 ( divide both sides by 2 )

b = 54

Thus a = 36 and b = 54

The sequence is therefore 36, 54, 72, 90 ,

a= 36

b= 54

Step-by-step explanation:

formula for AP is nth term= a+(n-1)d

3rd term=> a+2d=72

2nd term=> a+d=b

Comparing the two equations:

2d-d=72-b

d=72-b

4th term=> a+3d=a+b

d= 72-b

a+3(72-b)=a+b

a+216-3b=a+b

216-3b=b

216=4b

b=54

d=72-b

d=72-54

d=18

a+d= b

a=b-d

a=54-18

a=36

step-by-step explanation: